message
stringlengths 2
45.8k
| message_type
stringclasses 2
values | message_id
int64 0
1
| conversation_id
int64 254
108k
| cluster
float64 3
3
| __index_level_0__
int64 508
217k
|
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After a successful field test, Heidi is considering deploying a trap along some Corridor, possibly not the first one. She wants to avoid meeting the Daleks inside the Time Vortex, so for abundance of caution she considers placing the traps only along those Corridors that are not going to be used according to the current Daleks' plan β which is to use a minimum spanning tree of Corridors. Heidi knows that all energy requirements for different Corridors are now different, and that the Daleks have a single unique plan which they are intending to use.
Your task is to calculate the number E_{max}(c), which is defined in the same way as in the easy version β i.e., the largest e β€ 10^9 such that if we changed the energy of corridor c to e, the Daleks might use it β but now for every corridor that Heidi considers.
Input
The first line: number n of destinations, number m of Time Corridors (2 β€ n β€ 10^5, n - 1 β€ m β€ 10^6). The next m lines: destinations a, b and energy e (1 β€ a, b β€ n, a β b, 0 β€ e β€ 10^9).
No pair \\{a, b\} will repeat. The graph is guaranteed to be connected. All energy requirements e are distinct.
Output
Output m-(n-1) lines, each containing one integer: E_{max}(c_i) for the i-th Corridor c_i from the input that is not part of the current Daleks' plan (minimum spanning tree).
Example
Input
3 3
1 2 8
2 3 3
3 1 4
Output
4
Note
If m = n-1, then you need not output anything.
Submitted Solution:
```
def naiveSolve():
return
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
class UnionFind:
def __init__(self, n):
self.parent = list(range(n))
def find(self, a): #return parent of a. a and b are in same set if they have same parent
acopy = a
while a != self.parent[a]:
a = self.parent[a]
while acopy != a: #path compression
self.parent[acopy], acopy = a, self.parent[acopy]
return a
def union(self, a, b): #union a and b
self.parent[self.find(b)] = self.find(a)
def main():
n,m=readIntArr()
a=[]
b=[]
c=[]
for _ in range(m):
aa,bb,cc=readIntArr()
a.append(aa); b.append(bb); c.append(cc)
idxes=list(range(m))
idxes.sort(key=lambda i:c[i])
uf=UnionFind(n+1)
inA=[] # in MST
inB=[]
inC=[]
outA=[] # not in MST
outB=[]
for i in range(m):
aa=a[idxes[i]]
bb=b[idxes[i]]
cc=c[idxes[i]]
if uf.find(aa)!=uf.find(bb):
uf.union(aa,bb)
inA.append(aa)
inB.append(bb)
inC.append(cc)
else:
outA.append(aa)
outB.append(bb)
# find largest edge between each outA and outB
adj=[[] for _ in range(n+1)] # store (edge,cost)
assert len(inA)==n-1
for i in range(n-1):
u,v,c=inA[i],inB[i],inC[i]
adj[u].append((v,c))
adj[v].append((u,c))
@bootstrap
def dfs(node,p,cost):
up[node][0]=p
maxEdge[node][0]=cost
tin[node]=time[0]
time[0]+=1
for v,c in adj[node]:
if v!=p:
yield dfs(v,node,c)
tout[node]=time[0]
time[0]+=1
yield None
time=[0]
tin=[-1]*(n+1)
tout=[-1]*(n+1)
# binary lifting to find LCA
maxPow=0
while pow(2,maxPow)<n:
maxPow+=1
maxPow+=1
up=[[-1 for _ in range(maxPow)] for __ in range(n+1)]
maxEdge=[[-inf for _ in range(maxPow)] for __ in range(n+1)]
dfs(1,-1,-inf)
for i in range(1,maxPow):
for u in range(2,n+1):
if up[u][i-1]==-1 or up[up[u][i-1]][i-1]==-1: # reached beyond root
continue
up[u][i]=up[up[u][i-1]][i-1]
maxEdge[u][i]=max(maxEdge[u][i-1],maxEdge[up[u][i-1]][i-1])
def isAncestor(u,v): # True if u is ancestor of v
return tin[u]<=tin[v] and tout[u]>=tout[v]
def findMaxEdgeValue(u,v):
# traverse u to LCA, then v to LCA
res=0
if not isAncestor(u,v):
u2=u
while not isAncestor(up[u2][0],v): # still have to move up
for i in range(maxPow-1,-1,-1):
if up[u2][i]!=-1 and not isAncestor(up[u2][i],v):
res=max(res,maxEdge[u2][i])
u2=up[u2][i]
# print('u2:{}'.format(u2)) ##
# next level up is lca
res=max(res,maxEdge[u2][0])
if not isAncestor(v,u):
v2=v
while not isAncestor(up[v2][0],u): # still have to move up
for i in range(maxPow-1,-1,-1):
if up[v2][i]!=-1 and not isAncestor(up[v2][i],u):
res=max(res,maxEdge[v2][i])
v2=up[v2][i]
# print('v2:{} u:{}'.format(v2,u)) ##
# if v2==1:
# print(isAncestor(v,u)) ##
# break
# next level up is lca
res=max(res,maxEdge[v2][0])
return res
# for i in range(1,4):
# for j in range(1,4):
# if i!=j:
# print('i:{} j:{} anc:{}'.format(i,j,isAncestor(i,j)))
ans=[]
# print(outA,outB)
# for u in range(1,n+1):
# print('u:{} up:{} maxEdge:{}'.format(u,up[u],maxEdge[u])) ##
for i in range(len(outA)):
u,v=outA[i],outB[i]
ans.append(findMaxEdgeValue(u,v))
multiLineArrayPrint(ans)
return
import sys
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
# input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.
def oneLineArrayPrint(arr):
print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
def readIntArr():
return [int(x) for x in input().split()]
# def readFloatArr():
# return [float(x) for x in input().split()]
def makeArr(defaultValFactory,dimensionArr): # eg. makeArr(lambda:0,[n,m])
dv=defaultValFactory;da=dimensionArr
if len(da)==1:return [dv() for _ in range(da[0])]
else:return [makeArr(dv,da[1:]) for _ in range(da[0])]
def queryInteractive(r):
print('? {}'.format(r))
sys.stdout.flush()
return readIntArr()
def answerInteractive(adj,n):
print('!')
for u in range(1,n+1):
for v in adj[u]:
if v>u:
print('{} {}'.format(u,v))
sys.stdout.flush()
inf=float('inf')
MOD=10**9+7
# MOD=998244353
from math import gcd,floor,ceil
# from math import floor,ceil # for Python2
for _abc in range(1):
main()
```
|
instruction
| 0
| 102,976
| 3
| 205,952
|
No
|
output
| 1
| 102,976
| 3
| 205,953
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After a successful field test, Heidi is considering deploying a trap along some Corridor, possibly not the first one. She wants to avoid meeting the Daleks inside the Time Vortex, so for abundance of caution she considers placing the traps only along those Corridors that are not going to be used according to the current Daleks' plan β which is to use a minimum spanning tree of Corridors. Heidi knows that all energy requirements for different Corridors are now different, and that the Daleks have a single unique plan which they are intending to use.
Your task is to calculate the number E_{max}(c), which is defined in the same way as in the easy version β i.e., the largest e β€ 10^9 such that if we changed the energy of corridor c to e, the Daleks might use it β but now for every corridor that Heidi considers.
Input
The first line: number n of destinations, number m of Time Corridors (2 β€ n β€ 10^5, n - 1 β€ m β€ 10^6). The next m lines: destinations a, b and energy e (1 β€ a, b β€ n, a β b, 0 β€ e β€ 10^9).
No pair \\{a, b\} will repeat. The graph is guaranteed to be connected. All energy requirements e are distinct.
Output
Output m-(n-1) lines, each containing one integer: E_{max}(c_i) for the i-th Corridor c_i from the input that is not part of the current Daleks' plan (minimum spanning tree).
Example
Input
3 3
1 2 8
2 3 3
3 1 4
Output
4
Note
If m = n-1, then you need not output anything.
Submitted Solution:
```
def naiveSolve():
return
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
class UnionFind:
def __init__(self, n):
self.parent = list(range(n))
def find(self, a): #return parent of a. a and b are in same set if they have same parent
acopy = a
while a != self.parent[a]:
a = self.parent[a]
while acopy != a: #path compression
self.parent[acopy], acopy = a, self.parent[acopy]
return a
def union(self, a, b): #union a and b
self.parent[self.find(b)] = self.find(a)
def main():
n,m=readIntArr()
a=[]
b=[]
c=[]
for _ in range(m):
aa,bb,cc=readIntArr()
a.append(aa); b.append(bb); c.append(cc)
idxes=list(range(m))
idxes.sort(key=lambda i:c[i])
uf=UnionFind(n+1)
inA=[] # in MST
inB=[]
inC=[]
outA=[] # not in MST
outB=[]
for i in range(m):
aa=a[idxes[i]]
bb=b[idxes[i]]
cc=c[idxes[i]]
if uf.find(aa)!=uf.find(bb):
uf.union(aa,bb)
inA.append(aa)
inB.append(bb)
inC.append(cc)
else:
outA.append(aa)
outB.append(bb)
# find largest edge between each outA and outB
adj=[[] for _ in range(n+1)] # store (edge,cost)
for i in range(n-1):
u,v,c=inA[i],inB[i],inC[i]
adj[u].append((v,c))
adj[v].append((u,c))
@bootstrap
def dfs(node,p,cost):
up[node][0]=p
maxEdge[node][0]=cost
tin[node]=time[0]
time[0]+=1
for v,c in adj[node]:
if v!=p:
yield dfs(v,node,c)
tout[node]=time[0]
time[0]+=1
yield None
time=[0]
tin=[-1]*(n+1)
tout=[-1]*(n+1)
# binary lifting to find LCA
maxPow=0
while pow(2,maxPow)<n:
maxPow+=1
maxPow+=1
up=[[-1 for _ in range(maxPow)] for __ in range(n+1)]
maxEdge=[[-inf for _ in range(maxPow)] for __ in range(n+1)]
dfs(1,-1,-inf)
for i in range(1,maxPow):
for u in range(1,n+1):
up[u][i]=up[up[u][i-1]][i-1]
maxEdge[u][i]=max(maxEdge[u][i-1],maxEdge[up[u][i-1]][i-1])
def isAncestor(u,v): # True if u is ancestor of v
return tin[u]<=tin[v] and tout[u]>=tout[v]
def findMaxEdgeValue(u,v):
# traverse u to LCA, then v to LCA
res=0
if not isAncestor(u,v):
u2=u
while not isAncestor(up[u2][0],v): # still have to move up
for i in range(maxPow-1,-1,-1):
if up[u2][i]!=-1 and not isAncestor(up[u2][i],v):
res=max(res,maxEdge[u2][i])
u2=up[u2][i]
# print('u2:{}'.format(u2)) ##
# next level up is lca
res=max(res,maxEdge[u2][0])
if not isAncestor(v,u):
v2=v
while not isAncestor(up[v2][0],u): # still have to move up
for i in range(maxPow-1,-1,-1):
if up[v2][i]!=-1 and not isAncestor(up[v2][i],u):
res=max(res,maxEdge[v2][i])
v2=up[v2][i]
# print('v2:{} u:{}'.format(v2,u)) ##
# if v2==1:
# print(isAncestor(v,u)) ##
# break
# next level up is lca
res=max(res,maxEdge[v2][0])
return res
# for i in range(1,4):
# for j in range(1,4):
# if i!=j:
# print('i:{} j:{} anc:{}'.format(i,j,isAncestor(i,j)))
ans=[]
# print(outA,outB)
for i in range(len(outA)):
u,v=outA[i],outB[i]
ans.append(findMaxEdgeValue(u,v))
multiLineArrayPrint(ans)
return
import sys
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
# input=lambda: sys.stdin.readline().rstrip("\r\n") #FOR READING STRING/TEXT INPUTS.
def oneLineArrayPrint(arr):
print(' '.join([str(x) for x in arr]))
def multiLineArrayPrint(arr):
print('\n'.join([str(x) for x in arr]))
def multiLineArrayOfArraysPrint(arr):
print('\n'.join([' '.join([str(x) for x in y]) for y in arr]))
def readIntArr():
return [int(x) for x in input().split()]
# def readFloatArr():
# return [float(x) for x in input().split()]
def makeArr(defaultValFactory,dimensionArr): # eg. makeArr(lambda:0,[n,m])
dv=defaultValFactory;da=dimensionArr
if len(da)==1:return [dv() for _ in range(da[0])]
else:return [makeArr(dv,da[1:]) for _ in range(da[0])]
def queryInteractive(r):
print('? {}'.format(r))
sys.stdout.flush()
return readIntArr()
def answerInteractive(adj,n):
print('!')
for u in range(1,n+1):
for v in adj[u]:
if v>u:
print('{} {}'.format(u,v))
sys.stdout.flush()
inf=float('inf')
MOD=10**9+7
# MOD=998244353
from math import gcd,floor,ceil
# from math import floor,ceil # for Python2
for _abc in range(1):
main()
```
|
instruction
| 0
| 102,977
| 3
| 205,954
|
No
|
output
| 1
| 102,977
| 3
| 205,955
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Annie has gotten bored of winning every coding contest and farming unlimited rating. Today, she is going to farm potatoes instead.
Annie's garden is an infinite 2D plane. She has n potatoes to plant, and the i-th potato must be planted at (x_i,y_i). Starting at the point (0, 0), Annie begins walking, in one step she can travel one unit right or up (increasing her x or y coordinate by 1 respectively). At any point (X,Y) during her walk she can plant some potatoes at arbitrary points using her potato gun, consuming max(|X-x|,|Y-y|) units of energy in order to plant a potato at (x,y). Find the minimum total energy required to plant every potato.
Note that Annie may plant any number of potatoes from any point.
Input
The first line contains the integer n (1 β€ n β€ 800 000).
The next n lines contain two integers x_i and y_i (0 β€ x_i,y_i β€ 10^9), representing the location of the i-th potato. It is possible that some potatoes should be planted in the same location.
Output
Print the minimum total energy to plant all potatoes.
Examples
Input
2
1 1
2 2
Output
0
Input
2
1 1
2 0
Output
1
Input
3
5 5
7 7
4 9
Output
2
Input
10
5 1
4 0
9 6
0 2
10 1
9 10
3 10
0 10
8 9
1 5
Output
19
Input
10
1 1
2 2
2 0
4 2
4 0
2 0
0 2
4 0
4 2
5 1
Output
6
Note
In example 1, Annie can travel to each spot directly and plant a potato with no energy required.
In example 2, moving to (1,0), Annie plants the second potato using 1 energy. Next, she travels to (1,1) and plants the first potato with 0 energy.
Submitted Solution:
```
n=int(input())
lst_x=[]
lst_y=[]
for i in range(n):
x,y=map(int,input().split())
lst_x.append(x)
lst_y.append(y)
print(0)
```
|
instruction
| 0
| 103,124
| 3
| 206,248
|
No
|
output
| 1
| 103,124
| 3
| 206,249
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
My name is James diGriz, I'm the most clever robber and treasure hunter in the whole galaxy. There are books written about my adventures and songs about my operations, though you were able to catch me up in a pretty awkward moment.
I was able to hide from cameras, outsmart all the guards and pass numerous traps, but when I finally reached the treasure box and opened it, I have accidentally started the clockwork bomb! Luckily, I have met such kind of bombs before and I know that the clockwork mechanism can be stopped by connecting contacts with wires on the control panel of the bomb in a certain manner.
I see n contacts connected by n - 1 wires. Contacts are numbered with integers from 1 to n. Bomb has a security mechanism that ensures the following condition: if there exist k β₯ 2 contacts c1, c2, ..., ck forming a circuit, i. e. there exist k distinct wires between contacts c1 and c2, c2 and c3, ..., ck and c1, then the bomb immediately explodes and my story ends here. In particular, if two contacts are connected by more than one wire they form a circuit of length 2. It is also prohibited to connect a contact with itself.
On the other hand, if I disconnect more than one wire (i. e. at some moment there will be no more than n - 2 wires in the scheme) then the other security check fails and the bomb also explodes. So, the only thing I can do is to unplug some wire and plug it into a new place ensuring the fact that no circuits appear.
I know how I should put the wires in order to stop the clockwork. But my time is running out! Help me get out of this alive: find the sequence of operations each of which consists of unplugging some wire and putting it into another place so that the bomb is defused.
Input
The first line of the input contains n (2 β€ n β€ 500 000), the number of contacts.
Each of the following n - 1 lines contains two of integers xi and yi (1 β€ xi, yi β€ n, xi β yi) denoting the contacts currently connected by the i-th wire.
The remaining n - 1 lines contain the description of the sought scheme in the same format.
It is guaranteed that the starting and the ending schemes are correct (i. e. do not contain cicuits nor wires connecting contact with itself).
Output
The first line should contain k (k β₯ 0) β the minimum number of moves of unplugging and plugging back some wire required to defuse the bomb.
In each of the following k lines output four integers ai, bi, ci, di meaning that on the i-th step it is neccesary to unplug the wire connecting the contacts ai and bi and plug it to the contacts ci and di. Of course the wire connecting contacts ai and bi should be present in the scheme.
If there is no correct sequence transforming the existing scheme into the sought one, output -1.
Examples
Input
3
1 2
2 3
1 3
3 2
Output
1
1 2 1 3
Input
4
1 2
2 3
3 4
2 4
4 1
1 3
Output
3
1 2 1 3
4 3 4 1
2 3 2 4
Note
Picture with the clarification for the sample tests:
<image>
Submitted Solution:
```
#!/usr/bin/python3
# -*- config:utf-8 -*-
import sys
import re
# exception type for wrong input data
class IncorrectDataException(Exception):
pass
# transform graph without circles to tree
def createtree(cnt,data):
data=[[int(y) for y in x.split(' ')] for x in data]
edges={}
backtrace={}
for i in range(1,cnt+1):
edges[i]={}
backtrace[i]={}
for edge in data:
edges[edge[0]][edge[1]]=1
edges[edge[1]][edge[0]]=1
backtrace[edge[0]][edge[1]]=1
backtrace[edge[1]][edge[0]]=1
stack=[1]
while stack:
srcvert=stack.pop(-1)
for destvert in edges[srcvert].keys():
stack.append(destvert)
edges[destvert].pop(srcvert)
backtrace[srcvert].pop(destvert)
return edges,backtrace
# get source data
def getSrcData():
f = sys.stdin
buf = f.read()
sample=re.compile(r'^\d+\n(\d+ \d+\n)+',re.MULTILINE)
data=sample.match(buf)
if not (data):
raise IncorrectDataException("Incorrect source data: %s" % buf)
buf=data.group(0)
buf=buf.split('\n')
cnt=int(buf[0])
src=buf[1:cnt]
trg=buf[cnt:(cnt*2-1)]
src,srcbacktrace=createtree(cnt,src)
trg,trgbacktrace=createtree(cnt,trg)
trgbacktrace=None
return src,srcbacktrace,trg
# find transform src tree to trg tree
def getTransform(src,srcbacktrace,trg):
result=''
srcvert=1
stack=[1]
while stack:
srcvert=stack.pop(-1)
stack.reverse()
for destvert in trg[srcvert].keys():
stack.append(destvert)
if not destvert in src[srcvert]:
src4destvert=list(srcbacktrace[destvert].keys())[0] # get edge <src4destvert,destvert> in src tree
# change found edge for <srcvert,destvert>
# src[src4destvert][destvert]=0
# src[srcvert][destvert]=1
result+=str(src4destvert)+" "+str(destvert)+" "+str(srcvert)+" "+str(destvert)+"\n"
stack.reverse()
return result,src
# save result
def saveResult(result):
f = sys.stdout
f.write(str(result) + '\n')
f.flush()
# start
def run():
src,srcbacktrace,trg=getSrcData()
result,src=getTransform(src,srcbacktrace,trg)
cnt=result.count('\n')
result=str(cnt)+'\n'+result
saveResult(result)
if __name__ == '__main__':
run()
```
|
instruction
| 0
| 104,073
| 3
| 208,146
|
No
|
output
| 1
| 104,073
| 3
| 208,147
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
My name is James diGriz, I'm the most clever robber and treasure hunter in the whole galaxy. There are books written about my adventures and songs about my operations, though you were able to catch me up in a pretty awkward moment.
I was able to hide from cameras, outsmart all the guards and pass numerous traps, but when I finally reached the treasure box and opened it, I have accidentally started the clockwork bomb! Luckily, I have met such kind of bombs before and I know that the clockwork mechanism can be stopped by connecting contacts with wires on the control panel of the bomb in a certain manner.
I see n contacts connected by n - 1 wires. Contacts are numbered with integers from 1 to n. Bomb has a security mechanism that ensures the following condition: if there exist k β₯ 2 contacts c1, c2, ..., ck forming a circuit, i. e. there exist k distinct wires between contacts c1 and c2, c2 and c3, ..., ck and c1, then the bomb immediately explodes and my story ends here. In particular, if two contacts are connected by more than one wire they form a circuit of length 2. It is also prohibited to connect a contact with itself.
On the other hand, if I disconnect more than one wire (i. e. at some moment there will be no more than n - 2 wires in the scheme) then the other security check fails and the bomb also explodes. So, the only thing I can do is to unplug some wire and plug it into a new place ensuring the fact that no circuits appear.
I know how I should put the wires in order to stop the clockwork. But my time is running out! Help me get out of this alive: find the sequence of operations each of which consists of unplugging some wire and putting it into another place so that the bomb is defused.
Input
The first line of the input contains n (2 β€ n β€ 500 000), the number of contacts.
Each of the following n - 1 lines contains two of integers xi and yi (1 β€ xi, yi β€ n, xi β yi) denoting the contacts currently connected by the i-th wire.
The remaining n - 1 lines contain the description of the sought scheme in the same format.
It is guaranteed that the starting and the ending schemes are correct (i. e. do not contain cicuits nor wires connecting contact with itself).
Output
The first line should contain k (k β₯ 0) β the minimum number of moves of unplugging and plugging back some wire required to defuse the bomb.
In each of the following k lines output four integers ai, bi, ci, di meaning that on the i-th step it is neccesary to unplug the wire connecting the contacts ai and bi and plug it to the contacts ci and di. Of course the wire connecting contacts ai and bi should be present in the scheme.
If there is no correct sequence transforming the existing scheme into the sought one, output -1.
Examples
Input
3
1 2
2 3
1 3
3 2
Output
1
1 2 1 3
Input
4
1 2
2 3
3 4
2 4
4 1
1 3
Output
3
1 2 1 3
4 3 4 1
2 3 2 4
Note
Picture with the clarification for the sample tests:
<image>
Submitted Solution:
```
#!/usr/bin/python3
# -*- config:utf-8 -*-
import sys
import re
# exception type for wrong input data
class IncorrectDataException(Exception):
pass
# transform graph without circles to tree
def createtree(cnt,data,curvert):
data=[[int(y) for y in x.split(' ')] for x in data]
edges={}
backtrace={}
for i in range(1,cnt+1):
edges[i]={}
backtrace[i]={}
for edge in data:
edges[edge[0]][edge[1]]=1
edges[edge[1]][edge[0]]=1
backtrace[edge[0]][edge[1]]=1
backtrace[edge[1]][edge[0]]=1
if not curvert:
cnt=0
for i in edges.keys():
if len(edges[i])>cnt:
cnt=len(edges[i])
curvert=i
stack=[curvert]
while stack:
srcvert=stack.pop(-1)
for destvert in edges[srcvert].keys():
stack.append(destvert)
edges[destvert].pop(srcvert)
backtrace[srcvert].pop(destvert)
return edges,backtrace,curvert
# get source data
def getSrcData():
f = sys.stdin
buf = f.read()
sample=re.compile(r'^\d+\n(\d+ \d+\n)+',re.MULTILINE)
data=sample.match(buf)
if not (data):
raise IncorrectDataException("Incorrect source data: %s" % buf)
buf=data.group(0)
buf=buf.split('\n')
cnt=int(buf[0])
src=buf[1:cnt]
trg=buf[cnt:(cnt*2-1)]
trg,trgbacktrace,curvert=createtree(cnt,trg,0)
src,srcbacktrace,curvert=createtree(cnt,src,curvert)
trgbacktrace=None
return src,srcbacktrace,trg,curvert
# find transform src tree to trg tree
def getTransform(src,srcbacktrace,trg,curvert):
result=''
stack=[curvert]
while stack:
srcvert=stack.pop(-1)
stack.reverse()
for destvert in trg[srcvert].keys():
stack.append(destvert)
if not destvert in src[srcvert]:
src4destvert=list(srcbacktrace[destvert].keys())[0] # get edge <src4destvert,destvert> in src tree
# change found edge for <srcvert,destvert>
# src[src4destvert][destvert]=0
# src[srcvert][destvert]=1
result+=str(src4destvert)+" "+str(destvert)+" "+str(srcvert)+" "+str(destvert)+"\n"
stack.reverse()
return result,src
# save result
def saveResult(result):
f = sys.stdout
f.write(str(result) + '\n')
f.flush()
# start
def run():
src,srcbacktrace,trg,curvert=getSrcData()
result,src=getTransform(src,srcbacktrace,trg,curvert)
cnt=result.count('\n')
result=str(cnt)+'\n'+result
saveResult(result)
if __name__ == '__main__':
run()
```
|
instruction
| 0
| 104,074
| 3
| 208,148
|
No
|
output
| 1
| 104,074
| 3
| 208,149
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Mad scientist Mike does not use slow hard disks. His modification of a hard drive has not one, but n different heads that can read data in parallel.
When viewed from the side, Mike's hard drive is an endless array of tracks. The tracks of the array are numbered from left to right with integers, starting with 1. In the initial state the i-th reading head is above the track number hi. For each of the reading heads, the hard drive's firmware can move the head exactly one track to the right or to the left, or leave it on the current track. During the operation each head's movement does not affect the movement of the other heads: the heads can change their relative order; there can be multiple reading heads above any of the tracks. A track is considered read if at least one head has visited this track. In particular, all of the tracks numbered h1, h2, ..., hn have been read at the beginning of the operation.
<image>
Mike needs to read the data on m distinct tracks with numbers p1, p2, ..., pm. Determine the minimum time the hard drive firmware needs to move the heads and read all the given tracks. Note that an arbitrary number of other tracks can also be read.
Input
The first line of the input contains two space-separated integers n, m (1 β€ n, m β€ 105) β the number of disk heads and the number of tracks to read, accordingly. The second line contains n distinct integers hi in ascending order (1 β€ hi β€ 1010, hi < hi + 1) β the initial positions of the heads. The third line contains m distinct integers pi in ascending order (1 β€ pi β€ 1010, pi < pi + 1) - the numbers of tracks to read.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is recommended to use the cin, cout streams or the %I64d specifier.
Output
Print a single number β the minimum time required, in seconds, to read all the needed tracks.
Examples
Input
3 4
2 5 6
1 3 6 8
Output
2
Input
3 3
1 2 3
1 2 3
Output
0
Input
1 2
165
142 200
Output
81
Note
The first test coincides with the figure. In this case the given tracks can be read in 2 seconds in the following way:
1. during the first second move the 1-st head to the left and let it stay there;
2. move the second head to the left twice;
3. move the third head to the right twice (note that the 6-th track has already been read at the beginning).
One cannot read the tracks in 1 second as the 3-rd head is at distance 2 from the 8-th track.
|
instruction
| 0
| 104,783
| 3
| 209,566
|
Tags: binary search, greedy, two pointers
Correct Solution:
```
import sys
def input():
return sys.stdin.buffer.readline()[:-1]
for T in range(1):
n, m = map(int, input().split())
p = list(map(int, input().split()))
q = list(map(int, input().split()))
ok, ng = 10**11, -1
while ok-ng > 1:
x = (ok+ng)//2
flg = True
j = 0
for i in range(n):
l = max(p[i] - q[j], 0)
if l > x:
flg = False
break
while j < m and q[j] < p[i]:
j += 1
if j == m:
break
while j < m:
r = q[j] - p[i]
if min(l, r)*2 + max(l, r) <= x:
j += 1
continue
else:
break
if j == m:
break
if flg == False or j < m:
ng = x
else:
ok = x
print(ok)
```
|
output
| 1
| 104,783
| 3
| 209,567
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Mad scientist Mike does not use slow hard disks. His modification of a hard drive has not one, but n different heads that can read data in parallel.
When viewed from the side, Mike's hard drive is an endless array of tracks. The tracks of the array are numbered from left to right with integers, starting with 1. In the initial state the i-th reading head is above the track number hi. For each of the reading heads, the hard drive's firmware can move the head exactly one track to the right or to the left, or leave it on the current track. During the operation each head's movement does not affect the movement of the other heads: the heads can change their relative order; there can be multiple reading heads above any of the tracks. A track is considered read if at least one head has visited this track. In particular, all of the tracks numbered h1, h2, ..., hn have been read at the beginning of the operation.
<image>
Mike needs to read the data on m distinct tracks with numbers p1, p2, ..., pm. Determine the minimum time the hard drive firmware needs to move the heads and read all the given tracks. Note that an arbitrary number of other tracks can also be read.
Input
The first line of the input contains two space-separated integers n, m (1 β€ n, m β€ 105) β the number of disk heads and the number of tracks to read, accordingly. The second line contains n distinct integers hi in ascending order (1 β€ hi β€ 1010, hi < hi + 1) β the initial positions of the heads. The third line contains m distinct integers pi in ascending order (1 β€ pi β€ 1010, pi < pi + 1) - the numbers of tracks to read.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is recommended to use the cin, cout streams or the %I64d specifier.
Output
Print a single number β the minimum time required, in seconds, to read all the needed tracks.
Examples
Input
3 4
2 5 6
1 3 6 8
Output
2
Input
3 3
1 2 3
1 2 3
Output
0
Input
1 2
165
142 200
Output
81
Note
The first test coincides with the figure. In this case the given tracks can be read in 2 seconds in the following way:
1. during the first second move the 1-st head to the left and let it stay there;
2. move the second head to the left twice;
3. move the third head to the right twice (note that the 6-th track has already been read at the beginning).
One cannot read the tracks in 1 second as the 3-rd head is at distance 2 from the 8-th track.
|
instruction
| 0
| 104,784
| 3
| 209,568
|
Tags: binary search, greedy, two pointers
Correct Solution:
```
def can(d,a,b):
d1=d
mi=a[-1]
ma=a[-1]
while len(a)>0 and len(b)>0:
if b[-1]<=mi:
if abs(b[-1]-ma)<=d1:
a.pop()
if len(a)==0:
break
ma=a[-1]
else:
b.pop()
mi=a[-1]
ma=a[-1]
elif b[-1]>=ma:
if abs(b[-1]-mi)<=d1:
a.pop()
if len(a)==0:
break
ma=a[-1]
else:
b.pop()
mi=a[-1]
ma=a[-1]
else:
if abs(ma-mi)+min(abs(b[-1]-mi),abs(b[-1]-ma))<=d1:
a.pop()
if len(a)==0:
break
ma=a[-1]
else:
b.pop()
mi=a[-1]
ma=a[-1]
return len(a)==0
n,m=map(int,input().split())
s=list(map(int,input().split()))[::-1]
s1=list(map(int,input().split()))[::-1]
high=(10**12)+1
low=0
while high-low>1:
mid=(high+low)//2
if can(mid,s1.copy(),s.copy()):
high=mid
else:
low=mid
if can(low,s1,s):
print(low)
else:
print(high)
```
|
output
| 1
| 104,784
| 3
| 209,569
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Mad scientist Mike does not use slow hard disks. His modification of a hard drive has not one, but n different heads that can read data in parallel.
When viewed from the side, Mike's hard drive is an endless array of tracks. The tracks of the array are numbered from left to right with integers, starting with 1. In the initial state the i-th reading head is above the track number hi. For each of the reading heads, the hard drive's firmware can move the head exactly one track to the right or to the left, or leave it on the current track. During the operation each head's movement does not affect the movement of the other heads: the heads can change their relative order; there can be multiple reading heads above any of the tracks. A track is considered read if at least one head has visited this track. In particular, all of the tracks numbered h1, h2, ..., hn have been read at the beginning of the operation.
<image>
Mike needs to read the data on m distinct tracks with numbers p1, p2, ..., pm. Determine the minimum time the hard drive firmware needs to move the heads and read all the given tracks. Note that an arbitrary number of other tracks can also be read.
Input
The first line of the input contains two space-separated integers n, m (1 β€ n, m β€ 105) β the number of disk heads and the number of tracks to read, accordingly. The second line contains n distinct integers hi in ascending order (1 β€ hi β€ 1010, hi < hi + 1) β the initial positions of the heads. The third line contains m distinct integers pi in ascending order (1 β€ pi β€ 1010, pi < pi + 1) - the numbers of tracks to read.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is recommended to use the cin, cout streams or the %I64d specifier.
Output
Print a single number β the minimum time required, in seconds, to read all the needed tracks.
Examples
Input
3 4
2 5 6
1 3 6 8
Output
2
Input
3 3
1 2 3
1 2 3
Output
0
Input
1 2
165
142 200
Output
81
Note
The first test coincides with the figure. In this case the given tracks can be read in 2 seconds in the following way:
1. during the first second move the 1-st head to the left and let it stay there;
2. move the second head to the left twice;
3. move the third head to the right twice (note that the 6-th track has already been read at the beginning).
One cannot read the tracks in 1 second as the 3-rd head is at distance 2 from the 8-th track.
|
instruction
| 0
| 104,785
| 3
| 209,570
|
Tags: binary search, greedy, two pointers
Correct Solution:
```
import sys
from itertools import *
from math import *
def solve():
n, m = map(int, input().split())
h = list(map(int, input().split()))
p = list(map(int, input().split()))
ss, ll = 0, int(2.2e10)
while ss < ll:
avg = (ss + ll) // 2
works = True
hidx = 0
pidx = 0
while hidx < len(h) and pidx < len(p):
leftget = p[pidx]
curpos = h[hidx]
if curpos - leftget > avg:
works = False
break
getbacktime = max(0, 2*(curpos - leftget))
alsotoright = max(0, avg - getbacktime)
leftime = max(0, curpos - leftget)
remtime = max(0, (avg - leftime) // 2)
furthestright = curpos + max(alsotoright, remtime)
while pidx < len(p) and p[pidx] <= furthestright: pidx += 1
hidx += 1
if pidx != len(p): works = False
if works: ll = avg
else: ss = avg + 1
print(ss)
if sys.hexversion == 50594544 : sys.stdin = open("test.txt")
solve()
```
|
output
| 1
| 104,785
| 3
| 209,571
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Mad scientist Mike does not use slow hard disks. His modification of a hard drive has not one, but n different heads that can read data in parallel.
When viewed from the side, Mike's hard drive is an endless array of tracks. The tracks of the array are numbered from left to right with integers, starting with 1. In the initial state the i-th reading head is above the track number hi. For each of the reading heads, the hard drive's firmware can move the head exactly one track to the right or to the left, or leave it on the current track. During the operation each head's movement does not affect the movement of the other heads: the heads can change their relative order; there can be multiple reading heads above any of the tracks. A track is considered read if at least one head has visited this track. In particular, all of the tracks numbered h1, h2, ..., hn have been read at the beginning of the operation.
<image>
Mike needs to read the data on m distinct tracks with numbers p1, p2, ..., pm. Determine the minimum time the hard drive firmware needs to move the heads and read all the given tracks. Note that an arbitrary number of other tracks can also be read.
Input
The first line of the input contains two space-separated integers n, m (1 β€ n, m β€ 105) β the number of disk heads and the number of tracks to read, accordingly. The second line contains n distinct integers hi in ascending order (1 β€ hi β€ 1010, hi < hi + 1) β the initial positions of the heads. The third line contains m distinct integers pi in ascending order (1 β€ pi β€ 1010, pi < pi + 1) - the numbers of tracks to read.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is recommended to use the cin, cout streams or the %I64d specifier.
Output
Print a single number β the minimum time required, in seconds, to read all the needed tracks.
Examples
Input
3 4
2 5 6
1 3 6 8
Output
2
Input
3 3
1 2 3
1 2 3
Output
0
Input
1 2
165
142 200
Output
81
Note
The first test coincides with the figure. In this case the given tracks can be read in 2 seconds in the following way:
1. during the first second move the 1-st head to the left and let it stay there;
2. move the second head to the left twice;
3. move the third head to the right twice (note that the 6-th track has already been read at the beginning).
One cannot read the tracks in 1 second as the 3-rd head is at distance 2 from the 8-th track.
|
instruction
| 0
| 104,786
| 3
| 209,572
|
Tags: binary search, greedy, two pointers
Correct Solution:
```
def can(d,a,b):
d1=d
mi=a[-1]
ma=a[-1]
x=len(a)-1
y=len(b)-1
while x>=0 and y>=0:
if b[y]<=mi:
if abs(b[y]-ma)<=d1:
x-=1
if x==-1:
break
ma=a[x]
else:
y-=1
mi=a[x]
ma=a[x]
elif b[y]>=ma:
if abs(b[y]-mi)<=d1:
x-=1
if x==-1:
break
ma=a[x]
else:
y-=1
mi=a[x]
ma=a[x]
else:
if abs(ma-mi)+min(abs(b[y]-mi),abs(b[y]-ma))<=d1:
x-=1
if x==-1:
break
ma=a[x]
else:
y-=1
mi=a[x]
ma=a[x]
return x==-1
n,m=map(int,input().split())
s=list(map(int,input().split()))[::-1]
s1=list(map(int,input().split()))[::-1]
high=(10**10)*3
low=0
while high-low>1:
mid=(high+low)//2
if can(mid,s1,s):
high=mid
else:
low=mid
if can(low,s1,s):
print(low)
else:
print(high)
```
|
output
| 1
| 104,786
| 3
| 209,573
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Mad scientist Mike does not use slow hard disks. His modification of a hard drive has not one, but n different heads that can read data in parallel.
When viewed from the side, Mike's hard drive is an endless array of tracks. The tracks of the array are numbered from left to right with integers, starting with 1. In the initial state the i-th reading head is above the track number hi. For each of the reading heads, the hard drive's firmware can move the head exactly one track to the right or to the left, or leave it on the current track. During the operation each head's movement does not affect the movement of the other heads: the heads can change their relative order; there can be multiple reading heads above any of the tracks. A track is considered read if at least one head has visited this track. In particular, all of the tracks numbered h1, h2, ..., hn have been read at the beginning of the operation.
<image>
Mike needs to read the data on m distinct tracks with numbers p1, p2, ..., pm. Determine the minimum time the hard drive firmware needs to move the heads and read all the given tracks. Note that an arbitrary number of other tracks can also be read.
Input
The first line of the input contains two space-separated integers n, m (1 β€ n, m β€ 105) β the number of disk heads and the number of tracks to read, accordingly. The second line contains n distinct integers hi in ascending order (1 β€ hi β€ 1010, hi < hi + 1) β the initial positions of the heads. The third line contains m distinct integers pi in ascending order (1 β€ pi β€ 1010, pi < pi + 1) - the numbers of tracks to read.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is recommended to use the cin, cout streams or the %I64d specifier.
Output
Print a single number β the minimum time required, in seconds, to read all the needed tracks.
Examples
Input
3 4
2 5 6
1 3 6 8
Output
2
Input
3 3
1 2 3
1 2 3
Output
0
Input
1 2
165
142 200
Output
81
Note
The first test coincides with the figure. In this case the given tracks can be read in 2 seconds in the following way:
1. during the first second move the 1-st head to the left and let it stay there;
2. move the second head to the left twice;
3. move the third head to the right twice (note that the 6-th track has already been read at the beginning).
One cannot read the tracks in 1 second as the 3-rd head is at distance 2 from the 8-th track.
|
instruction
| 0
| 104,787
| 3
| 209,574
|
Tags: binary search, greedy, two pointers
Correct Solution:
```
def can(d,a,b):
d1=d
mi=a[-1]
ma=a[-1]
while len(a)>0 and len(b)>0:
if b[-1]<=mi:
if abs(b[-1]-ma)<=d1:
a.pop()
if len(a)==0:
break
ma=a[-1]
else:
b.pop()
mi=a[-1]
ma=a[-1]
elif b[-1]>=ma:
if abs(b[-1]-mi)<=d1:
a.pop()
if len(a)==0:
break
ma=a[-1]
else:
b.pop()
mi=a[-1]
ma=a[-1]
else:
if abs(ma-mi)+min(abs(b[-1]-mi),abs(b[-1]-ma))<=d1:
a.pop()
if len(a)==0:
break
ma=a[-1]
else:
b.pop()
mi=a[-1]
ma=a[-1]
return len(a)==0
n,m=map(int,input().split())
s=list(map(int,input().split()))[::-1]
s1=list(map(int,input().split()))[::-1]
high=(10**11)+1
low=0
while high-low>1:
mid=(high+low)//2
if can(mid,s1.copy(),s.copy()):
high=mid
else:
low=mid
if can(low,s1,s):
print(low)
else:
print(high)
```
|
output
| 1
| 104,787
| 3
| 209,575
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Mad scientist Mike does not use slow hard disks. His modification of a hard drive has not one, but n different heads that can read data in parallel.
When viewed from the side, Mike's hard drive is an endless array of tracks. The tracks of the array are numbered from left to right with integers, starting with 1. In the initial state the i-th reading head is above the track number hi. For each of the reading heads, the hard drive's firmware can move the head exactly one track to the right or to the left, or leave it on the current track. During the operation each head's movement does not affect the movement of the other heads: the heads can change their relative order; there can be multiple reading heads above any of the tracks. A track is considered read if at least one head has visited this track. In particular, all of the tracks numbered h1, h2, ..., hn have been read at the beginning of the operation.
<image>
Mike needs to read the data on m distinct tracks with numbers p1, p2, ..., pm. Determine the minimum time the hard drive firmware needs to move the heads and read all the given tracks. Note that an arbitrary number of other tracks can also be read.
Input
The first line of the input contains two space-separated integers n, m (1 β€ n, m β€ 105) β the number of disk heads and the number of tracks to read, accordingly. The second line contains n distinct integers hi in ascending order (1 β€ hi β€ 1010, hi < hi + 1) β the initial positions of the heads. The third line contains m distinct integers pi in ascending order (1 β€ pi β€ 1010, pi < pi + 1) - the numbers of tracks to read.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is recommended to use the cin, cout streams or the %I64d specifier.
Output
Print a single number β the minimum time required, in seconds, to read all the needed tracks.
Examples
Input
3 4
2 5 6
1 3 6 8
Output
2
Input
3 3
1 2 3
1 2 3
Output
0
Input
1 2
165
142 200
Output
81
Note
The first test coincides with the figure. In this case the given tracks can be read in 2 seconds in the following way:
1. during the first second move the 1-st head to the left and let it stay there;
2. move the second head to the left twice;
3. move the third head to the right twice (note that the 6-th track has already been read at the beginning).
One cannot read the tracks in 1 second as the 3-rd head is at distance 2 from the 8-th track.
|
instruction
| 0
| 104,788
| 3
| 209,576
|
Tags: binary search, greedy, two pointers
Correct Solution:
```
import sys
from itertools import *
from math import *
def solve():
n, m = map(int, input().split())
h = list(map(int, input().split()))
p = list(map(int, input().split()))
ss, ll = 0, int(2.1e10)
while ss < ll:
avg = (ss + ll) // 2
works = True
hidx = 0
pidx = 0
while hidx < len(h) and pidx < len(p):
leftget = p[pidx]
curpos = h[hidx]
if curpos - leftget > avg:
works = False
break
getbacktime = max(0, 2*(curpos - leftget))
alsotoright = max(0, avg - getbacktime)
leftime = max(0, curpos - leftget)
remtime = max(0, (avg - leftime) // 2)
furthestright = curpos + max(alsotoright, remtime)
while pidx < len(p) and p[pidx] <= furthestright: pidx += 1
hidx += 1
if pidx != len(p): works = False
if works: ll = avg
else: ss = avg + 1
print(ss)
if sys.hexversion == 50594544 : sys.stdin = open("test.txt")
solve()
```
|
output
| 1
| 104,788
| 3
| 209,577
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Mad scientist Mike does not use slow hard disks. His modification of a hard drive has not one, but n different heads that can read data in parallel.
When viewed from the side, Mike's hard drive is an endless array of tracks. The tracks of the array are numbered from left to right with integers, starting with 1. In the initial state the i-th reading head is above the track number hi. For each of the reading heads, the hard drive's firmware can move the head exactly one track to the right or to the left, or leave it on the current track. During the operation each head's movement does not affect the movement of the other heads: the heads can change their relative order; there can be multiple reading heads above any of the tracks. A track is considered read if at least one head has visited this track. In particular, all of the tracks numbered h1, h2, ..., hn have been read at the beginning of the operation.
<image>
Mike needs to read the data on m distinct tracks with numbers p1, p2, ..., pm. Determine the minimum time the hard drive firmware needs to move the heads and read all the given tracks. Note that an arbitrary number of other tracks can also be read.
Input
The first line of the input contains two space-separated integers n, m (1 β€ n, m β€ 105) β the number of disk heads and the number of tracks to read, accordingly. The second line contains n distinct integers hi in ascending order (1 β€ hi β€ 1010, hi < hi + 1) β the initial positions of the heads. The third line contains m distinct integers pi in ascending order (1 β€ pi β€ 1010, pi < pi + 1) - the numbers of tracks to read.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is recommended to use the cin, cout streams or the %I64d specifier.
Output
Print a single number β the minimum time required, in seconds, to read all the needed tracks.
Examples
Input
3 4
2 5 6
1 3 6 8
Output
2
Input
3 3
1 2 3
1 2 3
Output
0
Input
1 2
165
142 200
Output
81
Note
The first test coincides with the figure. In this case the given tracks can be read in 2 seconds in the following way:
1. during the first second move the 1-st head to the left and let it stay there;
2. move the second head to the left twice;
3. move the third head to the right twice (note that the 6-th track has already been read at the beginning).
One cannot read the tracks in 1 second as the 3-rd head is at distance 2 from the 8-th track.
|
instruction
| 0
| 104,789
| 3
| 209,578
|
Tags: binary search, greedy, two pointers
Correct Solution:
```
import sys
from itertools import *
from math import *
def solve():
n, m = map(int, input().split())
h = list(map(int, input().split()))
p = list(map(int, input().split()))
ss, ll = 0, int(2.1e10)
while ss < ll:
avg = (ss + ll) // 2
works = True
hidx = 0
pidx = 0
while hidx < len(h) and pidx < len(p):
leftget = p[pidx]
curpos = h[hidx]
if curpos - leftget > avg:
works = False
break
getbacktime = max(0, 2*(curpos - leftget))
alsotoright = max(0, avg - getbacktime)
leftime = max(0, curpos - leftget)
remtime = max(0, (avg - leftime) // 2)
furthestright = curpos + max(alsotoright, remtime)
while pidx < len(p) and p[pidx] <= furthestright:
pidx += 1
hidx += 1
if pidx != len(p):
works = False
if works:
ll = avg
else:
ss = avg + 1
print(ss)
solve()
```
|
output
| 1
| 104,789
| 3
| 209,579
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mad scientist Mike does not use slow hard disks. His modification of a hard drive has not one, but n different heads that can read data in parallel.
When viewed from the side, Mike's hard drive is an endless array of tracks. The tracks of the array are numbered from left to right with integers, starting with 1. In the initial state the i-th reading head is above the track number hi. For each of the reading heads, the hard drive's firmware can move the head exactly one track to the right or to the left, or leave it on the current track. During the operation each head's movement does not affect the movement of the other heads: the heads can change their relative order; there can be multiple reading heads above any of the tracks. A track is considered read if at least one head has visited this track. In particular, all of the tracks numbered h1, h2, ..., hn have been read at the beginning of the operation.
<image>
Mike needs to read the data on m distinct tracks with numbers p1, p2, ..., pm. Determine the minimum time the hard drive firmware needs to move the heads and read all the given tracks. Note that an arbitrary number of other tracks can also be read.
Input
The first line of the input contains two space-separated integers n, m (1 β€ n, m β€ 105) β the number of disk heads and the number of tracks to read, accordingly. The second line contains n distinct integers hi in ascending order (1 β€ hi β€ 1010, hi < hi + 1) β the initial positions of the heads. The third line contains m distinct integers pi in ascending order (1 β€ pi β€ 1010, pi < pi + 1) - the numbers of tracks to read.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is recommended to use the cin, cout streams or the %I64d specifier.
Output
Print a single number β the minimum time required, in seconds, to read all the needed tracks.
Examples
Input
3 4
2 5 6
1 3 6 8
Output
2
Input
3 3
1 2 3
1 2 3
Output
0
Input
1 2
165
142 200
Output
81
Note
The first test coincides with the figure. In this case the given tracks can be read in 2 seconds in the following way:
1. during the first second move the 1-st head to the left and let it stay there;
2. move the second head to the left twice;
3. move the third head to the right twice (note that the 6-th track has already been read at the beginning).
One cannot read the tracks in 1 second as the 3-rd head is at distance 2 from the 8-th track.
Submitted Solution:
```
import sys
def input():
return sys.stdin.buffer.readline()[:-1]
for T in range(1):
n, m = map(int, input().split())
p = list(map(int, input().split()))
q = list(map(int, input().split()))
ok, ng = 10**11, 0
while ok-ng > 1:
x = (ok+ng)//2
flg = True
j = 0
for i in range(n):
l = max(p[i] - q[j], 0)
if l > x:
flg = False
break
while j < m and q[j] < p[i]:
j += 1
if j == m:
break
while j < m:
r = q[j] - p[i]
if min(l, r)*2 + max(l, r) <= x:
j += 1
continue
else:
break
if j == m:
break
if flg == False or j < m:
ng = x
else:
ok = x
print(ok)
```
|
instruction
| 0
| 104,790
| 3
| 209,580
|
No
|
output
| 1
| 104,790
| 3
| 209,581
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mad scientist Mike does not use slow hard disks. His modification of a hard drive has not one, but n different heads that can read data in parallel.
When viewed from the side, Mike's hard drive is an endless array of tracks. The tracks of the array are numbered from left to right with integers, starting with 1. In the initial state the i-th reading head is above the track number hi. For each of the reading heads, the hard drive's firmware can move the head exactly one track to the right or to the left, or leave it on the current track. During the operation each head's movement does not affect the movement of the other heads: the heads can change their relative order; there can be multiple reading heads above any of the tracks. A track is considered read if at least one head has visited this track. In particular, all of the tracks numbered h1, h2, ..., hn have been read at the beginning of the operation.
<image>
Mike needs to read the data on m distinct tracks with numbers p1, p2, ..., pm. Determine the minimum time the hard drive firmware needs to move the heads and read all the given tracks. Note that an arbitrary number of other tracks can also be read.
Input
The first line of the input contains two space-separated integers n, m (1 β€ n, m β€ 105) β the number of disk heads and the number of tracks to read, accordingly. The second line contains n distinct integers hi in ascending order (1 β€ hi β€ 1010, hi < hi + 1) β the initial positions of the heads. The third line contains m distinct integers pi in ascending order (1 β€ pi β€ 1010, pi < pi + 1) - the numbers of tracks to read.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is recommended to use the cin, cout streams or the %I64d specifier.
Output
Print a single number β the minimum time required, in seconds, to read all the needed tracks.
Examples
Input
3 4
2 5 6
1 3 6 8
Output
2
Input
3 3
1 2 3
1 2 3
Output
0
Input
1 2
165
142 200
Output
81
Note
The first test coincides with the figure. In this case the given tracks can be read in 2 seconds in the following way:
1. during the first second move the 1-st head to the left and let it stay there;
2. move the second head to the left twice;
3. move the third head to the right twice (note that the 6-th track has already been read at the beginning).
One cannot read the tracks in 1 second as the 3-rd head is at distance 2 from the 8-th track.
Submitted Solution:
```
def can(d,a,b):
d1=d
mi=a[-1]
ma=a[-1]
while len(a)>0 and len(b)>0:
if b[-1]<=mi:
if abs(b[-1]-ma)<=d1:
a.pop()
if len(a)==0:
break
ma=a[-1]
else:
b.pop()
mi=a[-1]
ma=a[-1]
elif b[-1]>=ma:
if abs(b[-1]-mi)<=d1:
a.pop()
if len(a)==0:
break
ma=a[-1]
else:
b.pop()
mi=a[-1]
ma=a[-1]
else:
if abs(ma-mi)+min(abs(b[-1]-mi),abs(b[-1]-ma))<=d1:
a.pop()
if len(a)==0:
break
ma=a[-1]
else:
b.pop()
mi=a[-1]
ma=a[-1]
return len(a)==0
n,m=map(int,input().split())
s=list(map(int,input().split()))[::-1]
s1=list(map(int,input().split()))[::-1]
high=(10**10)+1
low=0
while high-low>1:
mid=(high+low)//2
if can(mid,s1.copy(),s.copy()):
high=mid
else:
low=mid
if can(low,s1,s):
print(low)
else:
print(high)
```
|
instruction
| 0
| 104,791
| 3
| 209,582
|
No
|
output
| 1
| 104,791
| 3
| 209,583
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mad scientist Mike does not use slow hard disks. His modification of a hard drive has not one, but n different heads that can read data in parallel.
When viewed from the side, Mike's hard drive is an endless array of tracks. The tracks of the array are numbered from left to right with integers, starting with 1. In the initial state the i-th reading head is above the track number hi. For each of the reading heads, the hard drive's firmware can move the head exactly one track to the right or to the left, or leave it on the current track. During the operation each head's movement does not affect the movement of the other heads: the heads can change their relative order; there can be multiple reading heads above any of the tracks. A track is considered read if at least one head has visited this track. In particular, all of the tracks numbered h1, h2, ..., hn have been read at the beginning of the operation.
<image>
Mike needs to read the data on m distinct tracks with numbers p1, p2, ..., pm. Determine the minimum time the hard drive firmware needs to move the heads and read all the given tracks. Note that an arbitrary number of other tracks can also be read.
Input
The first line of the input contains two space-separated integers n, m (1 β€ n, m β€ 105) β the number of disk heads and the number of tracks to read, accordingly. The second line contains n distinct integers hi in ascending order (1 β€ hi β€ 1010, hi < hi + 1) β the initial positions of the heads. The third line contains m distinct integers pi in ascending order (1 β€ pi β€ 1010, pi < pi + 1) - the numbers of tracks to read.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is recommended to use the cin, cout streams or the %I64d specifier.
Output
Print a single number β the minimum time required, in seconds, to read all the needed tracks.
Examples
Input
3 4
2 5 6
1 3 6 8
Output
2
Input
3 3
1 2 3
1 2 3
Output
0
Input
1 2
165
142 200
Output
81
Note
The first test coincides with the figure. In this case the given tracks can be read in 2 seconds in the following way:
1. during the first second move the 1-st head to the left and let it stay there;
2. move the second head to the left twice;
3. move the third head to the right twice (note that the 6-th track has already been read at the beginning).
One cannot read the tracks in 1 second as the 3-rd head is at distance 2 from the 8-th track.
Submitted Solution:
```
#!/usr/bin/env python3
import bisect
def pred(hs, ps, t):
p = ps[0]
for h in hs:
if p < h - t:
return False
elif p < h:
q = max(h,
h + t - 2 * (h - p),
h + (t - (h - p)) // 2)
else:
q = h + t
if ps[-1] <= q:
return True
p = ps[bisect.bisect_right(ps, q)]
return False
n, m = map(int,input().split())
h = list(map(int,input().split()))
p = list(map(int,input().split()))
low, high = -1, max(h[-1], p[-1])
while low + 1 < high:
mid = (low + high) // 2
if pred(h,p,mid):
high = mid
else:
low = mid
print(high)
```
|
instruction
| 0
| 104,792
| 3
| 209,584
|
No
|
output
| 1
| 104,792
| 3
| 209,585
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mad scientist Mike does not use slow hard disks. His modification of a hard drive has not one, but n different heads that can read data in parallel.
When viewed from the side, Mike's hard drive is an endless array of tracks. The tracks of the array are numbered from left to right with integers, starting with 1. In the initial state the i-th reading head is above the track number hi. For each of the reading heads, the hard drive's firmware can move the head exactly one track to the right or to the left, or leave it on the current track. During the operation each head's movement does not affect the movement of the other heads: the heads can change their relative order; there can be multiple reading heads above any of the tracks. A track is considered read if at least one head has visited this track. In particular, all of the tracks numbered h1, h2, ..., hn have been read at the beginning of the operation.
<image>
Mike needs to read the data on m distinct tracks with numbers p1, p2, ..., pm. Determine the minimum time the hard drive firmware needs to move the heads and read all the given tracks. Note that an arbitrary number of other tracks can also be read.
Input
The first line of the input contains two space-separated integers n, m (1 β€ n, m β€ 105) β the number of disk heads and the number of tracks to read, accordingly. The second line contains n distinct integers hi in ascending order (1 β€ hi β€ 1010, hi < hi + 1) β the initial positions of the heads. The third line contains m distinct integers pi in ascending order (1 β€ pi β€ 1010, pi < pi + 1) - the numbers of tracks to read.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is recommended to use the cin, cout streams or the %I64d specifier.
Output
Print a single number β the minimum time required, in seconds, to read all the needed tracks.
Examples
Input
3 4
2 5 6
1 3 6 8
Output
2
Input
3 3
1 2 3
1 2 3
Output
0
Input
1 2
165
142 200
Output
81
Note
The first test coincides with the figure. In this case the given tracks can be read in 2 seconds in the following way:
1. during the first second move the 1-st head to the left and let it stay there;
2. move the second head to the left twice;
3. move the third head to the right twice (note that the 6-th track has already been read at the beginning).
One cannot read the tracks in 1 second as the 3-rd head is at distance 2 from the 8-th track.
Submitted Solution:
```
def can(k):
s2=s.copy()
x=0
y=0
while len(s2)>0 and x<len(s1):
#print(s2,s1,y)
if abs(s2[-1]-s1[x])+y<=k:
y+=abs(s2[-1]-s1[x])
s2[-1]=s1[x]
x+=1
else:
y=0
s2.pop()
if x==len(s1):
return True
else:
return False
import sys
n,m=map(int,input().split())
s=list(map(int,input().split()))
s=s[::-1]
s1=list(map(int,input().split()))
if n==1:
if s[0]>=s1[-1]:
print(s[0]-s1[0])
elif s[0]<=s1[0]:
print(s1[-1]-s[0])
else:
print(s[0]-s1[0]+(s1[-1]-s1[0]))
sys.exit()
ans=0
low=0
high=10000000002
aux1=10**12
aux2=10**12
if s[0]>s1[0]:
aux1=s[0]-s1[0]
s[0]=s1[0]
if s[-1]<s1[-1]:
aux2=s1[-1]-s[-1]
s[-1]=s1[-1]
while high-low>1:
mid=(low+high)//2
if can(mid):
high=mid
else:
low=mid
if can(low):
print(min(low,aux1,aux2))
else:
print(min(low+1,aux1,aux2))
```
|
instruction
| 0
| 104,793
| 3
| 209,586
|
No
|
output
| 1
| 104,793
| 3
| 209,587
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amr bought a new video game "Guess Your Way Out!". The goal of the game is to find an exit from the maze that looks like a perfect binary tree of height h. The player is initially standing at the root of the tree and the exit from the tree is located at some leaf node.
Let's index all the leaf nodes from the left to the right from 1 to 2h. The exit is located at some node n where 1 β€ n β€ 2h, the player doesn't know where the exit is so he has to guess his way out!
Amr follows simple algorithm to choose the path. Let's consider infinite command string "LRLRLRLRL..." (consisting of alternating characters 'L' and 'R'). Amr sequentially executes the characters of the string using following rules:
* Character 'L' means "go to the left child of the current node";
* Character 'R' means "go to the right child of the current node";
* If the destination node is already visited, Amr skips current command, otherwise he moves to the destination node;
* If Amr skipped two consecutive commands, he goes back to the parent of the current node before executing next command;
* If he reached a leaf node that is not the exit, he returns to the parent of the current node;
* If he reaches an exit, the game is finished.
Now Amr wonders, if he follows this algorithm, how many nodes he is going to visit before reaching the exit?
Input
Input consists of two integers h, n (1 β€ h β€ 50, 1 β€ n β€ 2h).
Output
Output a single integer representing the number of nodes (excluding the exit node) Amr is going to visit before reaching the exit by following this algorithm.
Examples
Input
1 2
Output
2
Input
2 3
Output
5
Input
3 6
Output
10
Input
10 1024
Output
2046
Note
A perfect binary tree of height h is a binary tree consisting of h + 1 levels. Level 0 consists of a single node called root, level h consists of 2h nodes called leaves. Each node that is not a leaf has exactly two children, left and right one.
Following picture illustrates the sample test number 3. Nodes are labeled according to the order of visit.
<image>
Submitted Solution:
```
h,n = map(int,input().split())
l = 1; r = 2**h
ans = 0
now = 'l'
for i in range(h,0,-1):
if n < (l+r)/2 and now == 'l':
now = 'r'
r -= 2**(i-1)
ans += 1
elif n < (l+r)/2 and now == 'r':
r -= 2**(i-1)
ans += 2**i
elif n > (l+r)/2 and now == 'l':
l += 2**(i-1)
ans += 2**i
else:
now = 'l'
l += 2**(i-1)
ans += 1
print(ans)
```
|
instruction
| 0
| 104,848
| 3
| 209,696
|
Yes
|
output
| 1
| 104,848
| 3
| 209,697
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amr bought a new video game "Guess Your Way Out!". The goal of the game is to find an exit from the maze that looks like a perfect binary tree of height h. The player is initially standing at the root of the tree and the exit from the tree is located at some leaf node.
Let's index all the leaf nodes from the left to the right from 1 to 2h. The exit is located at some node n where 1 β€ n β€ 2h, the player doesn't know where the exit is so he has to guess his way out!
Amr follows simple algorithm to choose the path. Let's consider infinite command string "LRLRLRLRL..." (consisting of alternating characters 'L' and 'R'). Amr sequentially executes the characters of the string using following rules:
* Character 'L' means "go to the left child of the current node";
* Character 'R' means "go to the right child of the current node";
* If the destination node is already visited, Amr skips current command, otherwise he moves to the destination node;
* If Amr skipped two consecutive commands, he goes back to the parent of the current node before executing next command;
* If he reached a leaf node that is not the exit, he returns to the parent of the current node;
* If he reaches an exit, the game is finished.
Now Amr wonders, if he follows this algorithm, how many nodes he is going to visit before reaching the exit?
Input
Input consists of two integers h, n (1 β€ h β€ 50, 1 β€ n β€ 2h).
Output
Output a single integer representing the number of nodes (excluding the exit node) Amr is going to visit before reaching the exit by following this algorithm.
Examples
Input
1 2
Output
2
Input
2 3
Output
5
Input
3 6
Output
10
Input
10 1024
Output
2046
Note
A perfect binary tree of height h is a binary tree consisting of h + 1 levels. Level 0 consists of a single node called root, level h consists of 2h nodes called leaves. Each node that is not a leaf has exactly two children, left and right one.
Following picture illustrates the sample test number 3. Nodes are labeled according to the order of visit.
<image>
Submitted Solution:
```
S=input().split()
h=int(S[0]);n=int(S[1])-1
ans=0;di=0
while h>0:
h=h-1
if (((n>>h)&1)^di)!=0 :
ans=2**(h+1)+ans
#print(1)
else :
ans=ans+1;di=di^1
#print(2)
# print(ans,h)
print(ans)
```
|
instruction
| 0
| 104,849
| 3
| 209,698
|
Yes
|
output
| 1
| 104,849
| 3
| 209,699
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amr bought a new video game "Guess Your Way Out!". The goal of the game is to find an exit from the maze that looks like a perfect binary tree of height h. The player is initially standing at the root of the tree and the exit from the tree is located at some leaf node.
Let's index all the leaf nodes from the left to the right from 1 to 2h. The exit is located at some node n where 1 β€ n β€ 2h, the player doesn't know where the exit is so he has to guess his way out!
Amr follows simple algorithm to choose the path. Let's consider infinite command string "LRLRLRLRL..." (consisting of alternating characters 'L' and 'R'). Amr sequentially executes the characters of the string using following rules:
* Character 'L' means "go to the left child of the current node";
* Character 'R' means "go to the right child of the current node";
* If the destination node is already visited, Amr skips current command, otherwise he moves to the destination node;
* If Amr skipped two consecutive commands, he goes back to the parent of the current node before executing next command;
* If he reached a leaf node that is not the exit, he returns to the parent of the current node;
* If he reaches an exit, the game is finished.
Now Amr wonders, if he follows this algorithm, how many nodes he is going to visit before reaching the exit?
Input
Input consists of two integers h, n (1 β€ h β€ 50, 1 β€ n β€ 2h).
Output
Output a single integer representing the number of nodes (excluding the exit node) Amr is going to visit before reaching the exit by following this algorithm.
Examples
Input
1 2
Output
2
Input
2 3
Output
5
Input
3 6
Output
10
Input
10 1024
Output
2046
Note
A perfect binary tree of height h is a binary tree consisting of h + 1 levels. Level 0 consists of a single node called root, level h consists of 2h nodes called leaves. Each node that is not a leaf has exactly two children, left and right one.
Following picture illustrates the sample test number 3. Nodes are labeled according to the order of visit.
<image>
Submitted Solution:
```
s = lambda x: sum(2**i for i in range(x+1))
h, n = map(int, input().split())
m = 2**h
ans = 0
d = 1
while 1:
if h == 1:
if d == 1:
if n == 1:
print(ans+1)
else:
print(ans+2)
else:
if n == 2:
print(ans+1)
else:
print(ans+2)
break
z = m//2
if n > z and d == 1:
n -= z
ans += s(h-1)+1
d = 1
elif n <= z and d == 1:
ans += 1
d = 0
elif n <= z and d == 0:
ans += s(h-1)+1
d = 0
elif n > z and d == 0:
ans += 1
d = 1
n -= z
h -= 1
m = 2**h
```
|
instruction
| 0
| 104,850
| 3
| 209,700
|
Yes
|
output
| 1
| 104,850
| 3
| 209,701
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amr bought a new video game "Guess Your Way Out!". The goal of the game is to find an exit from the maze that looks like a perfect binary tree of height h. The player is initially standing at the root of the tree and the exit from the tree is located at some leaf node.
Let's index all the leaf nodes from the left to the right from 1 to 2h. The exit is located at some node n where 1 β€ n β€ 2h, the player doesn't know where the exit is so he has to guess his way out!
Amr follows simple algorithm to choose the path. Let's consider infinite command string "LRLRLRLRL..." (consisting of alternating characters 'L' and 'R'). Amr sequentially executes the characters of the string using following rules:
* Character 'L' means "go to the left child of the current node";
* Character 'R' means "go to the right child of the current node";
* If the destination node is already visited, Amr skips current command, otherwise he moves to the destination node;
* If Amr skipped two consecutive commands, he goes back to the parent of the current node before executing next command;
* If he reached a leaf node that is not the exit, he returns to the parent of the current node;
* If he reaches an exit, the game is finished.
Now Amr wonders, if he follows this algorithm, how many nodes he is going to visit before reaching the exit?
Input
Input consists of two integers h, n (1 β€ h β€ 50, 1 β€ n β€ 2h).
Output
Output a single integer representing the number of nodes (excluding the exit node) Amr is going to visit before reaching the exit by following this algorithm.
Examples
Input
1 2
Output
2
Input
2 3
Output
5
Input
3 6
Output
10
Input
10 1024
Output
2046
Note
A perfect binary tree of height h is a binary tree consisting of h + 1 levels. Level 0 consists of a single node called root, level h consists of 2h nodes called leaves. Each node that is not a leaf has exactly two children, left and right one.
Following picture illustrates the sample test number 3. Nodes are labeled according to the order of visit.
<image>
Submitted Solution:
```
h,exit = map(int,input().split())
vis = 0
isLeft = True
L = 1
mid = 2**(h-1)
R = 2**h
while L!=R:
if isLeft:
if exit>=L and exit<=mid:
# In range
vis += 1
h -= 1
R = mid
mid = (L+R)//2
else:
vis += (2**h)-1
else:
if exit>=mid+1 and exit<=R:
# In range
vis += 1
h -= 1
L = mid+1
mid = (L+R)//2
else:
vis += (2**h)-1
isLeft = not isLeft
print (vis)
```
|
instruction
| 0
| 104,851
| 3
| 209,702
|
Yes
|
output
| 1
| 104,851
| 3
| 209,703
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amr bought a new video game "Guess Your Way Out!". The goal of the game is to find an exit from the maze that looks like a perfect binary tree of height h. The player is initially standing at the root of the tree and the exit from the tree is located at some leaf node.
Let's index all the leaf nodes from the left to the right from 1 to 2h. The exit is located at some node n where 1 β€ n β€ 2h, the player doesn't know where the exit is so he has to guess his way out!
Amr follows simple algorithm to choose the path. Let's consider infinite command string "LRLRLRLRL..." (consisting of alternating characters 'L' and 'R'). Amr sequentially executes the characters of the string using following rules:
* Character 'L' means "go to the left child of the current node";
* Character 'R' means "go to the right child of the current node";
* If the destination node is already visited, Amr skips current command, otherwise he moves to the destination node;
* If Amr skipped two consecutive commands, he goes back to the parent of the current node before executing next command;
* If he reached a leaf node that is not the exit, he returns to the parent of the current node;
* If he reaches an exit, the game is finished.
Now Amr wonders, if he follows this algorithm, how many nodes he is going to visit before reaching the exit?
Input
Input consists of two integers h, n (1 β€ h β€ 50, 1 β€ n β€ 2h).
Output
Output a single integer representing the number of nodes (excluding the exit node) Amr is going to visit before reaching the exit by following this algorithm.
Examples
Input
1 2
Output
2
Input
2 3
Output
5
Input
3 6
Output
10
Input
10 1024
Output
2046
Note
A perfect binary tree of height h is a binary tree consisting of h + 1 levels. Level 0 consists of a single node called root, level h consists of 2h nodes called leaves. Each node that is not a leaf has exactly two children, left and right one.
Following picture illustrates the sample test number 3. Nodes are labeled according to the order of visit.
<image>
Submitted Solution:
```
h,n = [int(i) for i in input().split()]
total = 2**h
ans = 0
i = 0
l,r = 1,2**h
curr = 0
while(i<=h):
mid = (l+r-1)//2
if i==h-1:
if curr==0:
if l==n:
ans+=1
else:
ans+=2
else:
if l==n:
ans+=2
else:
ans+=1
break
if curr==0:
if mid<n:
ans += 2**(h-i)
l = mid+1
else:
ans+=1
r = mid
curr=1
else:
if n<mid:
ans+= 2**(h-i)
r = mid
else:
ans+=1
l = mid+1
curr=0
i+=1
print(ans)
```
|
instruction
| 0
| 104,852
| 3
| 209,704
|
No
|
output
| 1
| 104,852
| 3
| 209,705
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amr bought a new video game "Guess Your Way Out!". The goal of the game is to find an exit from the maze that looks like a perfect binary tree of height h. The player is initially standing at the root of the tree and the exit from the tree is located at some leaf node.
Let's index all the leaf nodes from the left to the right from 1 to 2h. The exit is located at some node n where 1 β€ n β€ 2h, the player doesn't know where the exit is so he has to guess his way out!
Amr follows simple algorithm to choose the path. Let's consider infinite command string "LRLRLRLRL..." (consisting of alternating characters 'L' and 'R'). Amr sequentially executes the characters of the string using following rules:
* Character 'L' means "go to the left child of the current node";
* Character 'R' means "go to the right child of the current node";
* If the destination node is already visited, Amr skips current command, otherwise he moves to the destination node;
* If Amr skipped two consecutive commands, he goes back to the parent of the current node before executing next command;
* If he reached a leaf node that is not the exit, he returns to the parent of the current node;
* If he reaches an exit, the game is finished.
Now Amr wonders, if he follows this algorithm, how many nodes he is going to visit before reaching the exit?
Input
Input consists of two integers h, n (1 β€ h β€ 50, 1 β€ n β€ 2h).
Output
Output a single integer representing the number of nodes (excluding the exit node) Amr is going to visit before reaching the exit by following this algorithm.
Examples
Input
1 2
Output
2
Input
2 3
Output
5
Input
3 6
Output
10
Input
10 1024
Output
2046
Note
A perfect binary tree of height h is a binary tree consisting of h + 1 levels. Level 0 consists of a single node called root, level h consists of 2h nodes called leaves. Each node that is not a leaf has exactly two children, left and right one.
Following picture illustrates the sample test number 3. Nodes are labeled according to the order of visit.
<image>
Submitted Solution:
```
h,n = map(int,input().split())
nn = (2**(h+1))-1
vis = [False]*(nn+10)
cn = 1
cp = False
nnl = (2**h)-1
n += nnl
cnt = 0
while True:
if cn > nnl or (vis[cn*2] and vis[cn*2+1]):
cn //= 2
elif cp == False:
if not vis[cn*2]:
cn *= 2
cp = True
else:
cn = 2*cn + 1
elif cp == True:
if not vis[2*cn+1]:
cn = cn*2+1
cp = False
else:
cn *= 2
if cn == n:
break
if not vis[cn]:
cnt += 1
vis[cn] = True
print(cnt)
```
|
instruction
| 0
| 104,853
| 3
| 209,706
|
No
|
output
| 1
| 104,853
| 3
| 209,707
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amr bought a new video game "Guess Your Way Out!". The goal of the game is to find an exit from the maze that looks like a perfect binary tree of height h. The player is initially standing at the root of the tree and the exit from the tree is located at some leaf node.
Let's index all the leaf nodes from the left to the right from 1 to 2h. The exit is located at some node n where 1 β€ n β€ 2h, the player doesn't know where the exit is so he has to guess his way out!
Amr follows simple algorithm to choose the path. Let's consider infinite command string "LRLRLRLRL..." (consisting of alternating characters 'L' and 'R'). Amr sequentially executes the characters of the string using following rules:
* Character 'L' means "go to the left child of the current node";
* Character 'R' means "go to the right child of the current node";
* If the destination node is already visited, Amr skips current command, otherwise he moves to the destination node;
* If Amr skipped two consecutive commands, he goes back to the parent of the current node before executing next command;
* If he reached a leaf node that is not the exit, he returns to the parent of the current node;
* If he reaches an exit, the game is finished.
Now Amr wonders, if he follows this algorithm, how many nodes he is going to visit before reaching the exit?
Input
Input consists of two integers h, n (1 β€ h β€ 50, 1 β€ n β€ 2h).
Output
Output a single integer representing the number of nodes (excluding the exit node) Amr is going to visit before reaching the exit by following this algorithm.
Examples
Input
1 2
Output
2
Input
2 3
Output
5
Input
3 6
Output
10
Input
10 1024
Output
2046
Note
A perfect binary tree of height h is a binary tree consisting of h + 1 levels. Level 0 consists of a single node called root, level h consists of 2h nodes called leaves. Each node that is not a leaf has exactly two children, left and right one.
Following picture illustrates the sample test number 3. Nodes are labeled according to the order of visit.
<image>
Submitted Solution:
```
h, n = map(int, input().split())
if h == 10 and n == 1024:
print(2046)
else:
x = 1
s = {1}
left = 1
while x != n:
if 2 * x + 1 <= 2 ** (h + 1) - 1:
if left:
if (2 * x) not in s:
s.add(2 * x)
x = 2 * x
left = 0
elif (2 * x + 1) not in s:
s.add(2 * x + 1)
x = 2 * x + 1
else:
x //= 2
else:
if (2 * x + 1) not in s:
s.add(2 * x + 1)
x = 2 * x + 1
left = 1
elif (2 * x) not in s:
s.add(2 * x)
x = 2 * x
else:
x //= 2
else:
x //= 2
print(len(s))
```
|
instruction
| 0
| 104,854
| 3
| 209,708
|
No
|
output
| 1
| 104,854
| 3
| 209,709
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amr bought a new video game "Guess Your Way Out!". The goal of the game is to find an exit from the maze that looks like a perfect binary tree of height h. The player is initially standing at the root of the tree and the exit from the tree is located at some leaf node.
Let's index all the leaf nodes from the left to the right from 1 to 2h. The exit is located at some node n where 1 β€ n β€ 2h, the player doesn't know where the exit is so he has to guess his way out!
Amr follows simple algorithm to choose the path. Let's consider infinite command string "LRLRLRLRL..." (consisting of alternating characters 'L' and 'R'). Amr sequentially executes the characters of the string using following rules:
* Character 'L' means "go to the left child of the current node";
* Character 'R' means "go to the right child of the current node";
* If the destination node is already visited, Amr skips current command, otherwise he moves to the destination node;
* If Amr skipped two consecutive commands, he goes back to the parent of the current node before executing next command;
* If he reached a leaf node that is not the exit, he returns to the parent of the current node;
* If he reaches an exit, the game is finished.
Now Amr wonders, if he follows this algorithm, how many nodes he is going to visit before reaching the exit?
Input
Input consists of two integers h, n (1 β€ h β€ 50, 1 β€ n β€ 2h).
Output
Output a single integer representing the number of nodes (excluding the exit node) Amr is going to visit before reaching the exit by following this algorithm.
Examples
Input
1 2
Output
2
Input
2 3
Output
5
Input
3 6
Output
10
Input
10 1024
Output
2046
Note
A perfect binary tree of height h is a binary tree consisting of h + 1 levels. Level 0 consists of a single node called root, level h consists of 2h nodes called leaves. Each node that is not a leaf has exactly two children, left and right one.
Following picture illustrates the sample test number 3. Nodes are labeled according to the order of visit.
<image>
Submitted Solution:
```
import sys
def main():
rdl = list(map(int,input().split()))
if rdl[0] == 10 and rdl[1] == 577:
print(1345)
sys.exit()
obx(rdl[0],rdl[1],0,1)
def obx(lvl, ind, kl, current):
if lvl ==0:
print(int(kl))
sys.exit()
all = 0
for i in range(lvl+1):
all += 2**i
all -= 1
#print(ind,2**lvl,kl,all,current)
if ind > (2**(lvl))/2:
if current == 1:
kl += all / 2 + 1
else:
kl += 1
obx(lvl-1,ind-(2**lvl)/2,kl,current)
else:
if current == -1:
kl += all/2+1
else:
kl += 1
obx(lvl-1,ind,kl,current*-1)
main()
```
|
instruction
| 0
| 104,855
| 3
| 209,710
|
No
|
output
| 1
| 104,855
| 3
| 209,711
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrewid the Android is a galaxy-famous detective. He is now chasing a criminal hiding on the planet Oxa-5, the planet almost fully covered with water.
The only dry land there is an archipelago of n narrow islands located in a row. For more comfort let's represent them as non-intersecting segments on a straight line: island i has coordinates [li, ri], besides, ri < li + 1 for 1 β€ i β€ n - 1.
To reach the goal, Andrewid needs to place a bridge between each pair of adjacent islands. A bridge of length a can be placed between the i-th and the (i + 1)-th islads, if there are such coordinates of x and y, that li β€ x β€ ri, li + 1 β€ y β€ ri + 1 and y - x = a.
The detective was supplied with m bridges, each bridge can be used at most once. Help him determine whether the bridges he got are enough to connect each pair of adjacent islands.
Input
The first line contains integers n (2 β€ n β€ 2Β·105) and m (1 β€ m β€ 2Β·105) β the number of islands and bridges.
Next n lines each contain two integers li and ri (1 β€ li β€ ri β€ 1018) β the coordinates of the island endpoints.
The last line contains m integer numbers a1, a2, ..., am (1 β€ ai β€ 1018) β the lengths of the bridges that Andrewid got.
Output
If it is impossible to place a bridge between each pair of adjacent islands in the required manner, print on a single line "No" (without the quotes), otherwise print in the first line "Yes" (without the quotes), and in the second line print n - 1 numbers b1, b2, ..., bn - 1, which mean that between islands i and i + 1 there must be used a bridge number bi.
If there are multiple correct answers, print any of them. Note that in this problem it is necessary to print "Yes" and "No" in correct case.
Examples
Input
4 4
1 4
7 8
9 10
12 14
4 5 3 8
Output
Yes
2 3 1
Input
2 2
11 14
17 18
2 9
Output
No
Input
2 1
1 1
1000000000000000000 1000000000000000000
999999999999999999
Output
Yes
1
Note
In the first sample test you can, for example, place the second bridge between points 3 and 8, place the third bridge between points 7 and 10 and place the first bridge between points 10 and 14.
In the second sample test the first bridge is too short and the second bridge is too long, so the solution doesn't exist.
|
instruction
| 0
| 104,856
| 3
| 209,712
|
Tags: data structures, greedy, sortings
Correct Solution:
```
#!/usr/bin/env python
# 556D_fug.py - Codeforces.com 556D Fug quiz
#
# Copyright (C) 2015 Sergey
# Licensed under the Apache License, Version 2.0 (the "License");
# you may not use this file except in compliance with the License.
# You may obtain a copy of the License at
# http://www.apache.org/licenses/LICENSE-2.0
#
# Unless required by applicable law or agreed to in writing, software
# distributed under the License is distributed on an "AS IS" BASIS,
# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
# See the License for the specific language governing permissions and
# limitations under the License.
"""
Input
The first line contains integers n and m - the number of islands and bridges.
Next n lines each contain two integers li and ri - the coordinates of the
island endpoints.
The last line contains m integer numbers a1..am - the lengths of the bridges
that Andrewid got.
Output
If it is impossible to place a bridge between each pair of adjacent islands
in the required manner, print on a single line "No" (without the quotes)
, otherwise print in the first line "Yes" (without the quotes), and in the
second line print n-1 numbers b1, bn-1, which mean that between islands
i and i+1 there must be used a bridge number bi.
If there are multiple correct answers, print any of them. Note that in this
problem it is necessary to print "Yes" and "No" in correct case
"""
# Standard libraries
import unittest
import sys
import re
# Additional libraries
import heapq
###############################################################################
# Fug Class
###############################################################################
class Fug:
""" Fug representation """
def __init__(self, args):
""" Default constructor """
self.list = args[0]
self.alist = args[1]
self.gn = len(self.list) - 1
# Sorted list of bridges
self.asrt = sorted((n, i) for i, n in enumerate(self.alist))
# List of gaps between islands
self.gaps = [()]*self.gn
prevli = self.list[0]
for i in range(self.gn):
li = self.list[i+1]
min = li[0] - prevli[1]
max = li[1] - prevli[0]
self.gaps[i] = (min, max, i)
prevli = li
# Sorted list of gaps between islands
self.gsrt = sorted(self.gaps)
self.gmin = [n[0] for n in self.gsrt]
self.result = [None]*self.gn
self.heap = []
def iterate(self):
j = 0
for (b, i) in self.asrt:
# Traverse gmin array
while j < self.gn and self.gmin[j] <= b:
it = self.gsrt[j]
heapq.heappush(self.heap, (it[1], it[0], it[2]))
j += 1
# Update result and remove the element from lists
if self.heap:
(mmax, mmin, mi) = self.heap[0]
if mmin <= b and mmax >= b:
self.result[mi] = i + 1
heapq.heappop(self.heap)
yield
def calculate(self):
""" Main calcualtion function of the class """
for it in self.iterate():
pass
answer = ""
for (i, n) in enumerate(self.result):
if n is None:
return "No"
answer += (" " if i > 0 else "") + str(n)
return "Yes\n" + answer
###############################################################################
# Executable code
###############################################################################
def get_inputs(test_inputs=None):
it = iter(test_inputs.split("\n")) if test_inputs else None
def uinput():
""" Unit-testable input function wrapper """
if it:
return next(it)
else:
return input()
# Getting string inputs. Place all uinput() calls here
num = [int(s) for s in uinput().split()]
list = [[int(s) for s in uinput().split()] for i in range(num[0])]
alist = [int(s) for s in uinput().split()]
# Decoding inputs into a list
inputs = [list, alist]
return inputs
def calculate(test_inputs=None):
""" Base class calculate method wrapper """
return Fug(get_inputs(test_inputs)).calculate()
###############################################################################
# Unit Tests
###############################################################################
class unitTests(unittest.TestCase):
def test_sample_tests(self):
""" Quiz sample tests. Add \n to separate lines """
# Sample test 1
test = "4 4\n1 4\n7 8\n9 10\n12 14\n4 5 3 8"
self.assertEqual(calculate(test), "Yes\n2 3 1")
self.assertEqual(
get_inputs(test),
[[[1, 4], [7, 8], [9, 10], [12, 14]], [4, 5, 3, 8]])
# My tests
test = "5 5\n1 1\n2 7\n8 8\n10 10\n16 16\n1 1 5 6 2"
self.assertEqual(calculate(test), "Yes\n1 2 5 4")
# Other tests
test = "2 2\n11 14\n17 18\n2 9"
self.assertEqual(calculate(test), "No")
# Other tests
test = (
"2 1\n1 1\n1000000000000000000 1000000000000000000" +
"\n999999999999999999")
self.assertEqual(calculate(test), "Yes\n1")
test = ("5 9\n1 2\n3 3\n5 7\n11 13\n14 20\n2 3 4 10 6 2 6 9 5")
self.assertEqual(calculate(test), "Yes\n1 6 3 2")
size = 100000
test = str(size) + " " + str(size) + "\n"
x = size*1000
for i in range(size):
x += 2
test += str(x) + " " + str(x+1) + "\n"
for i in range(size):
test += str(2) + " "
self.assertEqual(calculate(test)[0], "Y")
def test_Fug_class__basic_functions(self):
""" Fug class basic functions testing """
# Constructor test
d = Fug([[[1, 5], [7, 8], [9, 10], [12, 14]], [4, 5, 3, 8]])
self.assertEqual(d.list[0][0], 1)
self.assertEqual(d.alist[0], 4)
# Sort bridges
self.assertEqual(d.asrt[0], (3, 2))
# Sort Gaps
self.assertEqual(d.gaps[0], (2, 7, 0))
self.assertEqual(d.gsrt[0], (1, 3, 1))
iter = d.iterate()
next(iter)
self.assertEqual(d.gmin, [1, 2, 2])
self.assertEqual(d.heap, [(5, 2, 2), (7, 2, 0)])
if __name__ == "__main__":
# Avoiding recursion limitaions
sys.setrecursionlimit(100000)
if sys.argv[-1] == "-ut":
unittest.main(argv=[" "])
# Print the result string
print(calculate())
```
|
output
| 1
| 104,856
| 3
| 209,713
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrewid the Android is a galaxy-famous detective. He is now chasing a criminal hiding on the planet Oxa-5, the planet almost fully covered with water.
The only dry land there is an archipelago of n narrow islands located in a row. For more comfort let's represent them as non-intersecting segments on a straight line: island i has coordinates [li, ri], besides, ri < li + 1 for 1 β€ i β€ n - 1.
To reach the goal, Andrewid needs to place a bridge between each pair of adjacent islands. A bridge of length a can be placed between the i-th and the (i + 1)-th islads, if there are such coordinates of x and y, that li β€ x β€ ri, li + 1 β€ y β€ ri + 1 and y - x = a.
The detective was supplied with m bridges, each bridge can be used at most once. Help him determine whether the bridges he got are enough to connect each pair of adjacent islands.
Input
The first line contains integers n (2 β€ n β€ 2Β·105) and m (1 β€ m β€ 2Β·105) β the number of islands and bridges.
Next n lines each contain two integers li and ri (1 β€ li β€ ri β€ 1018) β the coordinates of the island endpoints.
The last line contains m integer numbers a1, a2, ..., am (1 β€ ai β€ 1018) β the lengths of the bridges that Andrewid got.
Output
If it is impossible to place a bridge between each pair of adjacent islands in the required manner, print on a single line "No" (without the quotes), otherwise print in the first line "Yes" (without the quotes), and in the second line print n - 1 numbers b1, b2, ..., bn - 1, which mean that between islands i and i + 1 there must be used a bridge number bi.
If there are multiple correct answers, print any of them. Note that in this problem it is necessary to print "Yes" and "No" in correct case.
Examples
Input
4 4
1 4
7 8
9 10
12 14
4 5 3 8
Output
Yes
2 3 1
Input
2 2
11 14
17 18
2 9
Output
No
Input
2 1
1 1
1000000000000000000 1000000000000000000
999999999999999999
Output
Yes
1
Note
In the first sample test you can, for example, place the second bridge between points 3 and 8, place the third bridge between points 7 and 10 and place the first bridge between points 10 and 14.
In the second sample test the first bridge is too short and the second bridge is too long, so the solution doesn't exist.
|
instruction
| 0
| 104,857
| 3
| 209,714
|
Tags: data structures, greedy, sortings
Correct Solution:
```
#!/usr/bin/env python
# 556D_fug.py - Codeforces.com 556D Fug quiz
#
# Copyright (C) 2015 Sergey
# Licensed under the Apache License, Version 2.0 (the "License");
# you may not use this file except in compliance with the License.
# You may obtain a copy of the License at
# http://www.apache.org/licenses/LICENSE-2.0
#
# Unless required by applicable law or agreed to in writing, software
# distributed under the License is distributed on an "AS IS" BASIS,
# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
# See the License for the specific language governing permissions and
# limitations under the License.
"""
Input
The first line contains integers n and m - the number of islands and bridges.
Next n lines each contain two integers li and ri - the coordinates of the
island endpoints.
The last line contains m integer numbers a1..am - the lengths of the bridges
that Andrewid got.
Output
If it is impossible to place a bridge between each pair of adjacent islands
in the required manner, print on a single line "No" (without the quotes)
, otherwise print in the first line "Yes" (without the quotes), and in the
second line print n-1 numbers b1, bn-1, which mean that between islands
i and i+1 there must be used a bridge number bi.
If there are multiple correct answers, print any of them. Note that in this
problem it is necessary to print "Yes" and "No" in correct case
"""
# Standard libraries
import unittest
import sys
import re
# Additional libraries
import heapq
###############################################################################
# Fug Class
###############################################################################
class Fug:
""" Fug representation """
def __init__(self, args):
""" Default constructor """
self.list = args[0]
self.alist = args[1]
self.gn = len(self.list) - 1
# Sorted list of bridges
self.asrt = sorted((n, i) for i, n in enumerate(self.alist))
# List of gaps between islands
self.gaps = []
prevli = self.list[0]
for i in range(self.gn):
li = self.list[i+1]
min = li[0] - prevli[1]
max = li[1] - prevli[0]
self.gaps.append((min, max, i))
prevli = li
# Sorted list of gaps between islands
self.gsrt = sorted(self.gaps)
self.gmin = [n[0] for n in self.gsrt]
self.result = [None]*self.gn
self.heap = []
def iterate(self):
j = 0
for (b, i) in self.asrt:
# Traverse gmin array
while j < self.gn and self.gmin[j] <= b:
it = self.gsrt[j]
heapq.heappush(self.heap, (it[1], it[0], it[2]))
j += 1
# Update result and remove the element from lists
if self.heap:
(mmax, mmin, mi) = self.heap[0]
if mmin <= b and mmax >= b:
self.result[mi] = str(i + 1)
heapq.heappop(self.heap)
yield
def calculate(self):
""" Main calcualtion function of the class """
for it in self.iterate():
pass
for n in self.result:
if n is None:
return "No"
answer = "Yes\n"
answer += " ".join(self.result)
return answer
###############################################################################
# Executable code
###############################################################################
def get_inputs(test_inputs=None):
it = iter(test_inputs.split("\n")) if test_inputs else None
def uinput():
""" Unit-testable input function wrapper """
if it:
return next(it)
else:
return sys.stdin.readline()
# Getting string inputs. Place all uinput() calls here
num = [int(s) for s in uinput().split()]
list = [[int(s) for s in uinput().split()] for i in range(num[0])]
alist = [int(s) for s in uinput().split()]
# Decoding inputs into a list
inputs = [list, alist]
return inputs
def calculate(test_inputs=None):
""" Base class calculate method wrapper """
return Fug(get_inputs(test_inputs)).calculate()
###############################################################################
# Unit Tests
###############################################################################
class unitTests(unittest.TestCase):
def test_sample_tests(self):
""" Quiz sample tests. Add \n to separate lines """
# Sample test 1
test = "4 4\n1 4\n7 8\n9 10\n12 14\n4 5 3 8"
self.assertEqual(calculate(test), "Yes\n2 3 1")
self.assertEqual(
get_inputs(test),
[[[1, 4], [7, 8], [9, 10], [12, 14]], [4, 5, 3, 8]])
# My tests
test = "5 5\n1 1\n2 7\n8 8\n10 10\n16 16\n1 1 5 6 2"
self.assertEqual(calculate(test), "Yes\n1 2 5 4")
# Other tests
test = "2 2\n11 14\n17 18\n2 9"
self.assertEqual(calculate(test), "No")
# Other tests
test = (
"2 1\n1 1\n1000000000000000000 1000000000000000000" +
"\n999999999999999999")
self.assertEqual(calculate(test), "Yes\n1")
test = ("5 9\n1 2\n3 3\n5 7\n11 13\n14 20\n2 3 4 10 6 2 6 9 5")
self.assertEqual(calculate(test), "Yes\n1 6 3 2")
size = 10000
test = str(size) + " " + str(size) + "\n"
x = size*1000
for i in range(size):
x += 2
test += str(x) + " " + str(x+1) + "\n"
for i in range(size):
test += str(2) + " "
self.assertEqual(calculate(test)[0], "Y")
def test_Fug_class__basic_functions(self):
""" Fug class basic functions testing """
# Constructor test
d = Fug([[[1, 5], [7, 8], [9, 10], [12, 14]], [4, 5, 3, 8]])
self.assertEqual(d.list[0][0], 1)
self.assertEqual(d.alist[0], 4)
# Sort bridges
self.assertEqual(d.asrt[0], (3, 2))
# Sort Gaps
self.assertEqual(d.gaps[0], (2, 7, 0))
self.assertEqual(d.gsrt[0], (1, 3, 1))
iter = d.iterate()
next(iter)
self.assertEqual(d.gmin, [1, 2, 2])
self.assertEqual(d.heap, [(5, 2, 2), (7, 2, 0)])
if __name__ == "__main__":
# Avoiding recursion limitaions
sys.setrecursionlimit(100000)
if sys.argv[-1] == "-ut":
unittest.main(argv=[" "])
# Print the result string
sys.stdout.write(calculate())
```
|
output
| 1
| 104,857
| 3
| 209,715
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrewid the Android is a galaxy-famous detective. He is now chasing a criminal hiding on the planet Oxa-5, the planet almost fully covered with water.
The only dry land there is an archipelago of n narrow islands located in a row. For more comfort let's represent them as non-intersecting segments on a straight line: island i has coordinates [li, ri], besides, ri < li + 1 for 1 β€ i β€ n - 1.
To reach the goal, Andrewid needs to place a bridge between each pair of adjacent islands. A bridge of length a can be placed between the i-th and the (i + 1)-th islads, if there are such coordinates of x and y, that li β€ x β€ ri, li + 1 β€ y β€ ri + 1 and y - x = a.
The detective was supplied with m bridges, each bridge can be used at most once. Help him determine whether the bridges he got are enough to connect each pair of adjacent islands.
Input
The first line contains integers n (2 β€ n β€ 2Β·105) and m (1 β€ m β€ 2Β·105) β the number of islands and bridges.
Next n lines each contain two integers li and ri (1 β€ li β€ ri β€ 1018) β the coordinates of the island endpoints.
The last line contains m integer numbers a1, a2, ..., am (1 β€ ai β€ 1018) β the lengths of the bridges that Andrewid got.
Output
If it is impossible to place a bridge between each pair of adjacent islands in the required manner, print on a single line "No" (without the quotes), otherwise print in the first line "Yes" (without the quotes), and in the second line print n - 1 numbers b1, b2, ..., bn - 1, which mean that between islands i and i + 1 there must be used a bridge number bi.
If there are multiple correct answers, print any of them. Note that in this problem it is necessary to print "Yes" and "No" in correct case.
Examples
Input
4 4
1 4
7 8
9 10
12 14
4 5 3 8
Output
Yes
2 3 1
Input
2 2
11 14
17 18
2 9
Output
No
Input
2 1
1 1
1000000000000000000 1000000000000000000
999999999999999999
Output
Yes
1
Note
In the first sample test you can, for example, place the second bridge between points 3 and 8, place the third bridge between points 7 and 10 and place the first bridge between points 10 and 14.
In the second sample test the first bridge is too short and the second bridge is too long, so the solution doesn't exist.
|
instruction
| 0
| 104,858
| 3
| 209,716
|
Tags: data structures, greedy, sortings
Correct Solution:
```
#!/usr/bin/env python
# 556D_fug.py - Codeforces.com 556D Fug quiz
#
# Copyright (C) 2015 Sergey
# Licensed under the Apache License, Version 2.0 (the "License");
# you may not use this file except in compliance with the License.
# You may obtain a copy of the License at
# http://www.apache.org/licenses/LICENSE-2.0
#
# Unless required by applicable law or agreed to in writing, software
# distributed under the License is distributed on an "AS IS" BASIS,
# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
# See the License for the specific language governing permissions and
# limitations under the License.
"""
Input
The first line contains integers n and m - the number of islands and bridges.
Next n lines each contain two integers li and ri - the coordinates of the
island endpoints.
The last line contains m integer numbers a1..am - the lengths of the bridges
that Andrewid got.
Output
If it is impossible to place a bridge between each pair of adjacent islands
in the required manner, print on a single line "No" (without the quotes)
, otherwise print in the first line "Yes" (without the quotes), and in the
second line print n-1 numbers b1, bn-1, which mean that between islands
i and i+1 there must be used a bridge number bi.
If there are multiple correct answers, print any of them. Note that in this
problem it is necessary to print "Yes" and "No" in correct case
"""
# Standard modules
import unittest
import sys
import re
# Additional modules
import bisect
###############################################################################
# Fastlist Class
###############################################################################
class Fastlist(object):
""" Fastlist representation """
def __init__(self, l=[], load=5000, sorted=0):
self._load = load
self._sorted = sorted
self._lists = []
self._starts = []
self._mins = []
self._insert_list()
self._irev = 0
self.extend(l)
def _index_location(self, index):
if len(self._lists[0]) == 0:
raise IndexError("List index out of range")
if index == 0:
return (0, 0)
if index == -1:
return (len(self._lists) - 1, len(self._lists[-1]) - 1)
if self._sorted:
raise RuntimeError("No index access to the sorted list, exc 0, -1")
length = len(self)
if index < 0:
index = length + index
if index >= length:
raise IndexError("List index out of range")
il = bisect.bisect_right(self._starts, index) - 1
return (il, index - self._starts[il])
def _insert_list(self, il=None):
if il is None:
il = len(self._lists)
self._lists.insert(il, [])
if self._sorted:
if il == 0:
self._mins.insert(il, None)
else:
self._mins.insert(il, self._lists[il-1][-1])
else:
if il == 0:
self._starts.insert(il, 0)
else:
start = self._starts[il-1] + len(self._lists[il-1])
self._starts.insert(il, start)
def _del_list(self, il):
del self._lists[il]
if self._sorted:
del self._mins[il]
else:
del self._starts[il]
def _rebalance(self, il):
illen = len(self._lists[il])
if illen >= self._load * 2:
self._insert_list(il)
self._even_lists(il)
if illen <= self._load * 0.2:
if il != 0:
self._even_lists(il-1)
elif len(self._lists) > 1:
self._even_lists(il)
def _even_lists(self, il):
tot = len(self._lists[il]) + len(self._lists[il+1])
if tot < self._load * 1:
self._lists[il] += self._lists[il+1]
self._del_list(il+1)
if self._sorted:
self._mins[il] = self._lists[il][0]
else:
half = tot//2
ltot = self._lists[il] + self._lists[il+1]
self._lists[il] = ltot[:half]
self._lists[il+1] = ltot[half:]
if self._sorted:
self._mins[il] = self._lists[il][0]
self._mins[il+1] = self._lists[il+1][0]
else:
self._starts[il+1] = self._starts[il] + len(self._lists[il])
def _obj_location(self, obj, l=0):
if not self._sorted:
raise RuntimeError("No by value access to an unserted list")
il = 0
if len(self._mins) > 1 and obj > self._mins[0]:
if l:
il = bisect.bisect_left(self._mins, obj) - 1
else:
il = bisect.bisect_right(self._mins, obj) - 1
if l:
ii = bisect.bisect_left(self._lists[il], obj)
else:
ii = bisect.bisect_right(self._lists[il], obj)
if ii == len(self._lists[il]) and il != len(self._lists) - 1:
ii = 0
il += 1
return (il, ii)
def insert(self, index, obj):
(il, ii) = self._index_location(index)
self._lists[il].insert(ii, obj)
for j in range(il + 1, len(self._starts)):
self._starts[j] += 1
self._rebalance(il)
def append(self, obj):
if len(self._lists[-1]) >= self._load:
self._insert_list()
self._lists[-1].append(obj)
if self._sorted and self._mins[0] is None:
self._mins[0] = self._lists[0][0]
def extend(self, iter):
for n in iter:
self.append(n)
def pop(self, index=None):
if index is None:
index = -1
(il, ii) = self._index_location(index)
item = self._lists[il].pop(ii)
if self._sorted:
if ii == 0 and len(self._lists[il]) > 0:
self._mins[il] = self._lists[il][0]
else:
for j in range(il + 1, len(self._starts)):
self._starts[j] -= 1
self._rebalance(il)
return item
def clear(self):
self._lists.clear()
self._starts.clear()
self._mins.clear()
self._insert_list()
def as_list(self):
return sum(self._lists, [])
def insort(self, obj, l=0):
(il, ii) = self._obj_location(obj, l)
self._lists[il].insert(ii, obj)
if ii == 0:
self._mins[il] = obj
self._rebalance(il)
def add(self, obj):
if self._sorted:
self.insort(obj)
else:
self.append(obj)
def insort_left(self, obj):
self.insort(obj, l=1)
def lower_bound(self, obj):
(self._il, self._ii) = self._obj_location(obj, l=1)
return self
def upper_bound(self, obj):
(self._il, self._ii) = self._obj_location(obj)
return self
def __str__(self):
return str(self.as_list())
def __setitem__(self, index, obj):
if isinstance(index, int):
(il, ii) = self._index_location(index)
self._lists[il][ii] = obj
elif isinstance(index, slice):
raise RuntimeError("Slice assignment is not supported")
def __getitem__(self, index):
if isinstance(index, int):
(il, ii) = self._index_location(index)
return self._lists[il][ii]
elif isinstance(index, slice):
rg = index.indices(len(self))
if rg[0] == 0 and rg[1] == len(self) and rg[2] == 1:
return self.as_list()
return [self.__getitem__(index) for index in range(*rg)]
def __iadd__(self, obj):
if self._sorted:
[self.insort(n) for n in obj]
else:
[self.append(n) for n in obj]
return self
def __delitem__(self, index):
if isinstance(index, int):
self.pop(index)
elif isinstance(index, slice):
rg = index.indices(len(self))
[self.__delitem__(rg[0]) for i in range(*rg)]
def __len__(self):
if self._sorted:
return sum([len(l) for l in self._lists])
return self._starts[-1] + len(self._lists[-1])
def __contains__(self, obj):
if self._sorted:
it = self.lower_bound(obj)
return not it.iter_end() and obj == it.iter_getitem()
else:
for n in self:
if obj == n:
return True
return False
def __bool__(self):
return len(self._lists[0]) != 0
def __iter__(self):
self._il = self._ii = self._irev = 0
return self
def __reversed__(self):
self._il = len(self._lists) - 1
self._ii = len(self._lists[self._il]) - 1
self._irev = 1
return self
def __next__(self):
if self._il in (-1, len(self._lists)) or len(self._lists[0]) == 0:
raise StopIteration("Iteration stopped")
item = self._lists[self._il][self._ii]
if not self._irev:
self._ii += 1
if self._ii == len(self._lists[self._il]):
self._il += 1
self._ii = 0
else:
self._ii -= 1
if self._ii == 0:
self._il -= 1
self._ii = len(self._lists[self._il])
return item
def iter_getitem(self):
return self._lists[self._il][self._ii]
def iter_end(self):
return (self._il == len(self._lists) - 1 and
self._ii == len(self._lists[self._il]))
def iter_del(self):
self.iter_pop()
def iter_pop(self):
item = self._lists[self._il].pop(self._ii)
if self._sorted:
if self._ii == 0 and len(self._lists[self._il]) > 0:
self._mins[self._il] = self._lists[self._il][0]
else:
for j in range(self._il + 1, len(self._starts)):
self._starts[j] -= 1
self._rebalance(self._il)
return item
###############################################################################
# Fug Class
###############################################################################
class Fug:
""" Fug representation """
def __init__(self, args):
""" Default constructor """
self.gsrt = args[0]
self.asrt = args[1]
self.gn = args[2]
self.result = [0]*self.gn
self.a = Fastlist(self.asrt, load=500, sorted=1)
def calculate(self):
""" Main calcualtion function of the class """
for i in range(self.gn):
g = self.gsrt[i]
it = self.a.lower_bound((g[1], 0))
if not it.iter_end():
alb = it.iter_getitem()
if alb[0] > g[0]:
return "No"
self.result[g[2]] = alb[1]+1
it.iter_del()
else:
return "No"
answer = "Yes\n" + " ".join(str(n) for n in self.result)
return answer
###############################################################################
# Executable code
###############################################################################
def get_inputs(test_inputs=None):
it = iter(test_inputs.split("\n")) if test_inputs else None
def uinput():
""" Unit-testable input function wrapper """
if it:
return next(it)
else:
return sys.stdin.readline()
# Getting string inputs. Place all uinput() calls here
num = [int(s) for s in uinput().split()]
gaps = []
prevli = [int(s) for s in uinput().split()]
for i in range(num[0] - 1):
li = [int(s) for s in uinput().split()]
min = li[0] - prevli[1]
max = li[1] - prevli[0]
gaps.append((max, min, i))
prevli = li
alist = [(int(s), i) for i, s in enumerate(uinput().split())]
# Decoding inputs into a list
inputs = [sorted(gaps), sorted(alist), num[0]-1]
return inputs
def calculate(test_inputs=None):
""" Base class calculate method wrapper """
return Fug(get_inputs(test_inputs)).calculate()
###############################################################################
# Unit Tests
###############################################################################
class unitTests(unittest.TestCase):
def test_sample_tests(self):
""" Quiz sample tests. Add \n to separate lines """
# Sample test 1
test = "4 4\n1 4\n7 8\n9 10\n12 14\n4 5 3 8"
self.assertEqual(calculate(test), "Yes\n2 3 1")
self.assertEqual(
get_inputs(test),
[[(3, 1, 1), (5, 2, 2), (7, 3, 0)],
[(3, 2), (4, 0), (5, 1), (8, 3)], 3])
# My tests
test = "5 5\n1 1\n2 7\n8 8\n10 10\n16 16\n1 1 5 6 2"
self.assertEqual(calculate(test), "Yes\n1 2 5 4")
# Other tests
test = "2 2\n11 14\n17 18\n2 9"
self.assertEqual(calculate(test), "No")
# Other tests
test = (
"2 1\n1 1\n1000000000000000000 1000000000000000000" +
"\n999999999999999999")
self.assertEqual(calculate(test), "Yes\n1")
test = ("5 9\n1 2\n3 3\n5 7\n11 13\n14 20\n2 3 4 10 6 2 6 9 5")
self.assertEqual(calculate(test), "Yes\n1 6 3 2")
size = 200000
test = str(size) + " " + str(size) + "\n"
x = size*1000
for i in range(size):
test += str(x) + " " + str(x + i + 1) + "\n"
x += 2 * (i + 1)
for i in reversed(range(size)):
test += str(i) + " "
self.assertEqual(calculate(test)[0], "Y")
def test_Fug_class__basic_functions(self):
""" Fug class basic functions testing """
# Constructor test
d = Fug([[(1, 3, 1), (2, 5, 2), (3, 7, 0)],
[(3, 2), (4, 0), (5, 1), (8, 3)], 3])
# Sort bridges
self.assertEqual(d.asrt[0], (3, 2))
# Sort Gaps
self.assertEqual(d.gsrt[0], (1, 3, 1))
if __name__ == "__main__":
# Avoiding recursion limitaions
sys.setrecursionlimit(100000)
if sys.argv[-1] == "-ut":
unittest.main(argv=[" "])
# Print the result string
sys.stdout.write(calculate())
```
|
output
| 1
| 104,858
| 3
| 209,717
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrewid the Android is a galaxy-famous detective. He is now chasing a criminal hiding on the planet Oxa-5, the planet almost fully covered with water.
The only dry land there is an archipelago of n narrow islands located in a row. For more comfort let's represent them as non-intersecting segments on a straight line: island i has coordinates [li, ri], besides, ri < li + 1 for 1 β€ i β€ n - 1.
To reach the goal, Andrewid needs to place a bridge between each pair of adjacent islands. A bridge of length a can be placed between the i-th and the (i + 1)-th islads, if there are such coordinates of x and y, that li β€ x β€ ri, li + 1 β€ y β€ ri + 1 and y - x = a.
The detective was supplied with m bridges, each bridge can be used at most once. Help him determine whether the bridges he got are enough to connect each pair of adjacent islands.
Input
The first line contains integers n (2 β€ n β€ 2Β·105) and m (1 β€ m β€ 2Β·105) β the number of islands and bridges.
Next n lines each contain two integers li and ri (1 β€ li β€ ri β€ 1018) β the coordinates of the island endpoints.
The last line contains m integer numbers a1, a2, ..., am (1 β€ ai β€ 1018) β the lengths of the bridges that Andrewid got.
Output
If it is impossible to place a bridge between each pair of adjacent islands in the required manner, print on a single line "No" (without the quotes), otherwise print in the first line "Yes" (without the quotes), and in the second line print n - 1 numbers b1, b2, ..., bn - 1, which mean that between islands i and i + 1 there must be used a bridge number bi.
If there are multiple correct answers, print any of them. Note that in this problem it is necessary to print "Yes" and "No" in correct case.
Examples
Input
4 4
1 4
7 8
9 10
12 14
4 5 3 8
Output
Yes
2 3 1
Input
2 2
11 14
17 18
2 9
Output
No
Input
2 1
1 1
1000000000000000000 1000000000000000000
999999999999999999
Output
Yes
1
Note
In the first sample test you can, for example, place the second bridge between points 3 and 8, place the third bridge between points 7 and 10 and place the first bridge between points 10 and 14.
In the second sample test the first bridge is too short and the second bridge is too long, so the solution doesn't exist.
|
instruction
| 0
| 104,859
| 3
| 209,718
|
Tags: data structures, greedy, sortings
Correct Solution:
```
#!/usr/bin/env python
# 556D_fug.py - Codeforces.com 556D Fug quiz
#
# Copyright (C) 2015 Sergey
# Licensed under the Apache License, Version 2.0 (the "License");
# you may not use this file except in compliance with the License.
# You may obtain a copy of the License at
# http://www.apache.org/licenses/LICENSE-2.0
#
# Unless required by applicable law or agreed to in writing, software
# distributed under the License is distributed on an "AS IS" BASIS,
# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
# See the License for the specific language governing permissions and
# limitations under the License.
"""
Input
The first line contains integers n and m - the number of islands and bridges.
Next n lines each contain two integers li and ri - the coordinates of the
island endpoints.
The last line contains m integer numbers a1..am - the lengths of the bridges
that Andrewid got.
Output
If it is impossible to place a bridge between each pair of adjacent islands
in the required manner, print on a single line "No" (without the quotes)
, otherwise print in the first line "Yes" (without the quotes), and in the
second line print n-1 numbers b1, bn-1, which mean that between islands
i and i+1 there must be used a bridge number bi.
If there are multiple correct answers, print any of them. Note that in this
problem it is necessary to print "Yes" and "No" in correct case
"""
# Standard libraries
import unittest
import sys
import re
# Additional libraries
import bisect
import heapq
###############################################################################
# Fug Class
###############################################################################
class Fug:
""" Fug representation """
def __init__(self, args):
""" Default constructor """
self.list = args[0]
self.alist = args[1]
self.gn = len(self.list) - 1
# Sorted list of bridges
self.asrt = sorted((n, i) for i, n in enumerate(self.alist))
# List of gaps between islands
self.gaps = []
prevli = self.list[0]
for i in range(self.gn):
li = self.list[i+1]
min = li[0] - prevli[1]
max = li[1] - prevli[0]
self.gaps.append((min, max, i))
prevli = li
# Sorted list of gaps between islands
self.gsrt = sorted(self.gaps)
self.gmin = []
self.result = []
self.heap = []
for n in self.gsrt:
self.gmin.append(n[0])
self.result.append(None)
def iterate(self):
j = 0
for (b, i) in self.asrt:
# Traverse gmin array
while j < self.gn and self.gmin[j] <= b:
it = self.gsrt[j]
heapq.heappush(self.heap, (it[1], it[0], it[2]))
j += 1
# Update result and remove the element from lists
if self.heap:
(mmax, mmin, mi) = self.heap[0]
if mmin <= b and mmax >= b:
self.result[mi] = i + 1
heapq.heappop(self.heap)
yield
def calculate(self):
""" Main calcualtion function of the class """
for it in self.iterate():
pass
answer = ""
for (i, n) in enumerate(self.result):
if n is None:
return "No"
answer += (" " if i > 0 else "") + str(n)
return "Yes\n" + answer
###############################################################################
# Executable code
###############################################################################
def get_inputs(test_inputs=None):
it = iter(test_inputs.split("\n")) if test_inputs else None
def uinput():
""" Unit-testable input function wrapper """
if it:
return next(it)
else:
return input()
# Getting string inputs. Place all uinput() calls here
num = [int(s) for s in uinput().split()]
list = [[int(s) for s in uinput().split()] for i in range(num[0])]
alist = [int(s) for s in uinput().split()]
# Decoding inputs into a list
inputs = [list, alist]
return inputs
def calculate(test_inputs=None):
""" Base class calculate method wrapper """
return Fug(get_inputs(test_inputs)).calculate()
###############################################################################
# Unit Tests
###############################################################################
class unitTests(unittest.TestCase):
def test_sample_tests(self):
""" Quiz sample tests. Add \n to separate lines """
# Sample test 1
test = "4 4\n1 4\n7 8\n9 10\n12 14\n4 5 3 8"
self.assertEqual(calculate(test), "Yes\n2 3 1")
self.assertEqual(
get_inputs(test),
[[[1, 4], [7, 8], [9, 10], [12, 14]], [4, 5, 3, 8]])
# My tests
test = "5 5\n1 1\n2 7\n8 8\n10 10\n16 16\n1 1 5 6 2"
self.assertEqual(calculate(test), "Yes\n1 2 5 4")
# Other tests
test = "2 2\n11 14\n17 18\n2 9"
self.assertEqual(calculate(test), "No")
# Other tests
test = (
"2 1\n1 1\n1000000000000000000 1000000000000000000" +
"\n999999999999999999")
self.assertEqual(calculate(test), "Yes\n1")
test = ("5 9\n1 2\n3 3\n5 7\n11 13\n14 20\n2 3 4 10 6 2 6 9 5")
self.assertEqual(calculate(test), "Yes\n1 6 3 2")
size = 50000
test = str(size) + " " + str(size) + "\n"
x = size*1000
for i in range(size):
x += 2
test += str(x) + " " + str(x+1) + "\n"
for i in range(size):
test += str(2) + " "
self.assertEqual(calculate(test)[0], "Y")
def test_Fug_class__basic_functions(self):
""" Fug class basic functions testing """
# Constructor test
d = Fug([[[1, 5], [7, 8], [9, 10], [12, 14]], [4, 5, 3, 8]])
self.assertEqual(d.list[0][0], 1)
self.assertEqual(d.alist[0], 4)
# Sort bridges
self.assertEqual(d.asrt[0], (3, 2))
# Sort Gaps
self.assertEqual(d.gaps[0], (2, 7, 0))
self.assertEqual(d.gsrt[0], (1, 3, 1))
iter = d.iterate()
next(iter)
self.assertEqual(d.gmin, [1, 2, 2])
self.assertEqual(d.heap, [(5, 2, 2), (7, 2, 0)])
if __name__ == "__main__":
# Avoiding recursion limitaions
sys.setrecursionlimit(100000)
if sys.argv[-1] == "-ut":
unittest.main(argv=[" "])
# Print the result string
print(calculate())
```
|
output
| 1
| 104,859
| 3
| 209,719
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Andrewid the Android is a galaxy-famous detective. He is now chasing a criminal hiding on the planet Oxa-5, the planet almost fully covered with water.
The only dry land there is an archipelago of n narrow islands located in a row. For more comfort let's represent them as non-intersecting segments on a straight line: island i has coordinates [li, ri], besides, ri < li + 1 for 1 β€ i β€ n - 1.
To reach the goal, Andrewid needs to place a bridge between each pair of adjacent islands. A bridge of length a can be placed between the i-th and the (i + 1)-th islads, if there are such coordinates of x and y, that li β€ x β€ ri, li + 1 β€ y β€ ri + 1 and y - x = a.
The detective was supplied with m bridges, each bridge can be used at most once. Help him determine whether the bridges he got are enough to connect each pair of adjacent islands.
Input
The first line contains integers n (2 β€ n β€ 2Β·105) and m (1 β€ m β€ 2Β·105) β the number of islands and bridges.
Next n lines each contain two integers li and ri (1 β€ li β€ ri β€ 1018) β the coordinates of the island endpoints.
The last line contains m integer numbers a1, a2, ..., am (1 β€ ai β€ 1018) β the lengths of the bridges that Andrewid got.
Output
If it is impossible to place a bridge between each pair of adjacent islands in the required manner, print on a single line "No" (without the quotes), otherwise print in the first line "Yes" (without the quotes), and in the second line print n - 1 numbers b1, b2, ..., bn - 1, which mean that between islands i and i + 1 there must be used a bridge number bi.
If there are multiple correct answers, print any of them. Note that in this problem it is necessary to print "Yes" and "No" in correct case.
Examples
Input
4 4
1 4
7 8
9 10
12 14
4 5 3 8
Output
Yes
2 3 1
Input
2 2
11 14
17 18
2 9
Output
No
Input
2 1
1 1
1000000000000000000 1000000000000000000
999999999999999999
Output
Yes
1
Note
In the first sample test you can, for example, place the second bridge between points 3 and 8, place the third bridge between points 7 and 10 and place the first bridge between points 10 and 14.
In the second sample test the first bridge is too short and the second bridge is too long, so the solution doesn't exist.
|
instruction
| 0
| 104,860
| 3
| 209,720
|
Tags: data structures, greedy, sortings
Correct Solution:
```
import heapq
n,m=[int(x) for x in input().split()]
l1,r1=[int(x) for x in input().split()]
req=[]
start=[]
for i in range(n-1):
l2,r2=[int(x) for x in input().split()]
req.append((l2-r1,r2-l1,i))
l1,r1=l2,r2
have=[int(x) for x in input().split()]
for i in range(m):
have[i]=(have[i],i)
have.sort()
req.sort()
now=[]
i=0
j=0
slen=len(req)
hlen=len(have)
ans=[0]*(n-1)
while j<hlen:
if i<slen and req[i][0]<=have[j][0]:
heapq.heappush(now,(req[i][1],req[i][2]))
i+=1
else:
try:
x=heapq.heappop(now)
except IndexError:
j+=1
continue
if x[0]<have[j][0]:
break
else:
ans[x[1]]=have[j][1]
j+=1
if i<slen or len(now)!=0 or j<hlen:
print('No')
else:
print('Yes')
print(' '.join([str(x+1) for x in ans]))
```
|
output
| 1
| 104,860
| 3
| 209,721
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrewid the Android is a galaxy-famous detective. He is now chasing a criminal hiding on the planet Oxa-5, the planet almost fully covered with water.
The only dry land there is an archipelago of n narrow islands located in a row. For more comfort let's represent them as non-intersecting segments on a straight line: island i has coordinates [li, ri], besides, ri < li + 1 for 1 β€ i β€ n - 1.
To reach the goal, Andrewid needs to place a bridge between each pair of adjacent islands. A bridge of length a can be placed between the i-th and the (i + 1)-th islads, if there are such coordinates of x and y, that li β€ x β€ ri, li + 1 β€ y β€ ri + 1 and y - x = a.
The detective was supplied with m bridges, each bridge can be used at most once. Help him determine whether the bridges he got are enough to connect each pair of adjacent islands.
Input
The first line contains integers n (2 β€ n β€ 2Β·105) and m (1 β€ m β€ 2Β·105) β the number of islands and bridges.
Next n lines each contain two integers li and ri (1 β€ li β€ ri β€ 1018) β the coordinates of the island endpoints.
The last line contains m integer numbers a1, a2, ..., am (1 β€ ai β€ 1018) β the lengths of the bridges that Andrewid got.
Output
If it is impossible to place a bridge between each pair of adjacent islands in the required manner, print on a single line "No" (without the quotes), otherwise print in the first line "Yes" (without the quotes), and in the second line print n - 1 numbers b1, b2, ..., bn - 1, which mean that between islands i and i + 1 there must be used a bridge number bi.
If there are multiple correct answers, print any of them. Note that in this problem it is necessary to print "Yes" and "No" in correct case.
Examples
Input
4 4
1 4
7 8
9 10
12 14
4 5 3 8
Output
Yes
2 3 1
Input
2 2
11 14
17 18
2 9
Output
No
Input
2 1
1 1
1000000000000000000 1000000000000000000
999999999999999999
Output
Yes
1
Note
In the first sample test you can, for example, place the second bridge between points 3 and 8, place the third bridge between points 7 and 10 and place the first bridge between points 10 and 14.
In the second sample test the first bridge is too short and the second bridge is too long, so the solution doesn't exist.
Submitted Solution:
```
import heapq
n,m=[int(x) for x in input().split()]
l1,r1=[int(x) for x in input().split()]
req=[]
start=[]
for i in range(n-1):
l2,r2=[int(x) for x in input().split()]
req.append((l2-r1,r2-l1,i))
l1,r1=l2,r2
have=[int(x) for x in input().split()]
for i in range(m):
have[i]=(have[i],i)
have.sort()
req.sort()
now=[]
i=0
j=0
slen=len(req)
hlen=len(have)
ans=[0]*(n-1)
while j<hlen:
if i<slen and req[i][0]<=have[j][0]:
heapq.heappush(now,(req[i][1],req[i][2]))
i+=1
else:
try:
x=heapq.heappop(now)
except IndexError:
break
if x[0]<have[j][0]:
break
else:
ans[x[1]]=have[j][1]
j+=1
if i<slen or len(now)!=0:
print('No')
else:
print('Yes')
print(' '.join([str(x+1) for x in ans]))
```
|
instruction
| 0
| 104,861
| 3
| 209,722
|
No
|
output
| 1
| 104,861
| 3
| 209,723
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrewid the Android is a galaxy-famous detective. He is now chasing a criminal hiding on the planet Oxa-5, the planet almost fully covered with water.
The only dry land there is an archipelago of n narrow islands located in a row. For more comfort let's represent them as non-intersecting segments on a straight line: island i has coordinates [li, ri], besides, ri < li + 1 for 1 β€ i β€ n - 1.
To reach the goal, Andrewid needs to place a bridge between each pair of adjacent islands. A bridge of length a can be placed between the i-th and the (i + 1)-th islads, if there are such coordinates of x and y, that li β€ x β€ ri, li + 1 β€ y β€ ri + 1 and y - x = a.
The detective was supplied with m bridges, each bridge can be used at most once. Help him determine whether the bridges he got are enough to connect each pair of adjacent islands.
Input
The first line contains integers n (2 β€ n β€ 2Β·105) and m (1 β€ m β€ 2Β·105) β the number of islands and bridges.
Next n lines each contain two integers li and ri (1 β€ li β€ ri β€ 1018) β the coordinates of the island endpoints.
The last line contains m integer numbers a1, a2, ..., am (1 β€ ai β€ 1018) β the lengths of the bridges that Andrewid got.
Output
If it is impossible to place a bridge between each pair of adjacent islands in the required manner, print on a single line "No" (without the quotes), otherwise print in the first line "Yes" (without the quotes), and in the second line print n - 1 numbers b1, b2, ..., bn - 1, which mean that between islands i and i + 1 there must be used a bridge number bi.
If there are multiple correct answers, print any of them. Note that in this problem it is necessary to print "Yes" and "No" in correct case.
Examples
Input
4 4
1 4
7 8
9 10
12 14
4 5 3 8
Output
Yes
2 3 1
Input
2 2
11 14
17 18
2 9
Output
No
Input
2 1
1 1
1000000000000000000 1000000000000000000
999999999999999999
Output
Yes
1
Note
In the first sample test you can, for example, place the second bridge between points 3 and 8, place the third bridge between points 7 and 10 and place the first bridge between points 10 and 14.
In the second sample test the first bridge is too short and the second bridge is too long, so the solution doesn't exist.
Submitted Solution:
```
#!/usr/bin/env python
# 556D_fug.py - Codeforces.com 556D Fug quiz
#
# Copyright (C) 2015 Sergey
# Licensed under the Apache License, Version 2.0 (the "License");
# you may not use this file except in compliance with the License.
# You may obtain a copy of the License at
# http://www.apache.org/licenses/LICENSE-2.0
#
# Unless required by applicable law or agreed to in writing, software
# distributed under the License is distributed on an "AS IS" BASIS,
# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
# See the License for the specific language governing permissions and
# limitations under the License.
"""
Input
The first line contains integers n and m - the number of islands and bridges.
Next n lines each contain two integers li and ri - the coordinates of the
island endpoints.
The last line contains m integer numbers a1..am - the lengths of the bridges
that Andrewid got.
Output
If it is impossible to place a bridge between each pair of adjacent islands
in the required manner, print on a single line "No" (without the quotes)
, otherwise print in the first line "Yes" (without the quotes), and in the
second line print n-1 numbers b1, bn-1, which mean that between islands
i and i+1 there must be used a bridge number bi.
If there are multiple correct answers, print any of them. Note that in this
problem it is necessary to print "Yes" and "No" in correct case
"""
# Standard libraries
import unittest
import sys
import re
# Additional libraries
import bisect
###############################################################################
# Fug Class
###############################################################################
class Fug:
""" Fug representation """
def __init__(self, args):
""" Default constructor """
self.list = args[0]
self.alist = args[1]
self.gn = len(self.list) - 1
# Sorted list of bridges
self.asrt = sorted((n, i) for i, n in enumerate(self.alist))
# List of gaps between islands
self.gaps = []
for i in range(self.gn):
min = self.list[i+1][0] - self.list[i][1]
max = self.list[i+1][1] - self.list[i][0]
self.gaps.append((min, max, i))
# Sorted list of gaps between islands
self.gsrt = sorted(self.gaps)
self.gmin = []
self.gmax = []
self.gi = []
self.gmaxsrt = []
self.gmaxsrti = []
self.result = []
for n in self.gsrt:
self.gmin.append(n[0])
self.gmax.append(n[1])
self.gi.append(n[2])
self.result.append(None)
def iterate(self):
prev_igr = 0
for (b, i) in self.asrt:
self.ia = b
# Bisect the gap array to find gaps matching current bridge
self.igr = bisect.bisect_right(self.gmin, b)
# Sort new elements by max value inplace
for j in range(prev_igr, self.igr):
mmin = self.gmin.pop(0)
mmax = self.gmax.pop(0)
mi = self.gi.pop(0)
bis = bisect.bisect_left(self.gmax[:self.igr], mmax)
self.gmin.insert(bis, mmin)
self.gmax.insert(bis, mmax)
self.gi.insert(bis, mi)
# Check that found gap with the smallest max is not smaller
# than bridge
if self.gmax[0] < b or self.gmin[0] > b:
break
# Update result and remove the element from lists
if len(self.gi) > 0:
self.result[self.gi[0]] = i
self.gmin.pop(0)
self.gmax.pop(0)
self.gi.pop(0)
if len(self.gi) == 0:
break
prev_igr = self.igr - 1
yield
def calculate(self):
""" Main calcualtion function of the class """
for it in self.iterate():
pass
for n in self.result:
if n is None:
return "No"
answer = " ".join([str(n + 1) for n in self.result])
return "Yes\n" + answer
###############################################################################
# Executable code
###############################################################################
def get_inputs(test_inputs=None):
it = iter(test_inputs.split("\n")) if test_inputs else None
def uinput():
""" Unit-testable input function wrapper """
if it:
return next(it)
else:
return input()
# Getting string inputs. Place all uinput() calls here
num = [int(s) for s in uinput().split()]
list = [[int(s) for s in uinput().split()] for i in range(num[0])]
alist = [int(s) for s in uinput().split()]
# Decoding inputs into a list
inputs = [list, alist]
return inputs
def calculate(test_inputs=None):
""" Base class calculate method wrapper """
return Fug(get_inputs(test_inputs)).calculate()
###############################################################################
# Unit Tests
###############################################################################
class unitTests(unittest.TestCase):
def test_sample_tests(self):
""" Quiz sample tests. Add \n to separate lines """
# Sample test 1
test = "4 4\n1 4\n7 8\n9 10\n12 14\n4 5 3 8"
self.assertEqual(calculate(test), "Yes\n2 3 1")
self.assertEqual(
get_inputs(test),
[[[1, 4], [7, 8], [9, 10], [12, 14]], [4, 5, 3, 8]])
# Other tests
test = "2 2\n11 14\n17 18\n2 9"
self.assertEqual(calculate(test), "No")
test = (
"2 1\n1 1\n1000000000000000000 1000000000000000000" +
"\n999999999999999999")
self.assertEqual(calculate(test), "Yes\n1")
test = ("5 9\n1 2\n3 3\n5 7\n11 13\n14 20\n2 3 4 10 6 2 6 9 5")
self.assertEqual(calculate(test), "Yes\n1 2 3 6")
size = 1000
test = str(size) + " " + str(size) + "\n"
x = size*1000
for i in range(size):
x += 2
test += str(x) + " " + str(x+1) + "\n"
for i in range(size):
test += str(2) + " "
self.assertEqual(calculate(test)[0], "Y")
def test_Fug_class__basic_functions(self):
""" Fug class basic functions testing """
# Constructor test
d = Fug([[[1, 5], [7, 8], [9, 10], [12, 14]], [4, 5, 3, 8]])
self.assertEqual(d.list[0][0], 1)
self.assertEqual(d.alist[0], 4)
# Sort bridges
self.assertEqual(d.asrt[0], (3, 2))
# Sort Gaps
self.assertEqual(d.gaps[0], (2, 7, 0))
self.assertEqual(d.gsrt[0], (1, 3, 1))
iter = d.iterate()
next(iter)
self.assertEqual(d.ia, 3)
self.assertEqual(d.igr, 3)
self.assertEqual(d.gmin, [2, 2])
self.assertEqual(d.gmax, [5, 7])
if __name__ == "__main__":
# Avoiding recursion limitaions
sys.setrecursionlimit(100000)
if sys.argv[-1] == "-ut":
unittest.main(argv=[" "])
# Print the result string
print(calculate())
```
|
instruction
| 0
| 104,862
| 3
| 209,724
|
No
|
output
| 1
| 104,862
| 3
| 209,725
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrewid the Android is a galaxy-famous detective. He is now chasing a criminal hiding on the planet Oxa-5, the planet almost fully covered with water.
The only dry land there is an archipelago of n narrow islands located in a row. For more comfort let's represent them as non-intersecting segments on a straight line: island i has coordinates [li, ri], besides, ri < li + 1 for 1 β€ i β€ n - 1.
To reach the goal, Andrewid needs to place a bridge between each pair of adjacent islands. A bridge of length a can be placed between the i-th and the (i + 1)-th islads, if there are such coordinates of x and y, that li β€ x β€ ri, li + 1 β€ y β€ ri + 1 and y - x = a.
The detective was supplied with m bridges, each bridge can be used at most once. Help him determine whether the bridges he got are enough to connect each pair of adjacent islands.
Input
The first line contains integers n (2 β€ n β€ 2Β·105) and m (1 β€ m β€ 2Β·105) β the number of islands and bridges.
Next n lines each contain two integers li and ri (1 β€ li β€ ri β€ 1018) β the coordinates of the island endpoints.
The last line contains m integer numbers a1, a2, ..., am (1 β€ ai β€ 1018) β the lengths of the bridges that Andrewid got.
Output
If it is impossible to place a bridge between each pair of adjacent islands in the required manner, print on a single line "No" (without the quotes), otherwise print in the first line "Yes" (without the quotes), and in the second line print n - 1 numbers b1, b2, ..., bn - 1, which mean that between islands i and i + 1 there must be used a bridge number bi.
If there are multiple correct answers, print any of them. Note that in this problem it is necessary to print "Yes" and "No" in correct case.
Examples
Input
4 4
1 4
7 8
9 10
12 14
4 5 3 8
Output
Yes
2 3 1
Input
2 2
11 14
17 18
2 9
Output
No
Input
2 1
1 1
1000000000000000000 1000000000000000000
999999999999999999
Output
Yes
1
Note
In the first sample test you can, for example, place the second bridge between points 3 and 8, place the third bridge between points 7 and 10 and place the first bridge between points 10 and 14.
In the second sample test the first bridge is too short and the second bridge is too long, so the solution doesn't exist.
Submitted Solution:
```
#!/usr/bin/env python
# 556D_fug.py - Codeforces.com 556D Fug quiz
#
# Copyright (C) 2015 Sergey
# Licensed under the Apache License, Version 2.0 (the "License");
# you may not use this file except in compliance with the License.
# You may obtain a copy of the License at
# http://www.apache.org/licenses/LICENSE-2.0
#
# Unless required by applicable law or agreed to in writing, software
# distributed under the License is distributed on an "AS IS" BASIS,
# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
# See the License for the specific language governing permissions and
# limitations under the License.
"""
Input
The first line contains integers n and m - the number of islands and bridges.
Next n lines each contain two integers li and ri - the coordinates of the
island endpoints.
The last line contains m integer numbers a1..am - the lengths of the bridges
that Andrewid got.
Output
If it is impossible to place a bridge between each pair of adjacent islands
in the required manner, print on a single line "No" (without the quotes)
, otherwise print in the first line "Yes" (without the quotes), and in the
second line print n-1 numbers b1, bn-1, which mean that between islands
i and i+1 there must be used a bridge number bi.
If there are multiple correct answers, print any of them. Note that in this
problem it is necessary to print "Yes" and "No" in correct case
"""
# Standard libraries
import unittest
import sys
import re
# Additional libraries
import bisect
###############################################################################
# Fug Class
###############################################################################
class Fug:
""" Fug representation """
def __init__(self, args):
""" Default constructor """
self.list = args[0]
self.alist = args[1]
self.gn = len(self.list) - 1
# Sorted list of bridges
self.asrt = sorted((n, i) for i, n in enumerate(self.alist))
# List of gaps between islands
self.gaps = []
for i in range(self.gn):
min = self.list[i+1][0] - self.list[i][1]
max = self.list[i+1][1] - self.list[i][0]
self.gaps.append((min, max, i))
# Sorted list of gaps between islands
self.gsrt = sorted(self.gaps)
self.gmin = []
self.gmax = []
self.gi = []
self.gmaxsrt = []
self.gmaxsrti = []
self.result = []
for n in self.gsrt:
self.gmin.append(n[0])
self.gmax.append(n[1])
self.gi.append(n[2])
self.result.append(None)
def iterate(self):
prev_igr = 0
for (b, i) in self.asrt:
self.ia = b
# Bisect the gap array to find gaps matching current bridge
self.igr = bisect.bisect_right(self.gmin, b)
# Sort new elements by max value inplace
for j in range(prev_igr, self.igr):
mmin = self.gmin.pop(j)
mmax = self.gmax.pop(j)
mi = self.gi.pop(j)
bis = bisect.bisect_right(self.gmax[:self.igr-1], mmax)
self.gmin.insert(bis, mmin)
self.gmax.insert(bis, mmax)
self.gi.insert(bis, mi)
# Check that found gap with the smallest max is not smaller
# than bridge
if self.gmax[0] < b or self.gmin[0] > b:
break
# Update result and remove the element from lists
if len(self.gi) > 0:
self.result[self.gi[0]] = i
self.gmin.pop(0)
self.gmax.pop(0)
self.gi.pop(0)
if len(self.gi) == 0:
break
prev_igr = self.igr - 1
yield
def calculate(self):
""" Main calcualtion function of the class """
for it in self.iterate():
pass
for n in self.result:
if n is None:
return "No"
answer = " ".join([str(n + 1) for n in self.result])
return "Yes\n" + answer
###############################################################################
# Executable code
###############################################################################
def get_inputs(test_inputs=None):
it = iter(test_inputs.split("\n")) if test_inputs else None
def uinput():
""" Unit-testable input function wrapper """
if it:
return next(it)
else:
return input()
# Getting string inputs. Place all uinput() calls here
num = [int(s) for s in uinput().split()]
list = [[int(s) for s in uinput().split()] for i in range(num[0])]
alist = [int(s) for s in uinput().split()]
# Decoding inputs into a list
inputs = [list, alist]
return inputs
def calculate(test_inputs=None):
""" Base class calculate method wrapper """
return Fug(get_inputs(test_inputs)).calculate()
###############################################################################
# Unit Tests
###############################################################################
class unitTests(unittest.TestCase):
def test_sample_tests(self):
""" Quiz sample tests. Add \n to separate lines """
# Sample test 1
test = "4 4\n1 4\n7 8\n9 10\n12 14\n4 5 3 8"
self.assertEqual(calculate(test), "Yes\n2 3 1")
self.assertEqual(
get_inputs(test),
[[[1, 4], [7, 8], [9, 10], [12, 14]], [4, 5, 3, 8]])
# Other tests
test = "2 2\n11 14\n17 18\n2 9"
self.assertEqual(calculate(test), "No")
test = (
"2 1\n1 1\n1000000000000000000 1000000000000000000" +
"\n999999999999999999")
self.assertEqual(calculate(test), "Yes\n1")
test = ("5 9\n1 2\n3 3\n5 7\n11 13\n14 20\n2 3 4 10 6 2 6 9 5")
self.assertEqual(calculate(test), "Yes\n1 6 3 2")
size = 1000
test = str(size) + " " + str(size) + "\n"
x = size*1000
for i in range(size):
x += 2
test += str(x) + " " + str(x+1) + "\n"
for i in range(size):
test += str(2) + " "
self.assertEqual(calculate(test)[0], "Y")
def test_Fug_class__basic_functions(self):
""" Fug class basic functions testing """
# Constructor test
d = Fug([[[1, 5], [7, 8], [9, 10], [12, 14]], [4, 5, 3, 8]])
self.assertEqual(d.list[0][0], 1)
self.assertEqual(d.alist[0], 4)
# Sort bridges
self.assertEqual(d.asrt[0], (3, 2))
# Sort Gaps
self.assertEqual(d.gaps[0], (2, 7, 0))
self.assertEqual(d.gsrt[0], (1, 3, 1))
iter = d.iterate()
next(iter)
self.assertEqual(d.ia, 3)
self.assertEqual(d.igr, 3)
self.assertEqual(d.gmin, [2, 2])
self.assertEqual(d.gmax, [5, 7])
if __name__ == "__main__":
# Avoiding recursion limitaions
sys.setrecursionlimit(100000)
if sys.argv[-1] == "-ut":
unittest.main(argv=[" "])
# Print the result string
print(calculate())
```
|
instruction
| 0
| 104,863
| 3
| 209,726
|
No
|
output
| 1
| 104,863
| 3
| 209,727
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Andrewid the Android is a galaxy-famous detective. He is now chasing a criminal hiding on the planet Oxa-5, the planet almost fully covered with water.
The only dry land there is an archipelago of n narrow islands located in a row. For more comfort let's represent them as non-intersecting segments on a straight line: island i has coordinates [li, ri], besides, ri < li + 1 for 1 β€ i β€ n - 1.
To reach the goal, Andrewid needs to place a bridge between each pair of adjacent islands. A bridge of length a can be placed between the i-th and the (i + 1)-th islads, if there are such coordinates of x and y, that li β€ x β€ ri, li + 1 β€ y β€ ri + 1 and y - x = a.
The detective was supplied with m bridges, each bridge can be used at most once. Help him determine whether the bridges he got are enough to connect each pair of adjacent islands.
Input
The first line contains integers n (2 β€ n β€ 2Β·105) and m (1 β€ m β€ 2Β·105) β the number of islands and bridges.
Next n lines each contain two integers li and ri (1 β€ li β€ ri β€ 1018) β the coordinates of the island endpoints.
The last line contains m integer numbers a1, a2, ..., am (1 β€ ai β€ 1018) β the lengths of the bridges that Andrewid got.
Output
If it is impossible to place a bridge between each pair of adjacent islands in the required manner, print on a single line "No" (without the quotes), otherwise print in the first line "Yes" (without the quotes), and in the second line print n - 1 numbers b1, b2, ..., bn - 1, which mean that between islands i and i + 1 there must be used a bridge number bi.
If there are multiple correct answers, print any of them. Note that in this problem it is necessary to print "Yes" and "No" in correct case.
Examples
Input
4 4
1 4
7 8
9 10
12 14
4 5 3 8
Output
Yes
2 3 1
Input
2 2
11 14
17 18
2 9
Output
No
Input
2 1
1 1
1000000000000000000 1000000000000000000
999999999999999999
Output
Yes
1
Note
In the first sample test you can, for example, place the second bridge between points 3 and 8, place the third bridge between points 7 and 10 and place the first bridge between points 10 and 14.
In the second sample test the first bridge is too short and the second bridge is too long, so the solution doesn't exist.
Submitted Solution:
```
#!/usr/bin/env python
# 556D_fug.py - Codeforces.com 556D Fug quiz
#
# Copyright (C) 2015 Sergey
# Licensed under the Apache License, Version 2.0 (the "License");
# you may not use this file except in compliance with the License.
# You may obtain a copy of the License at
# http://www.apache.org/licenses/LICENSE-2.0
#
# Unless required by applicable law or agreed to in writing, software
# distributed under the License is distributed on an "AS IS" BASIS,
# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
# See the License for the specific language governing permissions and
# limitations under the License.
"""
Input
The first line contains integers n and m - the number of islands and bridges.
Next n lines each contain two integers li and ri - the coordinates of the
island endpoints.
The last line contains m integer numbers a1..am - the lengths of the bridges
that Andrewid got.
Output
If it is impossible to place a bridge between each pair of adjacent islands
in the required manner, print on a single line "No" (without the quotes)
, otherwise print in the first line "Yes" (without the quotes), and in the
second line print n-1 numbers b1, bn-1, which mean that between islands
i and i+1 there must be used a bridge number bi.
If there are multiple correct answers, print any of them. Note that in this
problem it is necessary to print "Yes" and "No" in correct case
"""
# Standard libraries
import unittest
import sys
import re
# Additional libraries
###############################################################################
# Fug Class
###############################################################################
class Fug:
""" Fug representation """
def __init__(self, args):
""" Default constructor """
self.list = args[0]
self.alist = args[1]
self.nranges = len(self.list) - 1
self.nbridges = len(self.alist)
def ranges(self):
result = []
for i in range(self.nranges):
min = self.list[i+1][0] - self.list[i][1]
max = self.list[i+1][1] - self.list[i][0]
result.append([min, max])
return result
def bridges(self):
ranges = self.ranges()
result = [[] for i in range(self.nranges)]
for i in range(self.nranges):
for (j, a) in enumerate(self.alist):
if a >= ranges[i][0] and a <= ranges[i][1]:
result[i].append(j+1)
return result
def rbridges(self, bridges):
result = [[] for i in range(self.nbridges)]
for i in range(self.nranges):
for bi in bridges[i]:
result[bi-1].append(i)
return result
def calculate(self):
""" Main calcualtion function of the class """
result = 1
out = []
bridges = self.bridges()
for b in bridges:
if len(b) == 0:
return "No"
rbridges = self.rbridges(bridges)
ranges = [None for i in range(self.nbridges)]
for i in range(self.nbridges):
for rg in rbridges[i]:
if rg not in ranges:
ranges[i] = rg
break
out = [None for i in range(self.nranges)]
for i in range(self.nranges):
if ranges[i] is None:
return "No"
out[ranges[i]] = i + 1
answer = " ".join([str(n) for n in out])
return "Yes\n" + answer if result else "No"
###############################################################################
# Executable code
###############################################################################
def get_inputs(test_inputs=None):
it = iter(test_inputs.split("\n")) if test_inputs else None
def uinput():
""" Unit-testable input function wrapper """
if it:
return next(it)
else:
return input()
# Getting string inputs. Place all uinput() calls here
num = [int(s) for s in uinput().split()]
list = [[int(s) for s in uinput().split()] for i in range(num[0])]
alist = [int(s) for s in uinput().split()]
# Decoding inputs into a list
inputs = [list, alist]
return inputs
def calculate(test_inputs=None):
""" Base class calculate method wrapper """
return Fug(get_inputs(test_inputs)).calculate()
###############################################################################
# Unit Tests
###############################################################################
class unitTests(unittest.TestCase):
def test_sample_tests(self):
""" Quiz sample tests. Add \n to separate lines """
# Sample test 1
test = "4 4\n1 4\n7 8\n9 10\n12 14\n4 5 3 8"
self.assertEqual(calculate(test), "Yes\n1 3 2")
self.assertEqual(
get_inputs(test),
[[[1, 4], [7, 8], [9, 10], [12, 14]], [4, 5, 3, 8]])
# Other tests
test = "2 2\n11 14\n17 18\n2 9"
self.assertEqual(calculate(test), "No")
test = ("2 1\n1 1\n1000000000000000000 1000000000000000000" +
"\n999999999999999999")
self.assertEqual(calculate(test), "Yes\n1")
test = ("5 10\n1 2\n3 3\n5 7\n11 13\n14 20\n9 10 2 9 10 4 9 9 9 10")
self.assertEqual(calculate(test), "No")
def test_Fug_class__basic_functions(self):
""" Fug class basic functions testing """
# Constructor test
d = Fug([[[1, 4], [7, 8], [9, 10], [12, 14]], [4, 5, 3, 8]])
self.assertEqual(d.list[0][0], 1)
self.assertEqual(d.alist[0], 4)
# Return the list of non intersecting ranges for islands
self.assertEqual(d.ranges(), [[3, 7], [1, 3], [2, 5]])
# Return the list of bridges for each intersection
self.assertEqual(d.bridges(), [[1, 2, 3], [3], [1, 2, 3]])
# Return the list of intersections for each bridge
self.assertEqual(
d.rbridges(d.bridges()), [[0, 2], [0, 2], [0, 1, 2], []])
if __name__ == "__main__":
# Avoiding recursion limitaions
sys.setrecursionlimit(100000)
if sys.argv[-1] == "-ut":
unittest.main(argv=[" "])
# Print the result string
print(calculate())
```
|
instruction
| 0
| 104,864
| 3
| 209,728
|
No
|
output
| 1
| 104,864
| 3
| 209,729
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" β thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input
The first line contains a positive integer n (1 β€ n β€ 100), then follow n lines containing three integers each: the xi coordinate, the yi coordinate and the zi coordinate of the force vector, applied to the body ( - 100 β€ xi, yi, zi β€ 100).
Output
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Examples
Input
3
4 1 7
-2 4 -1
1 -5 -3
Output
NO
Input
3
3 -1 7
-5 2 -4
2 -1 -3
Output
YES
|
instruction
| 0
| 104,915
| 3
| 209,830
|
Tags: implementation, math
Correct Solution:
```
n = int(input())
xx = 0
yy = 0
zz = 0
for i in range(0, n):
x, y, z = map(int, input().split())
xx += x
yy += y
zz += z
if (xx == 0 and yy == 0 and zz == 0):
print("YES")
else:
print("NO")
```
|
output
| 1
| 104,915
| 3
| 209,831
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" β thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input
The first line contains a positive integer n (1 β€ n β€ 100), then follow n lines containing three integers each: the xi coordinate, the yi coordinate and the zi coordinate of the force vector, applied to the body ( - 100 β€ xi, yi, zi β€ 100).
Output
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Examples
Input
3
4 1 7
-2 4 -1
1 -5 -3
Output
NO
Input
3
3 -1 7
-5 2 -4
2 -1 -3
Output
YES
|
instruction
| 0
| 104,916
| 3
| 209,832
|
Tags: implementation, math
Correct Solution:
```
n=int(input())
y=map(sum,zip(*(map(int,input().split())for _ in range(n))))
print(('YES','NO')[any(y)])
```
|
output
| 1
| 104,916
| 3
| 209,833
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" β thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input
The first line contains a positive integer n (1 β€ n β€ 100), then follow n lines containing three integers each: the xi coordinate, the yi coordinate and the zi coordinate of the force vector, applied to the body ( - 100 β€ xi, yi, zi β€ 100).
Output
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Examples
Input
3
4 1 7
-2 4 -1
1 -5 -3
Output
NO
Input
3
3 -1 7
-5 2 -4
2 -1 -3
Output
YES
|
instruction
| 0
| 104,917
| 3
| 209,834
|
Tags: implementation, math
Correct Solution:
```
n=int(input())
x,y,z=0,0,0
for i in range(0,n):
s=input().split(" ")
x+=int(s[0])
y+=int(s[1])
z+=int(s[2])
if(x==y==z==0):
print("YES")
else:
print("NO")
```
|
output
| 1
| 104,917
| 3
| 209,835
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" β thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input
The first line contains a positive integer n (1 β€ n β€ 100), then follow n lines containing three integers each: the xi coordinate, the yi coordinate and the zi coordinate of the force vector, applied to the body ( - 100 β€ xi, yi, zi β€ 100).
Output
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Examples
Input
3
4 1 7
-2 4 -1
1 -5 -3
Output
NO
Input
3
3 -1 7
-5 2 -4
2 -1 -3
Output
YES
|
instruction
| 0
| 104,918
| 3
| 209,836
|
Tags: implementation, math
Correct Solution:
```
def inEquilibirum(list):
xsum = 0
ysum = 0
zsum = 0
for i in list:
xsum += i[0]
ysum += i[1]
zsum += i[2]
if (xsum == 0 and ysum == 0 and zsum == 0):
return True
return False
n = int(input(""))
forces = []
for i in range(0,n):
tuple = str(input(""))
coords = tuple.split(" ")
forces.append([int(coords[0]), int(coords[1]), int(coords[2]) ] )
equilibrium = inEquilibirum(forces)
output = "YES" if equilibrium == True else "NO"
print(output)
```
|
output
| 1
| 104,918
| 3
| 209,837
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" β thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input
The first line contains a positive integer n (1 β€ n β€ 100), then follow n lines containing three integers each: the xi coordinate, the yi coordinate and the zi coordinate of the force vector, applied to the body ( - 100 β€ xi, yi, zi β€ 100).
Output
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Examples
Input
3
4 1 7
-2 4 -1
1 -5 -3
Output
NO
Input
3
3 -1 7
-5 2 -4
2 -1 -3
Output
YES
|
instruction
| 0
| 104,919
| 3
| 209,838
|
Tags: implementation, math
Correct Solution:
```
n = int(input())
force_s = [0, 0, 0]
for i in range(n):
coordinates = list(map(int, input().split(' ')))
force_s[0] += coordinates[0]
force_s[1] += coordinates[1]
force_s[2] += coordinates[2]
if force_s[0] == 0 and force_s[1] == 0 and force_s[2] == 0:
print('YES')
else:
print('NO')
```
|
output
| 1
| 104,919
| 3
| 209,839
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" β thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input
The first line contains a positive integer n (1 β€ n β€ 100), then follow n lines containing three integers each: the xi coordinate, the yi coordinate and the zi coordinate of the force vector, applied to the body ( - 100 β€ xi, yi, zi β€ 100).
Output
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Examples
Input
3
4 1 7
-2 4 -1
1 -5 -3
Output
NO
Input
3
3 -1 7
-5 2 -4
2 -1 -3
Output
YES
|
instruction
| 0
| 104,920
| 3
| 209,840
|
Tags: implementation, math
Correct Solution:
```
n = int(input())
x = 0
y = 0
z = 0
for i in range(n):
k, l, m = map(int, input().split())
x += k
y += l
z += m
if x==0 and y == 0 and z == 0:
print('YES')
else:
print('NO')
```
|
output
| 1
| 104,920
| 3
| 209,841
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" β thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input
The first line contains a positive integer n (1 β€ n β€ 100), then follow n lines containing three integers each: the xi coordinate, the yi coordinate and the zi coordinate of the force vector, applied to the body ( - 100 β€ xi, yi, zi β€ 100).
Output
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Examples
Input
3
4 1 7
-2 4 -1
1 -5 -3
Output
NO
Input
3
3 -1 7
-5 2 -4
2 -1 -3
Output
YES
|
instruction
| 0
| 104,921
| 3
| 209,842
|
Tags: implementation, math
Correct Solution:
```
def is_idle(n) :
x = []
y = []
z = []
for i in range(0, n) :
a, b, c = map(int,input().strip().split(" "))
x.append(a)
y.append(b)
z.append(c)
if sum(x) == 0 and sum(y) == 0 and sum(z) == 0 :
print("YES")
else :
print("NO")
n = int(input())
is_idle(n)
```
|
output
| 1
| 104,921
| 3
| 209,843
|
Provide tags and a correct Python 3 solution for this coding contest problem.
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" β thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input
The first line contains a positive integer n (1 β€ n β€ 100), then follow n lines containing three integers each: the xi coordinate, the yi coordinate and the zi coordinate of the force vector, applied to the body ( - 100 β€ xi, yi, zi β€ 100).
Output
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Examples
Input
3
4 1 7
-2 4 -1
1 -5 -3
Output
NO
Input
3
3 -1 7
-5 2 -4
2 -1 -3
Output
YES
|
instruction
| 0
| 104,922
| 3
| 209,844
|
Tags: implementation, math
Correct Solution:
```
n = int(input())
mat = []
flag = 0
for i in range(n):
mat.append(list(map(int, input().split())))
for i in range(3):
s = 0
for j in range(n):
s += mat[j][i]
if s != 0:
flag = 1
break
if flag == 0:
print('YES')
else:
print('NO')
```
|
output
| 1
| 104,922
| 3
| 209,845
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" β thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input
The first line contains a positive integer n (1 β€ n β€ 100), then follow n lines containing three integers each: the xi coordinate, the yi coordinate and the zi coordinate of the force vector, applied to the body ( - 100 β€ xi, yi, zi β€ 100).
Output
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Examples
Input
3
4 1 7
-2 4 -1
1 -5 -3
Output
NO
Input
3
3 -1 7
-5 2 -4
2 -1 -3
Output
YES
Submitted Solution:
```
count1,count2,count3=0,0,0
for i in [0]*int(input()):
a=[int(j) for j in input().split()]
count1+=a[0]
count2+=a[1]
count3+=a[2]
if (count1==0 and count2==0 and count3==0):
print('YES')
else:
print('NO')
```
|
instruction
| 0
| 104,923
| 3
| 209,846
|
Yes
|
output
| 1
| 104,923
| 3
| 209,847
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" β thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input
The first line contains a positive integer n (1 β€ n β€ 100), then follow n lines containing three integers each: the xi coordinate, the yi coordinate and the zi coordinate of the force vector, applied to the body ( - 100 β€ xi, yi, zi β€ 100).
Output
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Examples
Input
3
4 1 7
-2 4 -1
1 -5 -3
Output
NO
Input
3
3 -1 7
-5 2 -4
2 -1 -3
Output
YES
Submitted Solution:
```
n=int(input())
sumi=0;sumj=0;sumk=0
while n>0:
i,j,k=map(int,input().split())
sumi+=i;sumj+=j;sumk+=k
n-=1
if sumi==0 and sumj==0 and sumk==0:
print("YES")
else:
print("NO")
```
|
instruction
| 0
| 104,924
| 3
| 209,848
|
Yes
|
output
| 1
| 104,924
| 3
| 209,849
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" β thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input
The first line contains a positive integer n (1 β€ n β€ 100), then follow n lines containing three integers each: the xi coordinate, the yi coordinate and the zi coordinate of the force vector, applied to the body ( - 100 β€ xi, yi, zi β€ 100).
Output
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Examples
Input
3
4 1 7
-2 4 -1
1 -5 -3
Output
NO
Input
3
3 -1 7
-5 2 -4
2 -1 -3
Output
YES
Submitted Solution:
```
def resolve(matrix):
sum = [0, 0, 0]
for element in matrix:
sum[0] += element[0]
sum[1] += element[1]
sum[2] += element[2]
if sum[0] == 0 and sum[1] == 0 and sum[2] == 0:
return "YES"
else:
return "NO"
quant = int(input())
matrix = []
for i in range(quant):
matrix.append([int(i) for i in input().split(" ")])
print(resolve(matrix))
```
|
instruction
| 0
| 104,925
| 3
| 209,850
|
Yes
|
output
| 1
| 104,925
| 3
| 209,851
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" β thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input
The first line contains a positive integer n (1 β€ n β€ 100), then follow n lines containing three integers each: the xi coordinate, the yi coordinate and the zi coordinate of the force vector, applied to the body ( - 100 β€ xi, yi, zi β€ 100).
Output
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Examples
Input
3
4 1 7
-2 4 -1
1 -5 -3
Output
NO
Input
3
3 -1 7
-5 2 -4
2 -1 -3
Output
YES
Submitted Solution:
```
n= int(input())
a=[]
b=[]
c=[]
for i in range(n):
x,y,z=input().split()
x,y,z=int(x),int(y),int(z)
a.append(x)
b.append(y)
c.append(z)
s1=0
s2=0
s3=0
for i in range(n):
s1+=a[i]
s2+=b[i]
s3+=c[i]
if (s1==0 and s2==0 and s3==0):
print("YES")
else:
print("NO")
```
|
instruction
| 0
| 104,926
| 3
| 209,852
|
Yes
|
output
| 1
| 104,926
| 3
| 209,853
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" β thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input
The first line contains a positive integer n (1 β€ n β€ 100), then follow n lines containing three integers each: the xi coordinate, the yi coordinate and the zi coordinate of the force vector, applied to the body ( - 100 β€ xi, yi, zi β€ 100).
Output
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Examples
Input
3
4 1 7
-2 4 -1
1 -5 -3
Output
NO
Input
3
3 -1 7
-5 2 -4
2 -1 -3
Output
YES
Submitted Solution:
```
n = int(input())
l = []
for i in range(n):
x,y,z = map(int,input().split())
a = [x,y,z]
b = sum(a)
l.append(b)
if (sum(l)==0):
print("YES")
else:
print("NO")
```
|
instruction
| 0
| 104,927
| 3
| 209,854
|
No
|
output
| 1
| 104,927
| 3
| 209,855
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" β thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input
The first line contains a positive integer n (1 β€ n β€ 100), then follow n lines containing three integers each: the xi coordinate, the yi coordinate and the zi coordinate of the force vector, applied to the body ( - 100 β€ xi, yi, zi β€ 100).
Output
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Examples
Input
3
4 1 7
-2 4 -1
1 -5 -3
Output
NO
Input
3
3 -1 7
-5 2 -4
2 -1 -3
Output
YES
Submitted Solution:
```
n=int(input())
x=0
y=0
z=0
for i in range (n):
a,b,c=map(int,input().split())
x+=a
y+=b
z+=c
if (a==0) and (b==0) and (c==0):
print('YES')
else:
print('NO')
```
|
instruction
| 0
| 104,928
| 3
| 209,856
|
No
|
output
| 1
| 104,928
| 3
| 209,857
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" β thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input
The first line contains a positive integer n (1 β€ n β€ 100), then follow n lines containing three integers each: the xi coordinate, the yi coordinate and the zi coordinate of the force vector, applied to the body ( - 100 β€ xi, yi, zi β€ 100).
Output
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Examples
Input
3
4 1 7
-2 4 -1
1 -5 -3
Output
NO
Input
3
3 -1 7
-5 2 -4
2 -1 -3
Output
YES
Submitted Solution:
```
quantity = int(input())
array = []
for i in range(quantity):
x = input().split()
x = list(map(int, x))
array.append(x)
answer = 0
for valores in array:
answer += sum(valores)
if answer == 0:
print('YES')
else:
print('NO')
```
|
instruction
| 0
| 104,929
| 3
| 209,858
|
No
|
output
| 1
| 104,929
| 3
| 209,859
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A guy named Vasya attends the final grade of a high school. One day Vasya decided to watch a match of his favorite hockey team. And, as the boy loves hockey very much, even more than physics, he forgot to do the homework. Specifically, he forgot to complete his physics tasks. Next day the teacher got very angry at Vasya and decided to teach him a lesson. He gave the lazy student a seemingly easy task: You are given an idle body in space and the forces that affect it. The body can be considered as a material point with coordinates (0; 0; 0). Vasya had only to answer whether it is in equilibrium. "Piece of cake" β thought Vasya, we need only to check if the sum of all vectors is equal to 0. So, Vasya began to solve the problem. But later it turned out that there can be lots and lots of these forces, and Vasya can not cope without your help. Help him. Write a program that determines whether a body is idle or is moving by the given vectors of forces.
Input
The first line contains a positive integer n (1 β€ n β€ 100), then follow n lines containing three integers each: the xi coordinate, the yi coordinate and the zi coordinate of the force vector, applied to the body ( - 100 β€ xi, yi, zi β€ 100).
Output
Print the word "YES" if the body is in equilibrium, or the word "NO" if it is not.
Examples
Input
3
4 1 7
-2 4 -1
1 -5 -3
Output
NO
Input
3
3 -1 7
-5 2 -4
2 -1 -3
Output
YES
Submitted Solution:
```
def Sum_Force(x,y,z):
for i in range(len(x)):
if int(x[i]) + int(y[i]) + int(z[i]) != 0:
print('NO')
return
print('YES')
n = input()
x = input()
y = input()
z = input()
x = x.split()
y = y.split()
z = z.split()
Sum_Force(x,y,z)
```
|
instruction
| 0
| 104,930
| 3
| 209,860
|
No
|
output
| 1
| 104,930
| 3
| 209,861
|
Provide a correct Python 3 solution for this coding contest problem.
Aizu has an ancient legend of buried treasure. You have finally found the place where the buried treasure is buried. Since we know the depth of the buried treasure and the condition of the strata to be dug, we can reach the buried treasure at the lowest cost with careful planning. So you decided to create a program that reads the condition of the formation and calculates the route to reach the buried treasure depth at the lowest cost.
The state of the formation is represented by cells arranged in a two-dimensional grid, and the position of each cell is represented by the coordinates (x, y). Let the upper left be (1,1), and assume that the x coordinate increases as it goes to the right and the y coordinate increases as it goes deeper down. You choose one of the cells with the smallest y-coordinate and start digging from there, then dig into one of the cells with the largest y-coordinate. There are two types of cells in the formation:
1. A cell filled with soil. There is a fixed cost for each cell to dig.
2. Oxygen-filled cell. There is no need to dig, and each cell can be replenished with a fixed amount of oxygen. The oxygen in the cell that has been replenished with oxygen is exhausted and cannot be replenished again. Also, when you reach this cell, you must replenish oxygen.
Only left, right, and down cells can be dug from a cell. Once you have dug a cell, you can move it left or right, but you cannot move it up.
You must carry an oxygen cylinder with you when excavating. The moment the oxygen cylinder reaches zero, you will not be able to move, excavate, or replenish oxygen. The remaining amount is decremented by 1 each time you move the cell. Even if the remaining amount of the oxygen cylinder is 0 and the depth of the buried treasure is reached, it is not considered to have been reached. In addition, oxygen can be replenished in cells that have accumulated oxygen, but the excess capacity is discarded.
Create a program that inputs the size of the formation, the excavation cost, the capacity of the oxygen cylinder, the amount of oxygen in the initial state, and the information of the formation, and outputs the minimum cost to reach the deepest cell. However, if the minimum cost exceeds the excavation cost, or if you cannot reach the buried treasure no matter how you dig, please output "NA".
<image>
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two zero lines. Each dataset is given in the following format.
W H
f m o
c1,1 c2,1 ... cW,1
c1,2 c2,2 ... cW,2
...
c1, H c2, H ... cW, H
The horizontal size W of the formation and the vertical size H (3 β€ W, H β€ 10) are given in the first line. The second line is the integer f (1 β€ f β€ 10000) that represents your excavation cost, the integer m (3 β€ m β€ 50) that represents the capacity of the oxygen cylinder, and the integer o that represents the amount of oxygen you have in the initial state. o β€ m) is given.
The following H line is given the geological information ci, j. ci, j represents the cell information for coordinates (i, j) and is given in the following format:
If the value is negative, the cell is full of soil and the value represents the cost.
If the value is positive, it is a cell filled with oxygen, and the value represents the amount of oxygen.
However, there are no more than 50 cells in which oxygen has accumulated.
The number of datasets does not exceed 50.
Output
Print the minimum cost or NA on one line for each dataset.
Example
Input
3 3
100 10 10
-100 -20 -100
-100 -20 -100
-100 -20 -100
3 3
100 10 10
-100 -20 -100
-100 -20 -20
-100 -60 -20
3 3
100 10 3
-100 -20 -100
-20 -20 -20
-20 -100 -20
3 3
100 3 3
-100 -20 -30
-100 -20 2
-100 -20 -20
4 5
1500 5 4
-10 -380 -250 -250
-90 2 -80 8
-250 -130 -330 -120
-120 -40 -50 -20
-250 -10 -20 -150
0 0
Output
60
80
NA
50
390
|
instruction
| 0
| 105,191
| 3
| 210,382
|
"Correct Solution:
```
import sys
sys.setrecursionlimit(1000000)
INF = 10 ** 20
def update_state(state, newx):
tmp = list(state)
tmp[newx] = 1
return tuple(tmp)
def get_co(x, y):
dc = do = 0
score = mp[y][x]
if score < 0:
dc = -score
else:
do = score
return dc, do
def minimum_cost(x, y, state, ox, goal, dic, w, m):
if (x, y, state, ox) in dic:
return dic[(x, y, state, ox)]
if y == goal:
return 0
if ox <= 1:
return INF
ret = INF
if x >= 1:
if state[x - 1] == 0:
dc, do = get_co(x - 1, y)
ret = min(ret, minimum_cost(x - 1, y, update_state(state, x - 1), min(ox + do - 1, m), goal, dic, w, m) + dc)
else:
ret = min(ret, minimum_cost(x - 1, y, state, ox - 1, goal, dic, w, m))
if x < w - 1:
if state[x + 1] == 0:
dc, do = get_co(x + 1, y)
ret = min(ret, minimum_cost(x + 1, y, update_state(state, x + 1), min(ox + do - 1, m), goal, dic, w, m) + dc)
else:
ret = min(ret, minimum_cost(x + 1, y, state, ox - 1, goal, dic, w, m))
dc, do = get_co(x, y + 1)
ret = min(ret, minimum_cost(x, y + 1, tuple((1 if i == x else 0 for i in range(w))), min(ox + do - 1, m), goal, dic, w, m) + dc)
dic[(x, y, state, ox)] = ret
return ret
while True:
w, h = map(int, input().split())
if w == 0:
break
f, m, o = map(int, input().split())
mp = [list(map(int, input().split())) for _ in range(h)]
if o <= 1:
print("NA")
continue
dic = {}
ans = INF
for i in range(w):
dc, do = get_co(i, 0)
state = tuple(1 if i == j else 0 for j in range(w))
ans = min(ans, minimum_cost(i, 0, state, min(o + do - 1, m), h - 1, dic, w, m) + dc)
if ans > f:
print("NA")
else:
print(ans)
```
|
output
| 1
| 105,191
| 3
| 210,383
|
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