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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an H \times W grid (H vertical, W horizontal), where each square contains a lowercase English letter. Specifically, the letter in the square at the i-th row and j-th column is equal to the j-th character in the string S_i.
Snuke can apply the following operation to this grid any number of times:
* Choose two different rows and swap them. Or, choose two different columns and swap them.
Snuke wants this grid to be symmetric. That is, for any 1 \leq i \leq H and 1 \leq j \leq W, the letter in the square at the i-th row and j-th column and the letter in the square at the (H + 1 - i)-th row and (W + 1 - j)-th column should be equal.
Determine if Snuke can achieve this objective.
Constraints
* 1 \leq H \leq 12
* 1 \leq W \leq 12
* |S_i| = W
* S_i consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
H W
S_1
S_2
:
S_H
Output
If Snuke can make the grid symmetric, print `YES`; if he cannot, print `NO`.
Examples
Input
2 3
arc
rac
Output
YES
Input
3 7
atcoder
regular
contest
Output
NO
Input
12 12
bimonigaloaf
faurwlkbleht
dexwimqxzxbb
lxdgyoifcxid
ydxiliocfdgx
nfoabgilamoi
ibxbdqmzxxwe
pqirylfrcrnf
wtehfkllbura
yfrnpflcrirq
wvcclwgiubrk
lkbrwgwuiccv
Output
YES
Submitted Solution:
```
from collections import defaultdict
def main():
h, w = map(int, input().split())
s = []
for _ in range(h):
row = []
s.append(row)
for c in input().strip():
row.append(ord(c))
c_row = defaultdict(int)
for row in s:
c_row[tuple(sorted(row))] += 1
if sum(c % 2 for c in c_row.values()) > h % 2:
print('NO')
return
c_col = defaultdict(int)
for col in zip(*s):
c_col[tuple(sorted(col))] += 1
if sum(c % 2 for c in c_col.values()) > w % 2:
print('NO')
return
print('YES')
main()
```
|
instruction
| 0
| 1,533
| 16
| 3,066
|
No
|
output
| 1
| 1,533
| 16
| 3,067
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an H \times W grid (H vertical, W horizontal), where each square contains a lowercase English letter. Specifically, the letter in the square at the i-th row and j-th column is equal to the j-th character in the string S_i.
Snuke can apply the following operation to this grid any number of times:
* Choose two different rows and swap them. Or, choose two different columns and swap them.
Snuke wants this grid to be symmetric. That is, for any 1 \leq i \leq H and 1 \leq j \leq W, the letter in the square at the i-th row and j-th column and the letter in the square at the (H + 1 - i)-th row and (W + 1 - j)-th column should be equal.
Determine if Snuke can achieve this objective.
Constraints
* 1 \leq H \leq 12
* 1 \leq W \leq 12
* |S_i| = W
* S_i consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
H W
S_1
S_2
:
S_H
Output
If Snuke can make the grid symmetric, print `YES`; if he cannot, print `NO`.
Examples
Input
2 3
arc
rac
Output
YES
Input
3 7
atcoder
regular
contest
Output
NO
Input
12 12
bimonigaloaf
faurwlkbleht
dexwimqxzxbb
lxdgyoifcxid
ydxiliocfdgx
nfoabgilamoi
ibxbdqmzxxwe
pqirylfrcrnf
wtehfkllbura
yfrnpflcrirq
wvcclwgiubrk
lkbrwgwuiccv
Output
YES
Submitted Solution:
```
import sys
import itertools
H,W=map(int,input().split())
mat=[]
for i in range(H):
s=list(input())
mat.append(s)
#print(mat)
perm_h_list=list(itertools.permutations(range(H)))
perm_w_list=list(itertools.permutations(range(W)))
#print(perm_h_list)
#print(perm_w_list)
for perm_h in perm_h_list:
for perm_w in perm_w_list:
sym=True
for i in range(H):
for j in range(W):
if mat[perm_h[i]][perm_w[j]] != mat[perm_h[H-1-i]][perm_w[W-1-j]]:
sym=False
break
if not sym:
break
else:
print("YES")
sys.exit()
else:
print("NO")
```
|
instruction
| 0
| 1,534
| 16
| 3,068
|
No
|
output
| 1
| 1,534
| 16
| 3,069
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an H \times W grid (H vertical, W horizontal), where each square contains a lowercase English letter. Specifically, the letter in the square at the i-th row and j-th column is equal to the j-th character in the string S_i.
Snuke can apply the following operation to this grid any number of times:
* Choose two different rows and swap them. Or, choose two different columns and swap them.
Snuke wants this grid to be symmetric. That is, for any 1 \leq i \leq H and 1 \leq j \leq W, the letter in the square at the i-th row and j-th column and the letter in the square at the (H + 1 - i)-th row and (W + 1 - j)-th column should be equal.
Determine if Snuke can achieve this objective.
Constraints
* 1 \leq H \leq 12
* 1 \leq W \leq 12
* |S_i| = W
* S_i consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
H W
S_1
S_2
:
S_H
Output
If Snuke can make the grid symmetric, print `YES`; if he cannot, print `NO`.
Examples
Input
2 3
arc
rac
Output
YES
Input
3 7
atcoder
regular
contest
Output
NO
Input
12 12
bimonigaloaf
faurwlkbleht
dexwimqxzxbb
lxdgyoifcxid
ydxiliocfdgx
nfoabgilamoi
ibxbdqmzxxwe
pqirylfrcrnf
wtehfkllbura
yfrnpflcrirq
wvcclwgiubrk
lkbrwgwuiccv
Output
YES
Submitted Solution:
```
from collections import Counter
def check(field):
center = None
prev = None
for i, row in enumerate(field):
if prev is None:
prev = row
continue
if prev == row:
prev = None
continue
if center is not None:
return False
center = prev
prev = row
if center is not None:
cnt = Counter(center)
cntcnt = Counter(c % 2 for c in cnt.values())
if cntcnt[1] > 1:
return False
return True
h, w = map(int, input().split())
field = []
for i in range(h):
field.append(list(map(ord, input())))
field1 = sorted(sorted(row) for row in field)
field2 = sorted(sorted(col) for col in zip(*field))
print('YES' if check(field1) and check(field2) else 'NO')
```
|
instruction
| 0
| 1,535
| 16
| 3,070
|
No
|
output
| 1
| 1,535
| 16
| 3,071
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is an H \times W grid (H vertical, W horizontal), where each square contains a lowercase English letter. Specifically, the letter in the square at the i-th row and j-th column is equal to the j-th character in the string S_i.
Snuke can apply the following operation to this grid any number of times:
* Choose two different rows and swap them. Or, choose two different columns and swap them.
Snuke wants this grid to be symmetric. That is, for any 1 \leq i \leq H and 1 \leq j \leq W, the letter in the square at the i-th row and j-th column and the letter in the square at the (H + 1 - i)-th row and (W + 1 - j)-th column should be equal.
Determine if Snuke can achieve this objective.
Constraints
* 1 \leq H \leq 12
* 1 \leq W \leq 12
* |S_i| = W
* S_i consists of lowercase English letters.
Input
Input is given from Standard Input in the following format:
H W
S_1
S_2
:
S_H
Output
If Snuke can make the grid symmetric, print `YES`; if he cannot, print `NO`.
Examples
Input
2 3
arc
rac
Output
YES
Input
3 7
atcoder
regular
contest
Output
NO
Input
12 12
bimonigaloaf
faurwlkbleht
dexwimqxzxbb
lxdgyoifcxid
ydxiliocfdgx
nfoabgilamoi
ibxbdqmzxxwe
pqirylfrcrnf
wtehfkllbura
yfrnpflcrirq
wvcclwgiubrk
lkbrwgwuiccv
Output
YES
Submitted Solution:
```
n, m = map(int,input().split())
d = {i: {chr(i): 0 for i in range(ord('a'), ord('z')+1)} for i in range(n)}
d2 = {i: {chr(i): 0 for i in range(ord('a'), ord('z')+1)} for i in range(m)}
for i in range(n):
s = input()
for j in s:
d[i][j] += 1
for j in range(len(s)):
d2[j][s[j]] += 1
k = [0 for i in range(n)]
k2 = [0 for i in range(m)]
f = True
for i in range(n):
for j in range(n):
if d[i] == d[j]:
k[i] += 1
for i in range(m):
for j in range(m):
if d2[i] == d2[j]:
k2[i] += 1
k2 = sorted(k2)
k = sorted(k)
if n % 2:
if len(k) > 1:
if k[1] < 2:
f = False
else:
if k[0] < 2:
f = False
if m % 2:
if len(k2) > 1:
if k2[1] < 2:
f = False
else:
if k2[0] < 2:
f = False
if f:
print('YES')
else:
print('NO')
```
|
instruction
| 0
| 1,536
| 16
| 3,072
|
No
|
output
| 1
| 1,536
| 16
| 3,073
|
Provide a correct Python 3 solution for this coding contest problem.
Snuke got an integer sequence from his mother, as a birthday present. The sequence has N elements, and the i-th of them is i. Snuke performs the following Q operations on this sequence. The i-th operation, described by a parameter q_i, is as follows:
* Take the first q_i elements from the sequence obtained by concatenating infinitely many copy of the current sequence, then replace the current sequence with those q_i elements.
After these Q operations, find how many times each of the integers 1 through N appears in the final sequence.
Constraints
* 1 ≦ N ≦ 10^5
* 0 ≦ Q ≦ 10^5
* 1 ≦ q_i ≦ 10^{18}
* All input values are integers.
Input
The input is given from Standard Input in the following format:
N Q
q_1
:
q_Q
Output
Print N lines. The i-th line (1 ≦ i ≦ N) should contain the number of times integer i appears in the final sequence after the Q operations.
Examples
Input
5 3
6
4
11
Output
3
3
3
2
0
Input
10 10
9
13
18
8
10
10
9
19
22
27
Output
7
4
4
3
3
2
2
2
0
0
|
instruction
| 0
| 1,576
| 16
| 3,152
|
"Correct Solution:
```
import bisect,sys
input=sys.stdin.readline
N,Q=map(int,input().split())
que=[(N,0)]
for i in range(Q):
q=int(input())
que.append((q,i+1))
que.sort(reverse=True)
ext=[]
while que:
q,id=que.pop()
if not ext:
ext.append((q,id))
else:
if ext[-1][1]<id:
if ext[-1][0]==q:
ext.pop()
ext.append((q,id))
Q=len(ext)
data=[1]*Q
data[0]=ext[0][0]
edge=[[] for i in range(Q)]
nowext=[ext[0][0]]
for i in range(1,Q):
q=ext[i][0]
rest=q
while True:
id=bisect.bisect_right(nowext,rest)
if id==0:
break
else:
edge[id-1].append((i,rest//nowext[id-1]))
rest%=nowext[id-1]
nowext.append(ext[i][0])
data[i]=rest
#print(edge)
dp=[1]*Q
for i in range(Q-2,-1,-1):
temp=0
for nv,c in edge[i]:
temp+=dp[nv]*c
dp[i]=temp
#print(dp)
#print(data)
minus=[0]*(ext[0][0]+1)
for i in range(Q):
minus[data[i]]+=dp[i]
base=sum(dp)
for i in range(1,N+1):
if i-1<=ext[0][0]:
base-=minus[i-1]
print(base)
```
|
output
| 1
| 1,576
| 16
| 3,153
|
Provide a correct Python 3 solution for this coding contest problem.
Snuke has a string x of length N. Initially, every character in x is `0`.
Snuke can do the following two operations any number of times in any order:
* Choose A consecutive characters in x and replace each of them with `0`.
* Choose B consecutive characters in x and replace each of them with `1`.
Find the number of different strings that x can be after Snuke finishes doing operations. This count can be enormous, so compute it modulo (10^9+7).
Constraints
* 1 \leq N \leq 5000
* 1 \leq A,B \leq N
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
Print the number of different strings that x can be after Snuke finishes doing operations, modulo (10^9+7).
Examples
Input
4 2 3
Output
11
Input
10 7 2
Output
533
Input
1000 100 10
Output
828178524
|
instruction
| 0
| 3,092
| 16
| 6,184
|
"Correct Solution:
```
MOD = 10 ** 9 + 7
n, a, b = map(int, input().split())
if a > b:
a, b = b, a
dp_1s = [0] * b
for l in range(a + 2, b):
dp = [[0, 0] for _ in range(l + 1)] # at i, 0/1 precedes
dp[1][1] = 1
for i in range(1, l):
dp[i + 1][0] = (dp[i + 1][0] + dp[i][0]) % MOD
dp[i + 1][1] = (dp[i + 1][1] + dp[i][0] + dp[i][1]) % MOD
if i + a < l:
dp[i + a][0] = (dp[i + a][0] + dp[i][1]) % MOD
dp_1s[l] = dp[l][1] - 1
dp_0s_edge = [0] * b
for l in range(a + 1, b):
dp = [[0, 0] for _ in range(l + 1)]
dp[a][0] = 1
for i in range(1, l):
dp[i + 1][0] = (dp[i + 1][0] + dp[i][0]) % MOD
dp[i + 1][1] = (dp[i + 1][1] + dp[i][0] + dp[i][1]) % MOD
if i + a < l:
dp[i + a][0] = (dp[i + a][0] + dp[i][1]) % MOD
dp_0s_edge[l] = dp[l][1]
# starting at i, 0s/1s precede, i.e., 1s/0s follow
# when 0s precede, len of preceding 0s is lt a
dp = [[0, 0] for _ in range(n + 1)]
dp[0] = [1, 1]
# starting with 0s whose len is gt or eq to a
for l in range(a + 1, b):
dp[l][1] = dp_0s_edge[l]
for i in range(n):
for j in range(i + 1, min(n + 1, i + b)):
dp[j][1] = (dp[j][1] + dp[i][0]) % MOD
for j in range(i + 1, min(n + 1, i + a)):
dp[j][0] = (dp[j][0] + dp[i][1]) % MOD
for l in range(a + 2, b):
if i + l <= n:
dp[i + l][1] = (dp[i + l][1] + dp[i][0] * dp_1s[l]) % MOD
# ending with 0s whose len is gt or eq to a
for l in range(a + 1, b):
dp[n][0] = (dp[n][0] + dp[n - l][0] * dp_0s_edge[l]) % MOD
print((pow(2, n, MOD) - sum(dp[n])) % MOD)
```
|
output
| 1
| 3,092
| 16
| 6,185
|
Provide a correct Python 3 solution for this coding contest problem.
Snuke has a string x of length N. Initially, every character in x is `0`.
Snuke can do the following two operations any number of times in any order:
* Choose A consecutive characters in x and replace each of them with `0`.
* Choose B consecutive characters in x and replace each of them with `1`.
Find the number of different strings that x can be after Snuke finishes doing operations. This count can be enormous, so compute it modulo (10^9+7).
Constraints
* 1 \leq N \leq 5000
* 1 \leq A,B \leq N
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
Print the number of different strings that x can be after Snuke finishes doing operations, modulo (10^9+7).
Examples
Input
4 2 3
Output
11
Input
10 7 2
Output
533
Input
1000 100 10
Output
828178524
|
instruction
| 0
| 3,093
| 16
| 6,186
|
"Correct Solution:
```
n,aa,bb = list(map(int, input().split()))
a = max(aa,bb)
b = min(aa,bb)
p = 1000000007
xs0 = [0]*(n+1)
os0 = [0]*(n+1)
os0[0] = 1
for i in range(b,a-1):
for j in range(b,i+1):
xs0[i] += os0[i-j]
xs0[i] %= p
for j in range(1,i+1):
os0[i] += xs0[i-j]
os0[i] %= p
#print(xs0[:50])
#print(os0[:50])
os = [0]*a
for i in range(b,a-2): #number-x
for j in range(2,a-i): # adding o
os[i+j] += xs0[i] * (j-1)
x = [0]*(n+1)
o = [0]*(n+1)
for i in range(b+1,a): # adding distination
for j in range(b,i): # length of xs
o[i] += xs0[j]
o[i] %= p
x[0] = 1
o[0] = 1
for i in range(1,n+1):
for j in range(1,min(b,i+1)):
x[i] += o[i-j]
for j in range(1,min(a,i+1)):
o[i] += x[i-j]
for j in range(b+2,min(a,i+1)):
o[i] += x[i-j]*os[j]
x[i] %= p
o[i] %= p
for i in range(b+1,a):
for j in range(b,i):
o[n] += xs0[j] * x[n-i]
o[i]%=p
ans = o[n] + x[n]
ans %= p
#print(ans)
beki2 = [1]*5001
for i in range(5000):
beki2[i+1] = beki2[i]*2
beki2[i+1] %= p
ans = beki2[n] - ans
ans += p
ans %= p
print(ans)
```
|
output
| 1
| 3,093
| 16
| 6,187
|
Provide a correct Python 3 solution for this coding contest problem.
Snuke has a string x of length N. Initially, every character in x is `0`.
Snuke can do the following two operations any number of times in any order:
* Choose A consecutive characters in x and replace each of them with `0`.
* Choose B consecutive characters in x and replace each of them with `1`.
Find the number of different strings that x can be after Snuke finishes doing operations. This count can be enormous, so compute it modulo (10^9+7).
Constraints
* 1 \leq N \leq 5000
* 1 \leq A,B \leq N
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
Print the number of different strings that x can be after Snuke finishes doing operations, modulo (10^9+7).
Examples
Input
4 2 3
Output
11
Input
10 7 2
Output
533
Input
1000 100 10
Output
828178524
|
instruction
| 0
| 3,094
| 16
| 6,188
|
"Correct Solution:
```
import math
#import sys
#input = sys.stdin.readline
def make_divisors(n):
divisors = []
for i in range(1, int(n**0.5)+1):
if n % i == 0:
divisors.append(i)
if i != n // i:
divisors.append(n//i)
# divisors.sort()
return divisors
def ValueToBits(x,digit):
res = [0 for i in range(digit)]
now = x
for i in range(digit):
res[i]=now%2
now = now >> 1
return res
def BitsToValue(arr):
n = len(arr)
ans = 0
for i in range(n):
ans+= arr[i] * 2**i
return ans
def ZipArray(a):
aa = [[a[i],i]for i in range(n)]
aa.sort(key = lambda x : x[0])
for i in range(n):
aa[i][0]=i+1
aa.sort(key = lambda x : x[1])
b=[aa[i][0] for i in range(len(a))]
return b
def ValueToArray10(x, digit):
ans = [0 for i in range(digit)]
now = x
for i in range(digit):
ans[digit-i-1] = now%10
now = now //10
return ans
def Zeros(a,b):
if(b<=-1):
return [0 for i in range(a)]
else:
return [[0 for i in range(b)] for i in range(a)]
def AddV2(v,w):
return [v[0]+w[0],v[1]+w[1]]
dir4 = [[1,0],[0,1],[-1,0],[0,-1]]
def clamp(x,y,z):
return max(y,min(z,x))
class Bit:
def __init__(self, n):
self.size = n
self.tree = [0] * (n + 1)
def sum(self, i):
s = 0
while i > 0:
s += self.tree[i]
i -= i & -i
return s
def add(self, i, x):
while i <= self.size:
self.tree[i] += x
i += i & -i
#
def Zaatsu(a):
a.sort()
now = a[0][0]
od = 0
for i in range(n):
if(now==a[i][0]):
a[i][0]=od
else:
now = a[i][0]
od+=1
a[i][0] = od
a.sort(key = lambda x : x[1])
return a
class UnionFind:
def __init__(self, n):
self.par = [i for i in range(n+1)]
self.rank = [0] * (n+1)
# 検索
def find(self, x):
if self.par[x] == x:
return x
else:
self.par[x] = self.find(self.par[x])
return self.par[x]
# 併合
def union(self, x, y):
x = self.find(x)
y = self.find(y)
if self.rank[x] < self.rank[y]:
self.par[x] = y
else:
self.par[y] = x
if self.rank[x] == self.rank[y]:
self.rank[x] += 1
# 同じ集合に属するか判定
def same_check(self, x, y):
return self.find(x) == self.find(y)
'''
def cmb(n, r, p):
if (r < 0) or (n < r):
return 0
r = min(r, n - r)
return fact[n] * factinv[r] * factinv[n-r] % p
p = 2
N = 10 ** 6 + 2
fact = [1, 1] # fact[n] = (n! mod p)
factinv = [1, 1] # factinv[n] = ((n!)^(-1) mod p)
inv = [0, 1] # factinv 計算用
for i in range(2, N + 1):
fact.append((fact[-1] * i) % p)
inv.append((-inv[p % i] * (p // i)) % p)
factinv.append((factinv[-1] * inv[-1]) % p)
'''
def rl(x):
return range(len(x))
# a = list(map(int, input().split()))
#################################################
#################################################
#################################################
#################################################
#11-
n,aa,bb = list(map(int, input().split()))
a = max(aa,bb)
b = min(aa,bb)
p = 1000000007
xs0 = [0]*(n+1)
os0 = [0]*(n+1)
os0[0] = 1
for i in range(b,a-1):
for j in range(b,i+1):
xs0[i] += os0[i-j]
xs0[i] %= p
for j in range(1,i+1):
os0[i] += xs0[i-j]
os0[i] %= p
#print(xs0[:50])
#print(os0[:50])
os = [0]*a
for i in range(b,a-2): #number-x
for j in range(2,a-i): # adding o
os[i+j] += xs0[i] * (j-1)
x = [0]*(n+1)
o = [0]*(n+1)
for i in range(b+1,a): # adding distination
for j in range(b,i): # length of xs
o[i] += xs0[j]
o[i] %= p
x[0] = 1
o[0] = 1
for i in range(1,n+1):
for j in range(1,min(b,i+1)):
x[i] += o[i-j]
for j in range(1,min(a,i+1)):
o[i] += x[i-j]
for j in range(b+2,min(a,i+1)):
o[i] += x[i-j]*os[j]
x[i] %= p
o[i] %= p
for i in range(b+1,a):
for j in range(b,i):
o[n] += xs0[j] * x[n-i]
o[i]%=p
ans = o[n] + x[n]
ans %= p
#print(ans)
beki2 = [1]*5001
for i in range(5000):
beki2[i+1] = beki2[i]*2
beki2[i+1] %= p
ans = beki2[n] - ans
ans += p
ans %= p
print(ans)
```
|
output
| 1
| 3,094
| 16
| 6,189
|
Provide a correct Python 3 solution for this coding contest problem.
Snuke has a string x of length N. Initially, every character in x is `0`.
Snuke can do the following two operations any number of times in any order:
* Choose A consecutive characters in x and replace each of them with `0`.
* Choose B consecutive characters in x and replace each of them with `1`.
Find the number of different strings that x can be after Snuke finishes doing operations. This count can be enormous, so compute it modulo (10^9+7).
Constraints
* 1 \leq N \leq 5000
* 1 \leq A,B \leq N
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
Print the number of different strings that x can be after Snuke finishes doing operations, modulo (10^9+7).
Examples
Input
4 2 3
Output
11
Input
10 7 2
Output
533
Input
1000 100 10
Output
828178524
|
instruction
| 0
| 3,095
| 16
| 6,190
|
"Correct Solution:
```
N,A,B=map(int,input().split())
mod=10**9+7
A,B=min(A,B),max(A,B)
data=[0]*B
data[0]=1
for i in range(1,B):
data[i]=(data[i-1]+sum(data[i-1-k] for k in range(A,i)))%mod
#print(data)
dp=[0]*(N+1)
dp[0]=1
imos=[0]*(N+1)
imos[0]=1
for i in range(1,N+1):
if i!=N:
for j in range(1,min(i,B)):
#dp[i]+=sum(dp[i-j-k]*data[j-1] for k in range(1,min(A,i-j+1)))
R=i-j-1
L=i-j-min(A,i-j+1)
if R>L:
if L>-1:
dp[i]+=(imos[R]-imos[L])*data[j-1]
else:
dp[i]+=imos[R]*data[j-1]
dp[i]%=mod
if B>i:
dp[i]+=data[i]
dp[i]%=mod
else:
for j in range(B):
#dp[i]+=sum(dp[i-j-k]*data[j] for k in range(1,min(A,i-j+1)))
R=i-j-1
L=i-j-min(A,i-j+1)
if R>L:
if L>-1:
dp[i]+=(imos[R]-imos[L])*data[j]
else:
dp[i]+=imos[R]*data[j]
dp[i]%=mod
#print(dp[i])
#dp[i]+=sum(dp[i-k] for k in range(1,A))
imos[i]=dp[i]
imos[i]+=imos[i-1]
imos[i]%=mod
#print(dp)
print((pow(2,N,mod)-dp[N])%mod)
```
|
output
| 1
| 3,095
| 16
| 6,191
|
Provide a correct Python 3 solution for this coding contest problem.
Snuke has a string x of length N. Initially, every character in x is `0`.
Snuke can do the following two operations any number of times in any order:
* Choose A consecutive characters in x and replace each of them with `0`.
* Choose B consecutive characters in x and replace each of them with `1`.
Find the number of different strings that x can be after Snuke finishes doing operations. This count can be enormous, so compute it modulo (10^9+7).
Constraints
* 1 \leq N \leq 5000
* 1 \leq A,B \leq N
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
Print the number of different strings that x can be after Snuke finishes doing operations, modulo (10^9+7).
Examples
Input
4 2 3
Output
11
Input
10 7 2
Output
533
Input
1000 100 10
Output
828178524
|
instruction
| 0
| 3,096
| 16
| 6,192
|
"Correct Solution:
```
def solve(n, a, b):
MOD = 10 ** 9 + 7
if a > b:
a, b = b, a
if a == 1:
return pow(2, n, MOD)
# 長さ i の区間であり、右端が '1' であり、
# はじめ '1' で塗りつぶされていたのを、
# 長さ a 以上の '0' で0回以上上書きすることで実現できる並びの個数
dp1 = [0] * (b + 1)
dp1_acc = [0] * (b + 1)
dp1[0] = 1
dp1_acc[0] = 1
for i in range(1, b + 1):
tmp = dp1[i - 1] # 末尾に1を付ける
if i - a - 1 >= 0:
tmp = (tmp + dp1_acc[i - a - 1]) % MOD # 末尾に 00..01 を付ける
dp1[i] = tmp
dp1_acc[i] = (dp1_acc[i - 1] + tmp) % MOD
# 派生情報
# dp1[i-1]: 長さ i の区間であり、「両端」が'1'であるものの個数
# dp1[i] - dp1[i-1]: 長さ i の区間であり、左端が'0'、右端が'1'（またはその逆）のものの個数
# dp2x[i]
# 長さ i の区間であり、末尾が x であり、
# 長さa以上の'0'も、長さb以上の'1'も含まない01の並びの個数
# ただし'1'は、dp1で求めたように、その内部をa個以上の'0'で置きかえたものも含む
dp20 = [0] * (n + 1)
dp21 = [0] * (n + 1)
dp21_acc = [0] * (n + 1)
dp20[0] = dp21[0] = dp21_acc[0] = 1
for i in range(1, n + 1):
t0 = dp21_acc[i - 1]
if i >= a:
t0 -= dp21_acc[i - a]
dp20[i] = t0 % MOD
t1 = 0
for j in range(1, min(i + 1, b)):
t1 += dp20[i - j] * dp1[j - 1]
# 左端が '111...' でb個以上取れないもので、さらに'0'で置きかえられた結果、最左端が'0'のもの
if i < b:
t1 += dp1[i] - dp1[i - 1]
t1 %= MOD
dp21[i] = t1
dp21_acc[i] = (dp21_acc[i - 1] + t1) % MOD
disable = dp20[n] + dp21[n]
# 最後が'111..'でb個以上取れないもので、
# さらに'0'で置きかえられた結果、右端が'0'のものが数えられていない
for i in range(1, b):
disable += dp20[n - i] * (dp1[i] - dp1[i - 1])
return (pow(2, n, MOD) - disable) % MOD
n, a, b = map(int, input().split())
print(solve(n, a, b))
```
|
output
| 1
| 3,096
| 16
| 6,193
|
Provide a correct Python 3 solution for this coding contest problem.
Snuke has a string x of length N. Initially, every character in x is `0`.
Snuke can do the following two operations any number of times in any order:
* Choose A consecutive characters in x and replace each of them with `0`.
* Choose B consecutive characters in x and replace each of them with `1`.
Find the number of different strings that x can be after Snuke finishes doing operations. This count can be enormous, so compute it modulo (10^9+7).
Constraints
* 1 \leq N \leq 5000
* 1 \leq A,B \leq N
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
Print the number of different strings that x can be after Snuke finishes doing operations, modulo (10^9+7).
Examples
Input
4 2 3
Output
11
Input
10 7 2
Output
533
Input
1000 100 10
Output
828178524
|
instruction
| 0
| 3,097
| 16
| 6,194
|
"Correct Solution:
```
"""
Writer: SPD_9X2
すべて0にする、すべて1にするは自明に可能
なので A >= B としてよい(A<Bの時は01を入れ替えて考えればよい)
最後に置いた場所の周囲がどんな状況か考える
0*Aがある場合、その周囲はすべての場合構成可能
ない場合は…？
1*Bがある場合を考える。長さB以上の1列は後からおけるからおｋだが…
長さB以上の1をすべて0に変えたときに、長さA以上の0があればおｋ
どう数え上げる？
ダメな場合を引くのがよさそう？
→dp[i][last][num] = i番目まで見て、lastを現在num個連続しておいている場合の通り数
ただし、dp[?][1][B-1]はdp[?][0][B]に伝播する
これまずいか
ブロックで考えるやり方はどうなんだろう
つまり、連鎖を断ち切れるのは長さB以下の1列だけ
0 or B以上連続する1の列のみで構成されたブロックX(Yに隣接する端は0である必要あり)
B未満の1列であるブロックYを交互に置いていく
dp[i][XorY] = i番目の文字を次に置けるとき、最後に置いたブロックがXorYの時の通り数
初期値は
dp[0][X] = 1
dp[0][Y] = 1 ←これ、かなり注意する必要がある左端が1でもいいから
同様に、Nに移行する時にXを置くときも右端を1にできるのに注意
B以上連続する1を0に変えるとすべて0になる文字列の数をまず求める
上と同様にブロックdp?
0のブロックは,1以上のどんな長さでも置ける
1のブロックは,B以上の長さ置ける
dp[0][Z] = 1
dp[0][O] = 1
"""
N,A,B = map(int,input().split())
fdp = [[0,0] for i in range(N+1)]
fdp[0] = [1,1]
mod = 10**9+7
if A < B:
tmp = B
B = A
A = tmp
for i in range(N):
for j in range(i+1,N+1):
fdp[j][0] += fdp[i][1]
fdp[j][0] %= mod
for j in range(i+B,N+1):
fdp[j][1] += fdp[i][0]
fdp[j][1] %= mod
#print (fdp)
dp = [[0,0] for i in range(N+1)]
dp[0] = [1,1]
for i in range(N):
for j in range(i+1,min(N+1,i+B)):
dp[j][1] += dp[i][0]
dp[j][1] %= mod
#ここかなり注意
for j in range(i+1,min(N+1,i+A)):
l = j-i
if i == 0 or j == N:
if l == 1:
dp[j][0] += dp[i][1]
dp[j][0] %= mod
else:
dp[j][0] += dp[i][1] * sum(fdp[l-1])
dp[j][0] %= mod
else:
if l <= 2:
dp[j][0] += dp[i][1]
dp[j][0] %= mod
else:
dp[j][0] += dp[i][1] * sum(fdp[l-2])
dp[j][0] %= mod
#print (dp)
print ((pow(2,N,mod)-sum(dp[N])) % mod)
```
|
output
| 1
| 3,097
| 16
| 6,195
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke has a string x of length N. Initially, every character in x is `0`.
Snuke can do the following two operations any number of times in any order:
* Choose A consecutive characters in x and replace each of them with `0`.
* Choose B consecutive characters in x and replace each of them with `1`.
Find the number of different strings that x can be after Snuke finishes doing operations. This count can be enormous, so compute it modulo (10^9+7).
Constraints
* 1 \leq N \leq 5000
* 1 \leq A,B \leq N
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
Print the number of different strings that x can be after Snuke finishes doing operations, modulo (10^9+7).
Examples
Input
4 2 3
Output
11
Input
10 7 2
Output
533
Input
1000 100 10
Output
828178524
Submitted Solution:
```
def combination_mod(n, r, mod):
# フェルマーの小定理
# nCr=n!(r!)^p-2((n-r)!)^p-2
n_ = 1
for i in range(1, n + 1):
n_ = (n_ * i) % mod
r_ = 1
for i in range(1, r + 1):
r_ = (r_ * i) % mod
nr_ = 1
for i in range(1, n - r + 1):
nr_ = (nr_ * i) % mod
power_r = pow(r_, mod - 2, mod)
power_nr = pow(nr_, mod - 2, mod)
return (n_ * power_r * power_nr) % mod
def main():
N,A,B=map(int,input().split())
mod=pow(10,9)+7
PatternA=0
PatternB=0
#PatternA
for a in range(N-(A-1)):
PatternA+=combination_mod(N-(A-1),a,mod)
#PatternB
for b in range(N-(B-1)):
PatternB+=combination_mod(N-(B-1),b,mod)
res=PatternA+PatternB
print(res+1)
if __name__=="__main__":
main()
```
|
instruction
| 0
| 3,098
| 16
| 6,196
|
No
|
output
| 1
| 3,098
| 16
| 6,197
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke has a string x of length N. Initially, every character in x is `0`.
Snuke can do the following two operations any number of times in any order:
* Choose A consecutive characters in x and replace each of them with `0`.
* Choose B consecutive characters in x and replace each of them with `1`.
Find the number of different strings that x can be after Snuke finishes doing operations. This count can be enormous, so compute it modulo (10^9+7).
Constraints
* 1 \leq N \leq 5000
* 1 \leq A,B \leq N
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
Print the number of different strings that x can be after Snuke finishes doing operations, modulo (10^9+7).
Examples
Input
4 2 3
Output
11
Input
10 7 2
Output
533
Input
1000 100 10
Output
828178524
Submitted Solution:
```
def combination_mod(n, r, mod):
# フェルマーの小定理
# nCr=n!(r!)^p-2((n-r)!)^p-2
n_ = 1
for i in range(1, n + 1):
n_ = (n_ * i) % mod
r_ = 1
for i in range(1, r + 1):
r_ = (r_ * i) % mod
nr_ = 1
for i in range(1, n - r + 1):
nr_ = (nr_ * i) % mod
power_r = pow(r_, mod - 2, mod)
power_nr = pow(nr_, mod - 2, mod)
return (n_ * power_r * power_nr) % mod
def main():
N,A,B=map(int,input().split())
mod=pow(10,9)+7
PatternA=0
PatternB=0
#PatternA
for a in range(1,N-A+2):
PatternA+=combination_mod(N-(A-1),a,mod)
#PatternB
for b in range(1,N-B+2):
PatternB+=combination_mod(N-(B-1),b,mod)
res=PatternA+PatternB
print(res+1)
if __name__=="__main__":
main()
```
|
instruction
| 0
| 3,099
| 16
| 6,198
|
No
|
output
| 1
| 3,099
| 16
| 6,199
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke has a string x of length N. Initially, every character in x is `0`.
Snuke can do the following two operations any number of times in any order:
* Choose A consecutive characters in x and replace each of them with `0`.
* Choose B consecutive characters in x and replace each of them with `1`.
Find the number of different strings that x can be after Snuke finishes doing operations. This count can be enormous, so compute it modulo (10^9+7).
Constraints
* 1 \leq N \leq 5000
* 1 \leq A,B \leq N
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
Print the number of different strings that x can be after Snuke finishes doing operations, modulo (10^9+7).
Examples
Input
4 2 3
Output
11
Input
10 7 2
Output
533
Input
1000 100 10
Output
828178524
Submitted Solution:
```
import math
#import sys
#input = sys.stdin.readline
def make_divisors(n):
divisors = []
for i in range(1, int(n**0.5)+1):
if n % i == 0:
divisors.append(i)
if i != n // i:
divisors.append(n//i)
# divisors.sort()
return divisors
def ValueToBits(x,digit):
res = [0 for i in range(digit)]
now = x
for i in range(digit):
res[i]=now%2
now = now >> 1
return res
def BitsToValue(arr):
n = len(arr)
ans = 0
for i in range(n):
ans+= arr[i] * 2**i
return ans
def ZipArray(a):
aa = [[a[i],i]for i in range(n)]
aa.sort(key = lambda x : x[0])
for i in range(n):
aa[i][0]=i+1
aa.sort(key = lambda x : x[1])
b=[aa[i][0] for i in range(len(a))]
return b
def ValueToArray10(x, digit):
ans = [0 for i in range(digit)]
now = x
for i in range(digit):
ans[digit-i-1] = now%10
now = now //10
return ans
def Zeros(a,b):
if(b<=-1):
return [0 for i in range(a)]
else:
return [[0 for i in range(b)] for i in range(a)]
def AddV2(v,w):
return [v[0]+w[0],v[1]+w[1]]
dir4 = [[1,0],[0,1],[-1,0],[0,-1]]
def clamp(x,y,z):
return max(y,min(z,x))
class Bit:
def __init__(self, n):
self.size = n
self.tree = [0] * (n + 1)
def sum(self, i):
s = 0
while i > 0:
s += self.tree[i]
i -= i & -i
return s
def add(self, i, x):
while i <= self.size:
self.tree[i] += x
i += i & -i
#
def Zaatsu(a):
a.sort()
now = a[0][0]
od = 0
for i in range(n):
if(now==a[i][0]):
a[i][0]=od
else:
now = a[i][0]
od+=1
a[i][0] = od
a.sort(key = lambda x : x[1])
return a
class UnionFind:
def __init__(self, n):
self.par = [i for i in range(n+1)]
self.rank = [0] * (n+1)
# 検索
def find(self, x):
if self.par[x] == x:
return x
else:
self.par[x] = self.find(self.par[x])
return self.par[x]
# 併合
def union(self, x, y):
x = self.find(x)
y = self.find(y)
if self.rank[x] < self.rank[y]:
self.par[x] = y
else:
self.par[y] = x
if self.rank[x] == self.rank[y]:
self.rank[x] += 1
# 同じ集合に属するか判定
def same_check(self, x, y):
return self.find(x) == self.find(y)
'''
def cmb(n, r, p):
if (r < 0) or (n < r):
return 0
r = min(r, n - r)
return fact[n] * factinv[r] * factinv[n-r] % p
p = 2
N = 10 ** 6 + 2
fact = [1, 1] # fact[n] = (n! mod p)
factinv = [1, 1] # factinv[n] = ((n!)^(-1) mod p)
inv = [0, 1] # factinv 計算用
for i in range(2, N + 1):
fact.append((fact[-1] * i) % p)
inv.append((-inv[p % i] * (p // i)) % p)
factinv.append((factinv[-1] * inv[-1]) % p)
'''
def rl(x):
return range(len(x))
# a = list(map(int, input().split()))
#################################################
#################################################
#################################################
#################################################
#11-
n,aa,bb = list(map(int, input().split()))
a = max(aa,bb)
b = min(aa,bb)
p = 1000000007
xs0 = [0]*(n+1)
os0 = [0]*(n+1)
os0[0] = 1
for i in range(b,a-1):
for j in range(b,i+1):
xs0[i] += os0[i-j]
xs0[i] %= p
for j in range(1,i+1):
os0[i] += xs0[i-j]
os0[i] %= p
#print(xs0[:50])
#print(os0[:50])
os = [0]*a
for i in range(b,a-2): #number-x
for j in range(2,a-i): # adding o
os[i+j] += xs0[i] * (j-1)
x = [0]*(n+1)
o = [0]*(n+1)
for i in range(b+1,a-1): # adding distination
for j in range(b,i): # length of xs
o[i] += xs0[j]
o[i] %= p
x[0] = 1
o[0] = 1
for i in range(1,n+1):
for j in range(1,min(b,i+1)):
x[i] += o[i-j]
for j in range(1,min(a,i+1)):
o[i] += x[i-j]
for j in range(b+2,min(a,i+1)):
o[i] += x[i-j]*os[j]
x[i] %= p
o[i] %= p
for i in range(b+1,a-1):
for j in range(b,i):
o[n] += xs0[j] * x[n-i]
o[i]%=p
ans = o[n] + x[n]
ans %= p
#print(ans)
beki2 = [1]*5001
for i in range(5000):
beki2[i+1] = beki2[i]*2
beki2[i+1] %= p
ans = beki2[n] - ans
ans += p
ans %= p
print(ans)
```
|
instruction
| 0
| 3,100
| 16
| 6,200
|
No
|
output
| 1
| 3,100
| 16
| 6,201
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke has a string x of length N. Initially, every character in x is `0`.
Snuke can do the following two operations any number of times in any order:
* Choose A consecutive characters in x and replace each of them with `0`.
* Choose B consecutive characters in x and replace each of them with `1`.
Find the number of different strings that x can be after Snuke finishes doing operations. This count can be enormous, so compute it modulo (10^9+7).
Constraints
* 1 \leq N \leq 5000
* 1 \leq A,B \leq N
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N A B
Output
Print the number of different strings that x can be after Snuke finishes doing operations, modulo (10^9+7).
Examples
Input
4 2 3
Output
11
Input
10 7 2
Output
533
Input
1000 100 10
Output
828178524
Submitted Solution:
```
n,aa,bb = list(map(int, input().split()))
a = max(aa,bb)
b = min(aa,bb)
p = 1000000007
def calc(a,b,p):
xs0 = [0]*(n+1)
os0 = [0]*(n+1)
os0[0] = 1
for i in range(b,a-1):
for j in range(b,i+1):
xs0[i] += os0[i-j]
xs0[i] %= p
for j in range(1,i+1):
os0[i] += xs0[i-j]
os0[i] %= p
#print(xs0[:50])
#print(os0[:50])
os = [0]*a
for i in range(b,a-2): #number-x
for j in range(2,a-i): # adding o
os[i+j] += xs0[i] * (j-1)
x = [0]*(n+1)
o = [0]*(n+1)
for i in range(b+1,a): # adding distination
for j in range(b,i): # length of xs
o[i] += xs0[j]
o[i] %= p
x[0] = 1
o[0] = 1
for i in range(1,n+1):
for j in range(1,min(b,i+1)):
x[i] += o[i-j]
for j in range(1,min(a,i+1)):
o[i] += x[i-j]
for j in range(b+2,min(a,i+1)):
o[i] += x[i-j]*os[j]
x[i] %= p
o[i] %= p
for i in range(b+1,a):
for j in range(b,i):
o[n] += xs0[j] * x[n-i]
o[i]%=p
ans = o[n] + x[n]
ans %= p
#print(ans)
beki2 = [1]*5001
for i in range(5000):
beki2[i+1] = beki2[i]*2
beki2[i+1] %= p
ans = beki2[n] - ans
ans += p
ans %= p
return ans
print(calc(a,b,p))
```
|
instruction
| 0
| 3,101
| 16
| 6,202
|
No
|
output
| 1
| 3,101
| 16
| 6,203
|
Provide a correct Python 3 solution for this coding contest problem.
Snuke is standing on a two-dimensional plane. In one operation, he can move by 1 in the positive x-direction, or move by 1 in the positive y-direction.
Let us define a function f(r, c) as follows:
* f(r,c) := (The number of paths from the point (0, 0) to the point (r, c) that Snuke can trace by repeating the operation above)
Given are integers r_1, r_2, c_1, and c_2. Find the sum of f(i, j) over all pair of integers (i, j) such that r_1 ≤ i ≤ r_2 and c_1 ≤ j ≤ c_2, and compute this value modulo (10^9+7).
Constraints
* 1 ≤ r_1 ≤ r_2 ≤ 10^6
* 1 ≤ c_1 ≤ c_2 ≤ 10^6
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
r_1 c_1 r_2 c_2
Output
Print the sum of f(i, j) modulo (10^9+7).
Examples
Input
1 1 2 2
Output
14
Input
314 159 2653 589
Output
602215194
|
instruction
| 0
| 3,102
| 16
| 6,204
|
"Correct Solution:
```
def solve():
r1, c1, r2, c2 = map(int, input().split())
MOD = 10**9+7
n = r2+c2+2
fac = [1]*(n+1)
rev = [1]*(n+1)
for i in range(1,n+1):
fac[i] = i*fac[i-1]%MOD
rev[i] = pow(fac[i], MOD-2, MOD)
comb = lambda a,b:(fac[a]*rev[a-b]*rev[b])%MOD
f = lambda r,c:comb(r+c+2, c+1)
ans = (f(r2, c2)-f(r2, c1-1)-f(r1-1, c2)+f(r1-1, c1-1))%MOD
print(ans)
if __name__ == "__main__":
solve()
```
|
output
| 1
| 3,102
| 16
| 6,205
|
Provide a correct Python 3 solution for this coding contest problem.
Snuke is standing on a two-dimensional plane. In one operation, he can move by 1 in the positive x-direction, or move by 1 in the positive y-direction.
Let us define a function f(r, c) as follows:
* f(r,c) := (The number of paths from the point (0, 0) to the point (r, c) that Snuke can trace by repeating the operation above)
Given are integers r_1, r_2, c_1, and c_2. Find the sum of f(i, j) over all pair of integers (i, j) such that r_1 ≤ i ≤ r_2 and c_1 ≤ j ≤ c_2, and compute this value modulo (10^9+7).
Constraints
* 1 ≤ r_1 ≤ r_2 ≤ 10^6
* 1 ≤ c_1 ≤ c_2 ≤ 10^6
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
r_1 c_1 r_2 c_2
Output
Print the sum of f(i, j) modulo (10^9+7).
Examples
Input
1 1 2 2
Output
14
Input
314 159 2653 589
Output
602215194
|
instruction
| 0
| 3,103
| 16
| 6,206
|
"Correct Solution:
```
class Factorial:
def __init__(self,n,mod):
self.f=f=[0]*(n+1)
f[0]=b=1
self.mod=mod
for i in range(1,n+1):f[i]=b=b*i%mod
self.inv=inv=[0]*(n+1)
inv[0]=b=pow(self.f[n],mod-2,mod)
for i in range(1,n+1):inv[i]=b=b*(n+1-i)%mod
self.inv.reverse()
def factorial(self,i):
return self.f[i]
def ifactorial(self,i):
return self.inv[i]
def comb(self,n,k):
return self.f[n]*self.inv[n-k]%self.mod*self.inv[k]%self.mod if n>=k else 0
M=10**9+7
a,b,c,d=map(int,input().split())
comb=Factorial(c+d+2,M).comb
f=lambda r,c:comb(c+r+2,r+1)-c-r-2
print((f(c,d)-f(c,b-1)-f(a-1,d)+f(a-1,b-1))%M)
```
|
output
| 1
| 3,103
| 16
| 6,207
|
Provide a correct Python 3 solution for this coding contest problem.
Snuke is standing on a two-dimensional plane. In one operation, he can move by 1 in the positive x-direction, or move by 1 in the positive y-direction.
Let us define a function f(r, c) as follows:
* f(r,c) := (The number of paths from the point (0, 0) to the point (r, c) that Snuke can trace by repeating the operation above)
Given are integers r_1, r_2, c_1, and c_2. Find the sum of f(i, j) over all pair of integers (i, j) such that r_1 ≤ i ≤ r_2 and c_1 ≤ j ≤ c_2, and compute this value modulo (10^9+7).
Constraints
* 1 ≤ r_1 ≤ r_2 ≤ 10^6
* 1 ≤ c_1 ≤ c_2 ≤ 10^6
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
r_1 c_1 r_2 c_2
Output
Print the sum of f(i, j) modulo (10^9+7).
Examples
Input
1 1 2 2
Output
14
Input
314 159 2653 589
Output
602215194
|
instruction
| 0
| 3,104
| 16
| 6,208
|
"Correct Solution:
```
r1, c1, r2, c2 = list(map(int, input().split()))
MOD = 10 ** 9 + 7
def modInverse(x, p):
# Fermat's little theorem
return pow(x, p - 2, p)
def nCr(n, r, p):
num = den = 1
for i in range(r):
num = (num * (n - i)) % p
den = (den * (i + 1)) % p
return num * modInverse(den, p) % p
def hockey(n, m):
# sum i=0 to m of nCr(n + i, i) == nCr(n + m + 1, m)
return nCr(n + m + 1, m, MOD)
tot1 = hockey(c2, r2 + 1) - hockey(c2, r1)
tot2 = hockey(c1 - 1, r2 + 1) - hockey(c1 - 1, r1)
print((tot1 - tot2) % MOD)
```
|
output
| 1
| 3,104
| 16
| 6,209
|
Provide a correct Python 3 solution for this coding contest problem.
Snuke is standing on a two-dimensional plane. In one operation, he can move by 1 in the positive x-direction, or move by 1 in the positive y-direction.
Let us define a function f(r, c) as follows:
* f(r,c) := (The number of paths from the point (0, 0) to the point (r, c) that Snuke can trace by repeating the operation above)
Given are integers r_1, r_2, c_1, and c_2. Find the sum of f(i, j) over all pair of integers (i, j) such that r_1 ≤ i ≤ r_2 and c_1 ≤ j ≤ c_2, and compute this value modulo (10^9+7).
Constraints
* 1 ≤ r_1 ≤ r_2 ≤ 10^6
* 1 ≤ c_1 ≤ c_2 ≤ 10^6
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
r_1 c_1 r_2 c_2
Output
Print the sum of f(i, j) modulo (10^9+7).
Examples
Input
1 1 2 2
Output
14
Input
314 159 2653 589
Output
602215194
|
instruction
| 0
| 3,105
| 16
| 6,210
|
"Correct Solution:
```
a1,b1,a2,b2=map(int,input().split())
import sys
sys.setrecursionlimit(2000000000)
p = 10 ** 9 + 7
N = a2+b2+2 # N は必要分だけ用意する
R = max(a2+1,b2+1)
fact = [1, 1] # fact[n] = (n! mod p)
factinv = [1, 1] # factinv[n] = ((n!)^(-1) mod p)
inv = [0, 1] # factinv 計算用
for i in range(2, N + 1):
fact.append((fact[-1] * i) % p)
for i in range(2, R + 1):
inv.append((-inv[p % i] * (p // i)) % p)
factinv.append((factinv[-1] * inv[-1]) % p)
def comb(n, r, p):
if (r < 0) or (n < r):
return 0
r = min(r, n - r)
return fact[n] * factinv[r] * factinv[n-r] % p
score=comb(a2+b2+2,a2+1,p)-comb(a2+b1+1,a2+1,p)-comb(a1+b2+1,a1,p)+comb(a1+b1,a1,p)
print(score%(10**9+7))
```
|
output
| 1
| 3,105
| 16
| 6,211
|
Provide a correct Python 3 solution for this coding contest problem.
Snuke is standing on a two-dimensional plane. In one operation, he can move by 1 in the positive x-direction, or move by 1 in the positive y-direction.
Let us define a function f(r, c) as follows:
* f(r,c) := (The number of paths from the point (0, 0) to the point (r, c) that Snuke can trace by repeating the operation above)
Given are integers r_1, r_2, c_1, and c_2. Find the sum of f(i, j) over all pair of integers (i, j) such that r_1 ≤ i ≤ r_2 and c_1 ≤ j ≤ c_2, and compute this value modulo (10^9+7).
Constraints
* 1 ≤ r_1 ≤ r_2 ≤ 10^6
* 1 ≤ c_1 ≤ c_2 ≤ 10^6
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
r_1 c_1 r_2 c_2
Output
Print the sum of f(i, j) modulo (10^9+7).
Examples
Input
1 1 2 2
Output
14
Input
314 159 2653 589
Output
602215194
|
instruction
| 0
| 3,106
| 16
| 6,212
|
"Correct Solution:
```
mod = 10**9+7
def comb_mod(n,r,mod,invs):
ans = 1
for i in range(r):
ans *= n-i
ans %= mod
ans *= invs[r]
ans %= mod
return ans
A, B, C, D = map(int, input().split())
invs = [1]*(C+2)
for i in range(1,C+2):
invs[i] = invs[i-1]*pow(i,mod-2,mod)
invs[i] %= mod
ans = comb_mod(C+D+2,C+1,mod,invs)-comb_mod(B+C+1,B,mod,invs)-comb_mod(A+D+1,A,mod,invs)+comb_mod(A+B,A,mod,invs)
print(ans%mod)
```
|
output
| 1
| 3,106
| 16
| 6,213
|
Provide a correct Python 3 solution for this coding contest problem.
Snuke is standing on a two-dimensional plane. In one operation, he can move by 1 in the positive x-direction, or move by 1 in the positive y-direction.
Let us define a function f(r, c) as follows:
* f(r,c) := (The number of paths from the point (0, 0) to the point (r, c) that Snuke can trace by repeating the operation above)
Given are integers r_1, r_2, c_1, and c_2. Find the sum of f(i, j) over all pair of integers (i, j) such that r_1 ≤ i ≤ r_2 and c_1 ≤ j ≤ c_2, and compute this value modulo (10^9+7).
Constraints
* 1 ≤ r_1 ≤ r_2 ≤ 10^6
* 1 ≤ c_1 ≤ c_2 ≤ 10^6
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
r_1 c_1 r_2 c_2
Output
Print the sum of f(i, j) modulo (10^9+7).
Examples
Input
1 1 2 2
Output
14
Input
314 159 2653 589
Output
602215194
|
instruction
| 0
| 3,107
| 16
| 6,214
|
"Correct Solution:
```
r1, c1, r2, c2 = map(int, input().split())
L = 10 ** 9 + 7
# 互除法
def get_euclidian (A, B):
if B == 1:
return (1)
else:
return (int((1 - A * get_euclidian (B, A % B)) / B))
# 階乗計算
F = [1]
for i in range(1, r2 + c2 + 3):
F.append((F[i - 1] * i) % L)
# 組み合わせ計算
def get_combi (n, r):
Euc = get_euclidian(L, (F[r] * F[n - r]) % L)
return ((F[n] * Euc) % L)
combi1 = get_combi(r2 + c2 + 2, r2 + 1)
combi2 = get_combi(r2 + c1 + 1, c1)
combi3 = get_combi(r1 + c2 + 1, r1)
combi4 = get_combi(r1 + c1, r1)
print ((int(combi1 - combi2 - combi3 + combi4) + L) % L)
```
|
output
| 1
| 3,107
| 16
| 6,215
|
Provide a correct Python 3 solution for this coding contest problem.
Snuke is standing on a two-dimensional plane. In one operation, he can move by 1 in the positive x-direction, or move by 1 in the positive y-direction.
Let us define a function f(r, c) as follows:
* f(r,c) := (The number of paths from the point (0, 0) to the point (r, c) that Snuke can trace by repeating the operation above)
Given are integers r_1, r_2, c_1, and c_2. Find the sum of f(i, j) over all pair of integers (i, j) such that r_1 ≤ i ≤ r_2 and c_1 ≤ j ≤ c_2, and compute this value modulo (10^9+7).
Constraints
* 1 ≤ r_1 ≤ r_2 ≤ 10^6
* 1 ≤ c_1 ≤ c_2 ≤ 10^6
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
r_1 c_1 r_2 c_2
Output
Print the sum of f(i, j) modulo (10^9+7).
Examples
Input
1 1 2 2
Output
14
Input
314 159 2653 589
Output
602215194
|
instruction
| 0
| 3,108
| 16
| 6,216
|
"Correct Solution:
```
r1,c1,r2,c2=map(int,input().split())
MOD=10**9+7
idx=r2+c2+3
perm=[1]*(idx+1)
for i in range(1,idx+1):
perm[i]=perm[i-1]*i
perm[i]%=MOD
def inv_mod(a):
return pow(a,MOD-2,MOD)
def comb(n,m,p=10**9+7):
if n < m : return 0
if n < 0 or m < 0:return 0
m = min(m, n-m)
top = bot = 1
for i in range(m):
top = top*(n-i) % p
bot = bot*(i+1) % p
bot = pow(bot, p-2, p)
return top*bot % p
def g(i,j):
return comb(i+j+2,i+1,MOD)
def s(n,m):
return comb(n+m+2, m+1, MOD)
res=g(r2,c2)-g(r2,c1-1)-g(r1-1,c2)+g(r1-1,c1-1)
res%=MOD
ans = s(r2,c2) - s(r2, c1-1) - s(r1-1, c2) + s(r1-1, c1-1)
ans %= MOD
print(res)
```
|
output
| 1
| 3,108
| 16
| 6,217
|
Provide a correct Python 3 solution for this coding contest problem.
Snuke is standing on a two-dimensional plane. In one operation, he can move by 1 in the positive x-direction, or move by 1 in the positive y-direction.
Let us define a function f(r, c) as follows:
* f(r,c) := (The number of paths from the point (0, 0) to the point (r, c) that Snuke can trace by repeating the operation above)
Given are integers r_1, r_2, c_1, and c_2. Find the sum of f(i, j) over all pair of integers (i, j) such that r_1 ≤ i ≤ r_2 and c_1 ≤ j ≤ c_2, and compute this value modulo (10^9+7).
Constraints
* 1 ≤ r_1 ≤ r_2 ≤ 10^6
* 1 ≤ c_1 ≤ c_2 ≤ 10^6
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
r_1 c_1 r_2 c_2
Output
Print the sum of f(i, j) modulo (10^9+7).
Examples
Input
1 1 2 2
Output
14
Input
314 159 2653 589
Output
602215194
|
instruction
| 0
| 3,109
| 16
| 6,218
|
"Correct Solution:
```
def main():
M=10**9+7
a,b,c,d=map(int,input().split())
n=c+d+2
fac=[0]*(n+1)
fac[0]=lt=1
for i in range(1,n+1):fac[i]=lt=lt*i%M
inv=lambda n:pow(fac[n],M-2,M)
f=lambda r,c:fac[r+c+2]*inv(c+1)*inv(r+1)-c-r-2
print((f(c,d)-f(c,b-1)-f(a-1,d)+f(a-1,b-1))%M)
main()
```
|
output
| 1
| 3,109
| 16
| 6,219
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke is standing on a two-dimensional plane. In one operation, he can move by 1 in the positive x-direction, or move by 1 in the positive y-direction.
Let us define a function f(r, c) as follows:
* f(r,c) := (The number of paths from the point (0, 0) to the point (r, c) that Snuke can trace by repeating the operation above)
Given are integers r_1, r_2, c_1, and c_2. Find the sum of f(i, j) over all pair of integers (i, j) such that r_1 ≤ i ≤ r_2 and c_1 ≤ j ≤ c_2, and compute this value modulo (10^9+7).
Constraints
* 1 ≤ r_1 ≤ r_2 ≤ 10^6
* 1 ≤ c_1 ≤ c_2 ≤ 10^6
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
r_1 c_1 r_2 c_2
Output
Print the sum of f(i, j) modulo (10^9+7).
Examples
Input
1 1 2 2
Output
14
Input
314 159 2653 589
Output
602215194
Submitted Solution:
```
def main():
M=10**9+7
r1,c1,r2,c2=map(int,input().split())
n=r2+c2+2
val=1
fac=[val]
append=fac.append
for i in range(1,n+1):
val=val*i%M
append(val)
pr1=pow(fac[r1],M-2,M)
pc1=pow(fac[c1],M-2,M)
pr2=pow(fac[r2+1],M-2,M)
pc2=pow(fac[c2+1],M-2,M)
a=fac[r2+c2+2]*pr2*pc2
a-=fac[r2+c1+1]*pr2*pc1
a-=fac[r1+c2+1]*pr1*pc2
a+=fac[r1+c1]*pr1*pc1
print(a%M)
main()
```
|
instruction
| 0
| 3,110
| 16
| 6,220
|
Yes
|
output
| 1
| 3,110
| 16
| 6,221
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke is standing on a two-dimensional plane. In one operation, he can move by 1 in the positive x-direction, or move by 1 in the positive y-direction.
Let us define a function f(r, c) as follows:
* f(r,c) := (The number of paths from the point (0, 0) to the point (r, c) that Snuke can trace by repeating the operation above)
Given are integers r_1, r_2, c_1, and c_2. Find the sum of f(i, j) over all pair of integers (i, j) such that r_1 ≤ i ≤ r_2 and c_1 ≤ j ≤ c_2, and compute this value modulo (10^9+7).
Constraints
* 1 ≤ r_1 ≤ r_2 ≤ 10^6
* 1 ≤ c_1 ≤ c_2 ≤ 10^6
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
r_1 c_1 r_2 c_2
Output
Print the sum of f(i, j) modulo (10^9+7).
Examples
Input
1 1 2 2
Output
14
Input
314 159 2653 589
Output
602215194
Submitted Solution:
```
m=10**9+7
def f(r,c):
s=t=1
for i in range(c):s=s*(r+c-i)%m;t=t*-~i%m
return s*pow(t,m-2,m)%m
a,b,c,d=map(int,input().split())
print((f(a,b)-f(c+1,b)-f(a,d+1)+f(c+1,d+1))%m)
```
|
instruction
| 0
| 3,111
| 16
| 6,222
|
Yes
|
output
| 1
| 3,111
| 16
| 6,223
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke is standing on a two-dimensional plane. In one operation, he can move by 1 in the positive x-direction, or move by 1 in the positive y-direction.
Let us define a function f(r, c) as follows:
* f(r,c) := (The number of paths from the point (0, 0) to the point (r, c) that Snuke can trace by repeating the operation above)
Given are integers r_1, r_2, c_1, and c_2. Find the sum of f(i, j) over all pair of integers (i, j) such that r_1 ≤ i ≤ r_2 and c_1 ≤ j ≤ c_2, and compute this value modulo (10^9+7).
Constraints
* 1 ≤ r_1 ≤ r_2 ≤ 10^6
* 1 ≤ c_1 ≤ c_2 ≤ 10^6
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
r_1 c_1 r_2 c_2
Output
Print the sum of f(i, j) modulo (10^9+7).
Examples
Input
1 1 2 2
Output
14
Input
314 159 2653 589
Output
602215194
Submitted Solution:
```
r1,c1,r2,c2 = map(int,input().split())
MOD = 10**9+7
MAXN = r2+c2+5
fac = [1,1] + [0]*MAXN
for i in range(2,MAXN+2):
fac[i] = fac[i-1] * i % MOD
def inv(n):
return pow(n,MOD-2,MOD)
def comb(n,r):
if n < r: return 0
if n < 0 or r < 0: return 0
return fac[n] * inv(fac[r]) * inv(fac[n-r]) % MOD
ans = comb(r2+c2+2,r2+1) - comb(r2+c1+1,c1) - comb(r1+c2+1,r1) + comb(r1+c1,r1)
ans %= MOD
print(ans)
```
|
instruction
| 0
| 3,112
| 16
| 6,224
|
Yes
|
output
| 1
| 3,112
| 16
| 6,225
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke is standing on a two-dimensional plane. In one operation, he can move by 1 in the positive x-direction, or move by 1 in the positive y-direction.
Let us define a function f(r, c) as follows:
* f(r,c) := (The number of paths from the point (0, 0) to the point (r, c) that Snuke can trace by repeating the operation above)
Given are integers r_1, r_2, c_1, and c_2. Find the sum of f(i, j) over all pair of integers (i, j) such that r_1 ≤ i ≤ r_2 and c_1 ≤ j ≤ c_2, and compute this value modulo (10^9+7).
Constraints
* 1 ≤ r_1 ≤ r_2 ≤ 10^6
* 1 ≤ c_1 ≤ c_2 ≤ 10^6
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
r_1 c_1 r_2 c_2
Output
Print the sum of f(i, j) modulo (10^9+7).
Examples
Input
1 1 2 2
Output
14
Input
314 159 2653 589
Output
602215194
Submitted Solution:
```
r1, c1, r2, c2 = map(int, input().split())
M = r2+c2+10#1000010
mod = 10**9+7
f = [1]*M
p = 1
for i in range(1, M):
p = p*i%mod
f[i] = p
#finv[i] = pow(p, mod-2, mod)
finv = [1]*M
p = finv[-1] = pow(f[-1], mod-2, mod)
for i in range(M-2, 0, -1):
p = p*(i+1)%mod
finv[i] = p
g = [0]*M
p = 0
for i in range(1, M):
g[i] = p = ((p+1)*2-1)%mod
def comb(n, k):
return f[n] * finv[n-k] * finv[k] % mod
def count(R,C):
ret = g[R+C+1] % mod
for c in range(C):
x1 = comb(R+c, c)
x2 = g[C-c]
ret = (ret-x1*x2) % mod
for r in range(R):
x1 = comb(C+r, r)
x2 = g[R-r]
ret = (ret-x1*x2) % mod
return ret
ans = count(r2, c2) + count(r1-1, c1-1) - count(r2, c1-1) - count(r1-1, c2)
ans %= mod
print(ans)
```
|
instruction
| 0
| 3,113
| 16
| 6,226
|
Yes
|
output
| 1
| 3,113
| 16
| 6,227
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke is standing on a two-dimensional plane. In one operation, he can move by 1 in the positive x-direction, or move by 1 in the positive y-direction.
Let us define a function f(r, c) as follows:
* f(r,c) := (The number of paths from the point (0, 0) to the point (r, c) that Snuke can trace by repeating the operation above)
Given are integers r_1, r_2, c_1, and c_2. Find the sum of f(i, j) over all pair of integers (i, j) such that r_1 ≤ i ≤ r_2 and c_1 ≤ j ≤ c_2, and compute this value modulo (10^9+7).
Constraints
* 1 ≤ r_1 ≤ r_2 ≤ 10^6
* 1 ≤ c_1 ≤ c_2 ≤ 10^6
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
r_1 c_1 r_2 c_2
Output
Print the sum of f(i, j) modulo (10^9+7).
Examples
Input
1 1 2 2
Output
14
Input
314 159 2653 589
Output
602215194
Submitted Solution:
```
MOD = 1000000007
r1, c1, r2, c2 = map(int, input().split())
dp = [[1 for _ in range(r2+1)] for _ in range(c2+1)]
for j in range(c2):
for i in range(r2):
dp[j+1][i+1] = dp[j+1][i] + dp[j][i+1]
ans = 0
for j in range(c1, c2+1):
ans += sum(dp[j][r1:r2+1]) % MOD
ans = ans % MOD
print(ans)
```
|
instruction
| 0
| 3,114
| 16
| 6,228
|
No
|
output
| 1
| 3,114
| 16
| 6,229
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke is standing on a two-dimensional plane. In one operation, he can move by 1 in the positive x-direction, or move by 1 in the positive y-direction.
Let us define a function f(r, c) as follows:
* f(r,c) := (The number of paths from the point (0, 0) to the point (r, c) that Snuke can trace by repeating the operation above)
Given are integers r_1, r_2, c_1, and c_2. Find the sum of f(i, j) over all pair of integers (i, j) such that r_1 ≤ i ≤ r_2 and c_1 ≤ j ≤ c_2, and compute this value modulo (10^9+7).
Constraints
* 1 ≤ r_1 ≤ r_2 ≤ 10^6
* 1 ≤ c_1 ≤ c_2 ≤ 10^6
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
r_1 c_1 r_2 c_2
Output
Print the sum of f(i, j) modulo (10^9+7).
Examples
Input
1 1 2 2
Output
14
Input
314 159 2653 589
Output
602215194
Submitted Solution:
```
import sys
import socket
hostnames = ['N551J', 'F551C', 'X553M']
input_file = 'f1.in'
if socket.gethostname() in hostnames:
sys.stdin = open(input_file)
def read_int_list():
return list(map(int, input().split()))
def read_int():
return int(input())
def read_str_list():
return input().split()
def read_str():
return input()
M = 10 ** 9 + 7
MAX = 10 ** 6 + 2
inv = [0] * MAX
inv[1] = 1
for i in range(2, MAX):
inv[i] = - inv[M % i] * (M // i)
inv[i] %= M
def f(r, c):
res = 1
for k in range(c + 1):
res *= r + c + 2 - k
res %= M
res *= inv[k + 1]
res %= M
res -= 1
res %= M
return res
N = 3
b = [[1] * N for _ in range(N)]
for i in range(1, N):
for j in range(1, N):
b[i][j] = (b[i - 1][j - 1] + b[i - 1][j]) % M
def f_exact_but_slow(r, c):
res = 0
for i in range(r + 1):
for j in range(c + 1):
res += b[i][j]
res %= M
return res
def solve():
r1, c1, r2, c2 = read_int_list()
res = f(r2, c2)
res -= f(r1 - 1, c2)
res -= f(r2, c1 - 1)
res += f(r1 - 1, c1 - 1)
res %= M
return res
def main():
res = solve()
print(res)
def find_mismatch():
for r in range(20):
for c in range(20):
expected = f_exact_but_slow(r, c)
output = f(r, c)
if output == expected:
continue
print('r, c, expected, output:', end=' ')
print(r, c, expected, output)
if __name__ == '__main__':
main()
# find_mismatch()
```
|
instruction
| 0
| 3,115
| 16
| 6,230
|
No
|
output
| 1
| 3,115
| 16
| 6,231
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke is standing on a two-dimensional plane. In one operation, he can move by 1 in the positive x-direction, or move by 1 in the positive y-direction.
Let us define a function f(r, c) as follows:
* f(r,c) := (The number of paths from the point (0, 0) to the point (r, c) that Snuke can trace by repeating the operation above)
Given are integers r_1, r_2, c_1, and c_2. Find the sum of f(i, j) over all pair of integers (i, j) such that r_1 ≤ i ≤ r_2 and c_1 ≤ j ≤ c_2, and compute this value modulo (10^9+7).
Constraints
* 1 ≤ r_1 ≤ r_2 ≤ 10^6
* 1 ≤ c_1 ≤ c_2 ≤ 10^6
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
r_1 c_1 r_2 c_2
Output
Print the sum of f(i, j) modulo (10^9+7).
Examples
Input
1 1 2 2
Output
14
Input
314 159 2653 589
Output
602215194
Submitted Solution:
```
import math
r1,c1,r2,c2=map(int,input().split())
div=10**9+7
def g(r,c):
sum=0
for i in range(c+1):
sum+= f(r,i)
for j in range(r+1):
sum+= f(j,c)
return int(sum)%div
def cmb(n, r, p):
if (r < 0) or (n < r):
return 0
r = min(r, n - r)
return fact[n] * factinv[r] * factinv[n-r] % p
p = 10 ** 9 + 7
N = 10 ** 6 # N は必要分だけ用意する
fact = [1, 1]
factinv = [1, 1]
inv = [0, 1]
for i in range(2, N+100):
fact.append((fact[-1] * i) % p)
inv.append((-inv[p % i] * (p // i)) % p)
factinv.append((factinv[-1] * inv[-1]) % p)
def f(r,c):
return cmb(r+c,c,p)
#def f(r,c):
# RC=math.factorial(r+c)
# R=math.factorial(r)
# C=math.factorial(c)
# ans=int(RC//(R*C))%div
# return ans
print(g(r2,c2)-g(r1-1,c2)-g(r2,c1-1)+g(r1-1,c1-1))
```
|
instruction
| 0
| 3,116
| 16
| 6,232
|
No
|
output
| 1
| 3,116
| 16
| 6,233
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke is standing on a two-dimensional plane. In one operation, he can move by 1 in the positive x-direction, or move by 1 in the positive y-direction.
Let us define a function f(r, c) as follows:
* f(r,c) := (The number of paths from the point (0, 0) to the point (r, c) that Snuke can trace by repeating the operation above)
Given are integers r_1, r_2, c_1, and c_2. Find the sum of f(i, j) over all pair of integers (i, j) such that r_1 ≤ i ≤ r_2 and c_1 ≤ j ≤ c_2, and compute this value modulo (10^9+7).
Constraints
* 1 ≤ r_1 ≤ r_2 ≤ 10^6
* 1 ≤ c_1 ≤ c_2 ≤ 10^6
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
r_1 c_1 r_2 c_2
Output
Print the sum of f(i, j) modulo (10^9+7).
Examples
Input
1 1 2 2
Output
14
Input
314 159 2653 589
Output
602215194
Submitted Solution:
```
import sys
input = sys.stdin.buffer.readline
def main():
a,b,c,d = map(int,input().split())
MOD = 10**9+7
N = c+d+5
fac = [0 for _ in range(N+1)]
fac[0],fac[1] = 1,1
for i in range(2,N+1):
fac[i] = (fac[i-1]*i)%MOD
def comb(x,y):
return (fac[x+y]*pow(fac[x],MOD-2,MOD)*pow(fac[y],MOD-2,MOD))%MOD
print(comb(c+1,d+1)-comb(c+1,b)-comb(d+1,a)+comb(a,b))
if __name__ == "__main__":
main()
```
|
instruction
| 0
| 3,117
| 16
| 6,234
|
No
|
output
| 1
| 3,117
| 16
| 6,235
|
Provide a correct Python 3 solution for this coding contest problem.
Snuke has decided to play with N cards and a deque (that is, a double-ended queue). Each card shows an integer from 1 through N, and the deque is initially empty.
Snuke will insert the cards at the beginning or the end of the deque one at a time, in order from 1 to N. Then, he will perform the following action N times: take out the card from the beginning or the end of the deque and eat it.
Afterwards, we will construct an integer sequence by arranging the integers written on the eaten cards, in the order they are eaten. Among the sequences that can be obtained in this way, find the number of the sequences such that the K-th element is 1. Print the answer modulo 10^{9} + 7.
Constraints
* 1 ≦ K ≦ N ≦ 2{,}000
Input
The input is given from Standard Input in the following format:
N K
Output
Print the answer modulo 10^{9} + 7.
Examples
Input
2 1
Output
1
Input
17 2
Output
262144
Input
2000 1000
Output
674286644
|
instruction
| 0
| 3,205
| 16
| 6,410
|
"Correct Solution:
```
def cmb(n, r, mod):#コンビネーションの高速計算
if ( r<0 or r>n ):
return 0
r = min(r, n-r)
return g1[n] * g2[r] * g2[n-r] % mod
mod = 10**9+7 #出力の制限
N = 5000
g1 = [1, 1] # 元テーブル
g2 = [1, 1] #逆元テーブル
inverse = [0, 1] #逆元テーブル計算用テーブル
for i in range( 2, N + 1 ):
g1.append( ( g1[-1] * i ) % mod )
inverse.append( ( -inverse[mod % i] * (mod//i) ) % mod )
g2.append( (g2[-1] * inverse[-1]) % mod )
N,K=map(int,input().split())
if N==1:
print(1)
elif K==1:
print(pow(2,N-2,mod))
elif K==N:
dp=[[0 for i in range(K+1)] for j in range(K)]
imos=[0]*(K+1)
dp[0][K]=1
imos[K]=1
for i in range(1,K):
for j in range(K-i,K+1):
if j==K-i:
dp[i][j]=(imos[K]-imos[j])%mod
else:
dp[i][j]=(dp[i-1][j]+imos[K]-imos[j])%mod
imos=[dp[i][j] for j in range(K+1)]
for j in range(1,K+1):
imos[j]+=imos[j-1]
imos[j]%=mod
print(dp[N-1][1])
else:
dp=[[0 for i in range(K+1)] for j in range(K)]
imos=[0]*(K+1)
dp[0][K]=1
imos[K]=1
for i in range(1,K):
for j in range(K-i,K+1):
if j==K-i:
dp[i][j]=(imos[K]-imos[j])%mod
else:
dp[i][j]=(dp[i-1][j]+imos[K]-imos[j])%mod
imos=[dp[i][j] for j in range(K+1)]
for j in range(1,K+1):
imos[j]+=imos[j-1]
imos[j]%=mod
ans=0
for M in range(N-K+1,N+1):
id=M-N+K
ans+=dp[K-1][id]*cmb(M-2,N-K-1,mod)
ans*=pow(2,N-K-1,mod)
print(ans%mod)
```
|
output
| 1
| 3,205
| 16
| 6,411
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Snuke has decided to play with N cards and a deque (that is, a double-ended queue). Each card shows an integer from 1 through N, and the deque is initially empty.
Snuke will insert the cards at the beginning or the end of the deque one at a time, in order from 1 to N. Then, he will perform the following action N times: take out the card from the beginning or the end of the deque and eat it.
Afterwards, we will construct an integer sequence by arranging the integers written on the eaten cards, in the order they are eaten. Among the sequences that can be obtained in this way, find the number of the sequences such that the K-th element is 1. Print the answer modulo 10^{9} + 7.
Constraints
* 1 ≦ K ≦ N ≦ 2{,}000
Input
The input is given from Standard Input in the following format:
N K
Output
Print the answer modulo 10^{9} + 7.
Examples
Input
2 1
Output
1
Input
17 2
Output
262144
Input
2000 1000
Output
674286644
Submitted Solution:
```
import sys
import numpy as np
import numba
from numba import njit, b1, i4, i8
from numba.types import Omitted
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
MOD = 1_000_000_007
@njit((i8, i8), cache=True)
def main(N, K):
K -= 1
U = 5000
C = np.zeros((U, U), np.int64)
C[0, 0] = 1
for n in range(1, U):
C[n] += C[n - 1]
C[n, 1:] += C[n - 1, :-1]
C[n] %= MOD
dp = np.zeros((N + 10, N + 10), np.int64)
dp[0, 0] = 1
for n in range(1, N + 10):
dp[n, n] = (dp[n - 1, n - 1] + dp[n - 1, n]) % MOD
for m in range(n + 1, N + 10):
dp[n, m] = (dp[n - 1, m] + dp[n, m - 1]) % MOD
for n in range(N + 9, 0, -1):
dp[n] -= dp[n - 1]
dp[n] %= MOD
ans = 0
if K == N - 1:
return dp[:N, N - 1].sum() % MOD
for r in range(N - K, N + 1):
x = C[r - 2, N - K - 2]
a = N - r
b = K - a
if b == 0:
x = x * (dp[:a + 1, a].sum() % MOD) % MOD
else:
x *= (dp[1:a + 2, a + 1] * C[b - 1:a + b, b - 1][::-1] %
MOD).sum() % MOD
x %= MOD
ans += x
ans %= MOD
for i in range(N - K - 2):
ans = ans * 2 % MOD
return ans % MOD
N, K = map(int, read().split())
print(main(N, K))
```
|
instruction
| 0
| 3,206
| 16
| 6,412
|
No
|
output
| 1
| 3,206
| 16
| 6,413
|
Provide a correct Python 3 solution for this coding contest problem.
One day, Snuke was given a permutation of length N, a_1, a_2, ..., a_N, from his friend.
Find the following:
<image>
Constraints
* 1 ≦ N ≦ 200,000
* (a_1, a_2, ..., a_N) is a permutation of (1, 2, ..., N).
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the answer.
Note that the answer may not fit into a 32-bit integer.
Examples
Input
3
2 1 3
Output
9
Input
4
1 3 2 4
Output
19
Input
8
5 4 8 1 2 6 7 3
Output
85
|
instruction
| 0
| 3,207
| 16
| 6,414
|
"Correct Solution:
```
class BIT:
"0-indexed apparently"
def __init__(self, n):
self.n = n
self.data = [0] * (n+1)
def point_add(self, index, value):
index += 1
while index <= self.n:
self.data[index] += value
index += index & -index
def folded(self, r):
"[0, r)"
ret = 0
while r > 0:
ret += self.data[r]
r -= r & -r
return ret
def bisect_left(self, value):
index = 0
k = 1 << ((self.n + 1).bit_length() - 1)
while k > 0:
if index + k <= self.n and self.data[index + k] < value:
value -= self.data[index + k]
index += k
k >>= 1
return index + 1
N = int(input())
As = list(map(int, input().split()))
A_inv = [0] * N
for i, A in enumerate(As):
A_inv[A-1] = i
bit = BIT(N + 2)
bit.point_add(0, 1)
bit.point_add(N+1, 1)
ans = 0
for A, index in enumerate(A_inv):
A += 1
index += 1
v = bit.folded(index)
ans += A * (bit.bisect_left(v+1) - index - 1) * (index - bit.bisect_left(v) + 1)
#print(A, ans)
bit.point_add(index, 1)
print(ans)
```
|
output
| 1
| 3,207
| 16
| 6,415
|
Provide a correct Python 3 solution for this coding contest problem.
One day, Snuke was given a permutation of length N, a_1, a_2, ..., a_N, from his friend.
Find the following:
<image>
Constraints
* 1 ≦ N ≦ 200,000
* (a_1, a_2, ..., a_N) is a permutation of (1, 2, ..., N).
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the answer.
Note that the answer may not fit into a 32-bit integer.
Examples
Input
3
2 1 3
Output
9
Input
4
1 3 2 4
Output
19
Input
8
5 4 8 1 2 6 7 3
Output
85
|
instruction
| 0
| 3,208
| 16
| 6,416
|
"Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
l = []
r = []
s = []
for i, x in enumerate(a):
left = i
while len(s) > 0 and x < s[-1][0]:
left = s[-1][1]
s.pop()
l.append(left)
s.append([x, left])
s = []
for i in range(n-1, -1, -1):
right = i
x = a[i]
while len(s) > 0 and x < s[-1][0]:
right = s[-1][1]
s.pop()
r.append(right)
s.append([x, right])
r = r[::-1]
ans = 0
for i, (x, left, right) in enumerate(zip(a,l,r)):
ans += x * (right+1-i) * (i+1-left)
print(ans)
```
|
output
| 1
| 3,208
| 16
| 6,417
|
Provide a correct Python 3 solution for this coding contest problem.
One day, Snuke was given a permutation of length N, a_1, a_2, ..., a_N, from his friend.
Find the following:
<image>
Constraints
* 1 ≦ N ≦ 200,000
* (a_1, a_2, ..., a_N) is a permutation of (1, 2, ..., N).
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the answer.
Note that the answer may not fit into a 32-bit integer.
Examples
Input
3
2 1 3
Output
9
Input
4
1 3 2 4
Output
19
Input
8
5 4 8 1 2 6 7 3
Output
85
|
instruction
| 0
| 3,209
| 16
| 6,418
|
"Correct Solution:
```
from bisect import*
class BTreeNode:
def __init__(self):self.key,self.child=[],[]
class BTree:
def __init__(self):self.root=BTreeNode()
def search_higher(self,key):
ptr=self.root
ret=None
while ptr.child:
i=bisect(ptr.key,key)
if i!=len(ptr.key):ret=ptr.key[i]
ptr=ptr.child[i]
i=bisect(ptr.key,key)
if i!=len(ptr.key):ret=ptr.key[i]
return ret
def search_lower(self,key):
ptr=self.root
ret=None
while ptr.child:
i=bisect_left(ptr.key,key)
if i:ret=ptr.key[i-1]
ptr=ptr.child[i]
i=bisect_left(ptr.key,key)
if i:ret=ptr.key[i-1]
return ret
def insert(self,key):
def insert_rec(ptr,key):
b_size=512
if not ptr.child:
insort(ptr.key,key)
if len(ptr.key)==b_size*2-1:
ret=BTreeNode()
ret.key=ptr.key[:b_size]
ptr.key=ptr.key[b_size:]
return ret
else:
i=bisect(ptr.key,key)
tmp=insert_rec(ptr.child[i],key)
if tmp:
ptr.key.insert(i,tmp.key.pop())
ptr.child.insert(i,tmp)
if len(ptr.child)==b_size*2:
ret=BTreeNode()
ret.child=ptr.child[:b_size]
ptr.child=ptr.child[b_size:]
ret.key=ptr.key[:b_size]
ptr.key=ptr.key[b_size:]
return ret
tmp=insert_rec(self.root,key)
if tmp:
root=BTreeNode()
root.key=[tmp.key.pop()]
root.child=[tmp,self.root]
self.root=root
def main():
n,*a=map(int,open(0).read().split())
l=[0]*n
for i,v in enumerate(a,1):l[v-1]=i
t=BTree()
t.insert(0)
t.insert(n+1)
c=0
for i,v in enumerate(l,1):
c+=(t.search_higher(v)-v)*(v-t.search_lower(v))*i
t.insert(v)
print(c)
main()
```
|
output
| 1
| 3,209
| 16
| 6,419
|
Provide a correct Python 3 solution for this coding contest problem.
One day, Snuke was given a permutation of length N, a_1, a_2, ..., a_N, from his friend.
Find the following:
<image>
Constraints
* 1 ≦ N ≦ 200,000
* (a_1, a_2, ..., a_N) is a permutation of (1, 2, ..., N).
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the answer.
Note that the answer may not fit into a 32-bit integer.
Examples
Input
3
2 1 3
Output
9
Input
4
1 3 2 4
Output
19
Input
8
5 4 8 1 2 6 7 3
Output
85
|
instruction
| 0
| 3,210
| 16
| 6,420
|
"Correct Solution:
```
N = int(input())
A = [int(x) for x in input().split()]
B = [0] * (N + 1)
for i in range(N):
B[A[i]] = i + 1
L = list(range(N + 2))
R = list(range(N + 2))
cnt = 0
for i in range(N, 0, -1):
l = L[B[i]]
r = R[B[i]]
cnt += i * (B[i] - l + 1) * (r - B[i] + 1)
L[r + 1] = l
R[l - 1] = r
print(cnt)
```
|
output
| 1
| 3,210
| 16
| 6,421
|
Provide a correct Python 3 solution for this coding contest problem.
One day, Snuke was given a permutation of length N, a_1, a_2, ..., a_N, from his friend.
Find the following:
<image>
Constraints
* 1 ≦ N ≦ 200,000
* (a_1, a_2, ..., a_N) is a permutation of (1, 2, ..., N).
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the answer.
Note that the answer may not fit into a 32-bit integer.
Examples
Input
3
2 1 3
Output
9
Input
4
1 3 2 4
Output
19
Input
8
5 4 8 1 2 6 7 3
Output
85
|
instruction
| 0
| 3,211
| 16
| 6,422
|
"Correct Solution:
```
N = int(input())
A = list(map(int, input().split()))
idx = {A[i]: i for i in range(N)}
ans = 0
idx_l, idx_r = list(range(N + 2)), list(range(N + 2))
for a in range(N, 0, -1):
i = idx[a]
l, r = idx_l[i], idx_r[i]
ans += a * (i - l + 1) * (r - i + 1)
idx_l[r + 1], idx_r[l - 1] = l, r
print(ans)
```
|
output
| 1
| 3,211
| 16
| 6,423
|
Provide a correct Python 3 solution for this coding contest problem.
One day, Snuke was given a permutation of length N, a_1, a_2, ..., a_N, from his friend.
Find the following:
<image>
Constraints
* 1 ≦ N ≦ 200,000
* (a_1, a_2, ..., a_N) is a permutation of (1, 2, ..., N).
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the answer.
Note that the answer may not fit into a 32-bit integer.
Examples
Input
3
2 1 3
Output
9
Input
4
1 3 2 4
Output
19
Input
8
5 4 8 1 2 6 7 3
Output
85
|
instruction
| 0
| 3,212
| 16
| 6,424
|
"Correct Solution:
```
N = int(input())
a = list(map(int, input().split()))
left = [0] * (N + 1)
right = [0] * (N + 1)
# 左から
buf = list()
for i in range(N):
while len(buf) > 0 and a[buf[-1]] > a[i]:
buf.pop()
if len(buf) > 0:
left[i] = buf[-1]
else:
left[i] = -1
buf.append(i)
# 右から
buf = list()
for i in reversed(range(N)):
while len(buf) > 0 and a[buf[-1]] > a[i]:
buf.pop()
if len(buf) > 0:
right[i] = buf[-1]
else:
right[i] = N
buf.append(i)
ans = 0
for i in range(N):
ans += a[i] * (i - left[i]) * (right[i] - i)
print(ans)
```
|
output
| 1
| 3,212
| 16
| 6,425
|
Provide a correct Python 3 solution for this coding contest problem.
One day, Snuke was given a permutation of length N, a_1, a_2, ..., a_N, from his friend.
Find the following:
<image>
Constraints
* 1 ≦ N ≦ 200,000
* (a_1, a_2, ..., a_N) is a permutation of (1, 2, ..., N).
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the answer.
Note that the answer may not fit into a 32-bit integer.
Examples
Input
3
2 1 3
Output
9
Input
4
1 3 2 4
Output
19
Input
8
5 4 8 1 2 6 7 3
Output
85
|
instruction
| 0
| 3,213
| 16
| 6,426
|
"Correct Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
"""
・大きい数から挿入していく
・連結成分の両端を管理する
"""
N,*A = map(int,read().split())
A = [0] + A + [0]
ind = [0] * (N+1)
for i,x in enumerate(A):
ind[x] = i
left = list(range(1,len(A)+1))
right = list(range(-1,len(A)-1))
answer = 0
for i in ind[:0:-1]:
# i番目に数を挿入
l = left[i-1]; r = right[i+1]
left[r] = l; right[l] = r
x = i - l + 1; y = r - i + 1
answer += A[i] * (x * y)
print(answer)
```
|
output
| 1
| 3,213
| 16
| 6,427
|
Provide a correct Python 3 solution for this coding contest problem.
One day, Snuke was given a permutation of length N, a_1, a_2, ..., a_N, from his friend.
Find the following:
<image>
Constraints
* 1 ≦ N ≦ 200,000
* (a_1, a_2, ..., a_N) is a permutation of (1, 2, ..., N).
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the answer.
Note that the answer may not fit into a 32-bit integer.
Examples
Input
3
2 1 3
Output
9
Input
4
1 3 2 4
Output
19
Input
8
5 4 8 1 2 6 7 3
Output
85
|
instruction
| 0
| 3,214
| 16
| 6,428
|
"Correct Solution:
```
import sys
class Node:
def __init__(self, key, height):
self.key = key #ノードの木
self.height = height #このノードを根とする部分木の高さ
self.left = None
self.right = None
def size(self, n): return 0 if n is None else n.height
def bias(self): #左の方が高いと正、右が高いと負の値を返す
return self.size(self.left) - self.size(self.right)
#木の高さの計算
def calcSize(self):
self.height = 1 + max(self.size(self.left), self.size(self.right))
class AVLTree:
def __init__(self):
self.root = None #根
self.change = False #修正フラグ
self.lmax = None #左部分木のキーの最大値
###############
#回転操作, 修正操作
###############
def rotateL(self, n): #ノードnの左回転
r = n.right; rl = r.left
r.left = n; n.right = rl
r.left.calcSize()
r.calcSize()
return r
def rotateR(self, n):
l = n.left; lr = l.right
l.right = n; n.left = lr
l.right.calcSize()
l.calcSize()
return l
def rotateLR(self, n): #二重回転;左回転→右回転
n.left = self.rotateL(n.left)
return self.rotateR(n)
def rotateRL(self, n):
n.right = self.rotateR(n.right)
return self.rotateL(n)
def balanceL(self, n):
if not self.change: return n
h = n.height
if n.bias() == 2:
if n.left.bias() >= 0: n = self.rotateR(n)
else: n = self.rotateLR(n)
else: n.calcSize()
self.change = (h != n.height)
return n
def balanceR(self, n):
if not self.change: return n
h = n.height
if n.bias() == -2:
if n.right.bias() <= 0: n = self.rotateL(n)
else: n = self.rotateRL(n)
else: n.calcSize()
self.change = (h != n.height)
return n
###############
#Nodeの追加
###############
def insert(self, key): self.root = self.insert_sub(self.root, key)
def insert_sub(self, t, key): #新たなノードの挿入。初期位置は根。
if t is None:
self.change = True
return Node(key, 1)
if key < t.key:
t.left = self.insert_sub(t.left, key)
return self.balanceL(t)
elif key > t.key:
t.right = self.insert_sub(t.right, key)
return self.balanceR(t)
else:
self.change = False
t.value = value
return t
###############
#Nodeの削除
###############
def delete(self, key): self.root = self.delete_sub(self.root, key)
def delete_sub(self, t, key):
if t is None:
self.change = False
return None
if key < t.key:
t.left = self.delete_sub(t.left, key)
return self.balanceR(t)
elif key > t.key:
t.right - self.delete_sub(t.right, key)
return self.balanceL(t)
else:
if t.left is None:
self.change = True
return t.right
else:
t.left = self.delete_max(t.left)
t.key = self.lmax
t.value = self.value
return self.balanceR(t)
def delete_max(self, n):
if n.right is None: #nの右部分木が存在しない場合は左部分木を昇格させる
self.change = True
self.lmax = n.key
self.value = n.value
return n.left
else:
n.right = self.delete_max(n.right)
return self.balanceL(n)
###############
#Nodeの探索
###############
def search(self, key, lower, higher):
t = self.root
lb, hb = lower, higher
while t:
if key < t.key:
hb = t.key
t = t.left
else:
lb = t.key
t = t.right
return lb, hb
def solve(): #デバッグ用コード
N = int(input())
A = [int(a) for a in input().split()]
Ai = dict()
for i, a in enumerate(A): Ai[a] = i
Tree = AVLTree()
ini = Ai[1]
Ans = 1 * (ini + 1) * (N - ini)
Tree.insert(ini)
for i in range(2, N + 1):
ai = Ai[i]
l, r = Tree.search(ai, -1, N)
Ans += i * (ai - l) * (r - ai)
Tree.insert(ai)
print(Ans)
return 0
if __name__ == "__main__":
solve()
```
|
output
| 1
| 3,214
| 16
| 6,429
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day, Snuke was given a permutation of length N, a_1, a_2, ..., a_N, from his friend.
Find the following:
<image>
Constraints
* 1 ≦ N ≦ 200,000
* (a_1, a_2, ..., a_N) is a permutation of (1, 2, ..., N).
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the answer.
Note that the answer may not fit into a 32-bit integer.
Examples
Input
3
2 1 3
Output
9
Input
4
1 3 2 4
Output
19
Input
8
5 4 8 1 2 6 7 3
Output
85
Submitted Solution:
```
N = int(input())
a = list(map(int, input().split()))
# indexをソート
b = [0] * N
for i in range(N):
b[a[i] - 1] = i
# 初期化 端っこ対策としてひとつ余計に取っている
left = [0] * (N + 1)
right = [0] * (N + 1)
for i in range(N + 1):
left[i] = i-1
right[i] = i+1
# 更新
for idx in reversed(b):
left[right[idx]] = left[idx]
right[left[idx]] = right[idx]
ans = 0
for i in range(N):
ans += a[i] * (i - left[i]) * (right[i] - i)
print(ans)
```
|
instruction
| 0
| 3,215
| 16
| 6,430
|
Yes
|
output
| 1
| 3,215
| 16
| 6,431
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day, Snuke was given a permutation of length N, a_1, a_2, ..., a_N, from his friend.
Find the following:
<image>
Constraints
* 1 ≦ N ≦ 200,000
* (a_1, a_2, ..., a_N) is a permutation of (1, 2, ..., N).
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the answer.
Note that the answer may not fit into a 32-bit integer.
Examples
Input
3
2 1 3
Output
9
Input
4
1 3 2 4
Output
19
Input
8
5 4 8 1 2 6 7 3
Output
85
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
l = [0] * (n + 1)
l2 = list(range(n + 2))
l3 = list(range(n + 2))
ans = 0
for i in range(n):
l[a[i]] = i
for i in range(1, n + 1)[::-1]:
x, y = l2[l[i]], l3[l[i]]
ans += (l[i] - x + 1) * (y - l[i] + 1) * i
l2[y + 1], l3[x - 1] = x, y
print(ans)
```
|
instruction
| 0
| 3,216
| 16
| 6,432
|
Yes
|
output
| 1
| 3,216
| 16
| 6,433
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day, Snuke was given a permutation of length N, a_1, a_2, ..., a_N, from his friend.
Find the following:
<image>
Constraints
* 1 ≦ N ≦ 200,000
* (a_1, a_2, ..., a_N) is a permutation of (1, 2, ..., N).
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the answer.
Note that the answer may not fit into a 32-bit integer.
Examples
Input
3
2 1 3
Output
9
Input
4
1 3 2 4
Output
19
Input
8
5 4 8 1 2 6 7 3
Output
85
Submitted Solution:
```
from bisect import*
class BTreeNode:
def __init__(self):self.key,self.child=[],[]
class BTree:
def __init__(self):self.root=BTreeNode()
def search_higher(self,key):
ptr=self.root
ret=None
while ptr.child:
i=bisect(ptr.key,key)
if i!=len(ptr.key):ret=ptr.key[i]
ptr=ptr.child[i]
i=bisect(ptr.key,key)
if i!=len(ptr.key):ret=ptr.key[i]
return ret
def search_lower(self,key):
ptr=self.root
ret=None
while ptr.child:
i=bisect_left(ptr.key,key)
if i:ret=ptr.key[i-1]
ptr=ptr.child[i]
i=bisect_left(ptr.key,key)
if i:ret=ptr.key[i-1]
return ret
def insert(self,key):
def insert_rec(ptr):
b_size=256
if not ptr.child:
insort(ptr.key,key)
if len(ptr.key)==b_size*2-1:
ret=BTreeNode()
ret.key=ptr.key[:b_size]
ptr.key=ptr.key[b_size:]
return ret
else:
i=bisect(ptr.key,key)
tmp=insert_rec(ptr.child[i])
if tmp:
ptr.key.insert(i,tmp.key.pop())
ptr.child.insert(i,tmp)
if len(ptr.child)==b_size*2:
ret=BTreeNode()
ret.child=ptr.child[:b_size]
ptr.child=ptr.child[b_size:]
ret.key=ptr.key[:b_size]
ptr.key=ptr.key[b_size:]
return ret
tmp=insert_rec(self.root)
if tmp:
root=BTreeNode()
root.key=[tmp.key.pop()]
root.child=[tmp,self.root]
self.root=root
def main():
n,*a=map(int,open(0).read().split())
l=[0]*n
for i,v in enumerate(a,1):l[v-1]=i
t=BTree()
t.insert(0)
t.insert(n+1)
c=0
for i,v in enumerate(l,1):
c+=(t.search_higher(v)-v)*(v-t.search_lower(v))*i
t.insert(v)
print(c)
main()
```
|
instruction
| 0
| 3,217
| 16
| 6,434
|
Yes
|
output
| 1
| 3,217
| 16
| 6,435
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day, Snuke was given a permutation of length N, a_1, a_2, ..., a_N, from his friend.
Find the following:
<image>
Constraints
* 1 ≦ N ≦ 200,000
* (a_1, a_2, ..., a_N) is a permutation of (1, 2, ..., N).
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the answer.
Note that the answer may not fit into a 32-bit integer.
Examples
Input
3
2 1 3
Output
9
Input
4
1 3 2 4
Output
19
Input
8
5 4 8 1 2 6 7 3
Output
85
Submitted Solution:
```
# でつoO(YOU PLAY WITH THE CARDS YOU'RE DEALT..)
import sys
def main(N, A):
ans = 0
Ai = [(a, i + 1) for i, a in enumerate(A)]
Ai.sort(key=lambda x: x[0])
st = SegmentTree(n=N + 2, f=lambda x, y: x + y)
st.set_val(0, 1)
st.set_val(N + 1, 1)
for a, i in Ai:
lc = st.query(0, i)
l = st.bisect_left(lambda x: x < lc)
r = st.bisect_left(lambda x: x < lc + 1)
ans += a * (i - l) * (r - i)
st.set_val(i, 1)
print(ans)
class SegmentTree:
def __init__(self, n=None, f=max, identity_factory=int, initial_values=None):
assert(n or initial_values)
size = n if n else len(initial_values)
d = [identity_factory() for _ in range(2 * size + 1)]
self.__n, self.__d, self.__f, self.__e = size, d, f, identity_factory
if initial_values:
for i, v in enumerate(initial_values): d[size + i] = v
for i in range(size - 1, 0, -1): d[i] = f(d[i << 1], d[i << 1 | 1])
def get_val(self, index):
return self.__d[index + self.__n]
def set_val(self, index, new_value):
i, d, f = index + self.__n, self.__d, self.__f
if d[i] == new_value: return
d[i], i = new_value, i >> 1
while i:
d[i], i = f(d[i << 1], d[i << 1 | 1]), i >> 1
def modify(self, index, value):
self.set_val(index, self.__f(self.__d[index + self.__n], value))
def query(self, from_inclusive, to_exclusive):
ans = self.__e()
if to_exclusive <= from_inclusive: return ans
l, r, d, f = from_inclusive + self.__n, to_exclusive + self.__n, self.__d, self.__f
while l < r:
if l & 1: ans, l = f(ans, d[l]), l + 1
if r & 1: ans, r = f(d[r - 1], ans), r - 1
l, r = l >> 1, r >> 1
return ans
def bisect_left(self, func):
'''func()がFalseになるもっとも左のindexを探す
'''
i, j, n, f, d, v = self.__n, self.__n + self.__n, self.__n, self.__f, self.__d, self.__e()
while i < j:
if i & 1:
nv = f(v, d[i])
if not func(nv): break
v, i = nv, i + 1
i, j = i >> 1, j >> 1
while i < n:
nv = f(v, d[i << 1])
if func(nv): v, i = nv, i << 1 | 1
else: i = i << 1
return i - n
if __name__ == '__main__':
input = sys.stdin.readline
N = int(input())
*A, = map(int, input().split())
main(N, A)
```
|
instruction
| 0
| 3,218
| 16
| 6,436
|
Yes
|
output
| 1
| 3,218
| 16
| 6,437
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day, Snuke was given a permutation of length N, a_1, a_2, ..., a_N, from his friend.
Find the following:
<image>
Constraints
* 1 ≦ N ≦ 200,000
* (a_1, a_2, ..., a_N) is a permutation of (1, 2, ..., N).
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the answer.
Note that the answer may not fit into a 32-bit integer.
Examples
Input
3
2 1 3
Output
9
Input
4
1 3 2 4
Output
19
Input
8
5 4 8 1 2 6 7 3
Output
85
Submitted Solution:
```
# -*- coding: utf-8 -*-
import bisect
import heapq
import math
import random
import sys
from collections import Counter, defaultdict, deque
from decimal import ROUND_CEILING, ROUND_HALF_UP, Decimal
from functools import lru_cache, reduce
from itertools import combinations, combinations_with_replacement, product, permutations
from operator import add, mul, sub
sys.setrecursionlimit(10000)
def read_int():
return int(input())
def read_int_n():
return list(map(int, input().split()))
def read_float():
return float(input())
def read_float_n():
return list(map(float, input().split()))
def read_str():
return input().strip()
def read_str_n():
return list(map(str, input().split()))
def error_print(*args):
print(*args, file=sys.stderr)
def mt(f):
import time
def wrap(*args, **kwargs):
s = time.time()
ret = f(*args, **kwargs)
e = time.time()
error_print(e - s, 'sec')
return ret
return wrap
@mt
def slv(N, A):
def f(A):
n = len(A)
if not n:
return 0
if n == 1:
return A[0]
i = A.index(min(A))
return A[i]*(i+1)*(n-i) + f(A[:i]) + f(A[i+1:])
return f(A)
def main():
N = read_int()
A = read_int_n()
print(slv(N, A))
if __name__ == '__main__':
main()
```
|
instruction
| 0
| 3,219
| 16
| 6,438
|
No
|
output
| 1
| 3,219
| 16
| 6,439
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day, Snuke was given a permutation of length N, a_1, a_2, ..., a_N, from his friend.
Find the following:
<image>
Constraints
* 1 ≦ N ≦ 200,000
* (a_1, a_2, ..., a_N) is a permutation of (1, 2, ..., N).
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the answer.
Note that the answer may not fit into a 32-bit integer.
Examples
Input
3
2 1 3
Output
9
Input
4
1 3 2 4
Output
19
Input
8
5 4 8 1 2 6 7 3
Output
85
Submitted Solution:
```
def inpl(): return [int(i) for i in input().split()]
def minisum(x):
if not x:
return 0
mini = x.index(min(x))
d = len(x)
if d == 1:
return x[0]
result = 0
if mini > 0:
result += minisum(x[:mini])
if mini < d-1:
result += minisum(x[mini+1:])
return result + (d-mini)*(mini+1)*x[mini]
N = input()
a = inpl()
print(minisum(a))
```
|
instruction
| 0
| 3,220
| 16
| 6,440
|
No
|
output
| 1
| 3,220
| 16
| 6,441
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day, Snuke was given a permutation of length N, a_1, a_2, ..., a_N, from his friend.
Find the following:
<image>
Constraints
* 1 ≦ N ≦ 200,000
* (a_1, a_2, ..., a_N) is a permutation of (1, 2, ..., N).
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the answer.
Note that the answer may not fit into a 32-bit integer.
Examples
Input
3
2 1 3
Output
9
Input
4
1 3 2 4
Output
19
Input
8
5 4 8 1 2 6 7 3
Output
85
Submitted Solution:
```
import sys
sys.setrecursionlimit(10**7)
INF = 10 ** 18
MOD = 10 ** 9 + 7
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x) - 1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def II(): return int(sys.stdin.readline())
def SI(): return input()
from collections import Counter
from bisect import bisect_right
def main():
N = II()
A = LI()
idxs = [0] * (N + 1)
for i, a in enumerate(A, 1):
idxs[a] = i
lowers = [0, N + 1]
ans = 0
for a, i in enumerate(idxs[1:], 1):
j = bisect_right(lowers, i)
l, r = lowers[j - 1], lowers[j]
lowers.insert(j, i)
ans += (i - l) * (r - i) * a
return ans
print(main())
```
|
instruction
| 0
| 3,221
| 16
| 6,442
|
No
|
output
| 1
| 3,221
| 16
| 6,443
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day, Snuke was given a permutation of length N, a_1, a_2, ..., a_N, from his friend.
Find the following:
<image>
Constraints
* 1 ≦ N ≦ 200,000
* (a_1, a_2, ..., a_N) is a permutation of (1, 2, ..., N).
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the answer.
Note that the answer may not fit into a 32-bit integer.
Examples
Input
3
2 1 3
Output
9
Input
4
1 3 2 4
Output
19
Input
8
5 4 8 1 2 6 7 3
Output
85
Submitted Solution:
```
from bisect import bisect_left, insort_left
N = int(input())
A = list(map(int, input().split()))
# 各数の出現場所を持ってソート: 1-indexed
A_with_position = sorted([[a, i] for i, a in enumerate(A, start=1)])
# 出現したもののindex
appeared_index = [0, N + 1]
ans = 0
for a, i in A_with_position:
li = bisect_left(appeared_index, i) - 1
ri = bisect_left(appeared_index, i)
lx, rx = appeared_index[li], appeared_index[ri]
ans += a * (i - lx) * (rx - i)
insort_left(appeared_index, i)
print(ans)
```
|
instruction
| 0
| 3,222
| 16
| 6,444
|
No
|
output
| 1
| 3,222
| 16
| 6,445
|
Provide a correct Python 3 solution for this coding contest problem.
There is a sequence of length N: A_1, A_2, ..., A_N. Initially, this sequence is a permutation of 1, 2, ..., N.
On this sequence, Snuke can perform the following operation:
* Choose K consecutive elements in the sequence. Then, replace the value of each chosen element with the minimum value among the chosen elements.
Snuke would like to make all the elements in this sequence equal by repeating the operation above some number of times. Find the minimum number of operations required. It can be proved that, Under the constraints of this problem, this objective is always achievable.
Constraints
* 2 \leq K \leq N \leq 100000
* A_1, A_2, ..., A_N is a permutation of 1, 2, ..., N.
Input
Input is given from Standard Input in the following format:
N K
A_1 A_2 ... A_N
Output
Print the minimum number of operations required.
Examples
Input
4 3
2 3 1 4
Output
2
Input
3 3
1 2 3
Output
1
Input
8 3
7 3 1 8 4 6 2 5
Output
4
|
instruction
| 0
| 5,645
| 16
| 11,290
|
"Correct Solution:
```
import math
n,k=map(int,input().split());input();print(math.ceil((n-1)/(k-1)))
```
|
output
| 1
| 5,645
| 16
| 11,291
|
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