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|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.
Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.
Input
The first line contains integer q (1 β€ q β€ 1000), the number of handle change requests.
Next q lines contain the descriptions of the requests, one per line.
Each query consists of two non-empty strings old and new, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings old and new are distinct. The lengths of the strings do not exceed 20.
The requests are given chronologically. In other words, by the moment of a query there is a single person with handle old, and handle new is not used and has not been used by anyone.
Output
In the first line output the integer n β the number of users that changed their handles at least once.
In the next n lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, old and new, separated by a space, meaning that before the user had handle old, and after all the requests are completed, his handle is new. You may output lines in any order.
Each user who changes the handle must occur exactly once in this description.
Examples
Input
5
Misha ILoveCodeforces
Vasya Petrov
Petrov VasyaPetrov123
ILoveCodeforces MikeMirzayanov
Petya Ivanov
Output
3
Petya Ivanov
Misha MikeMirzayanov
Vasya VasyaPetrov123
Submitted Solution:
```
n = int(input())
p = []
for i in range(n):
x, y = map(str, input().split(' '))
for j in range(len(p)):
if(p[j][1] == x):
p[j][1] = y
else:
p.append([x,y])
if(p == []):
p.append([x,y])
for i in range(len(p)):
print(p[i][0] + ' ' + p[i][1])
```
|
instruction
| 0
| 12,871
| 11
| 25,742
|
No
|
output
| 1
| 12,871
| 11
| 25,743
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.
Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.
Input
The first line contains integer q (1 β€ q β€ 1000), the number of handle change requests.
Next q lines contain the descriptions of the requests, one per line.
Each query consists of two non-empty strings old and new, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings old and new are distinct. The lengths of the strings do not exceed 20.
The requests are given chronologically. In other words, by the moment of a query there is a single person with handle old, and handle new is not used and has not been used by anyone.
Output
In the first line output the integer n β the number of users that changed their handles at least once.
In the next n lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, old and new, separated by a space, meaning that before the user had handle old, and after all the requests are completed, his handle is new. You may output lines in any order.
Each user who changes the handle must occur exactly once in this description.
Examples
Input
5
Misha ILoveCodeforces
Vasya Petrov
Petrov VasyaPetrov123
ILoveCodeforces MikeMirzayanov
Petya Ivanov
Output
3
Petya Ivanov
Misha MikeMirzayanov
Vasya VasyaPetrov123
Submitted Solution:
```
n = int(input())
a, b = [], []
for i in range(n):
x, y = input().split()
if x not in b:
a.append(x); b.append(y)
else:
b[b.index(x)] = y
for i in range(len(a)):
print(a[i], b[i])
```
|
instruction
| 0
| 12,872
| 11
| 25,744
|
No
|
output
| 1
| 12,872
| 11
| 25,745
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.
Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.
Input
The first line contains integer q (1 β€ q β€ 1000), the number of handle change requests.
Next q lines contain the descriptions of the requests, one per line.
Each query consists of two non-empty strings old and new, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings old and new are distinct. The lengths of the strings do not exceed 20.
The requests are given chronologically. In other words, by the moment of a query there is a single person with handle old, and handle new is not used and has not been used by anyone.
Output
In the first line output the integer n β the number of users that changed their handles at least once.
In the next n lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, old and new, separated by a space, meaning that before the user had handle old, and after all the requests are completed, his handle is new. You may output lines in any order.
Each user who changes the handle must occur exactly once in this description.
Examples
Input
5
Misha ILoveCodeforces
Vasya Petrov
Petrov VasyaPetrov123
ILoveCodeforces MikeMirzayanov
Petya Ivanov
Output
3
Petya Ivanov
Misha MikeMirzayanov
Vasya VasyaPetrov123
Submitted Solution:
```
from math import sqrt,ceil
def mi():return map(int,input().split())
def li():return list(mi())
def ii():return int(input())
def si():return input()
rank=[]
#dsu(disjoint set unit) -> :)
def find(x,parent):
if(x==parent[x]):
return x
parent[x]=find(parent[x],parent)
return parent[x]
def union(x,y,parent):
x1=find(x,parent)
y1=find(y,parent)
if(x1==y1):
return
if(rank[x1]>=rank[y1]):
parent[y]=x1
rank[x1]+=1
else:
parent[x]=y1
rank[y1]+=1
def main():
q=ii()
old={}
a=[]
for i in range(q):
s1,s2=map(str,input().split())
old[s2]=s1
a.append(s2)
for i in range(q):
for j in range(q):
if(a[j]==old[a[j]]):
continue
if(a[i]==old[a[j]]):
old[a[j]]=old[a[i]]
a[j]=a[i]
old[a[j]]=a[i]
b=[]
for i in old.keys():
if i!=old[i]:
b.append(i)
print(len(b))
for i in b:
print(old[i],i)
if __name__=="__main__":
main()
```
|
instruction
| 0
| 12,873
| 11
| 25,746
|
No
|
output
| 1
| 12,873
| 11
| 25,747
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Misha hacked the Codeforces site. Then he decided to let all the users change their handles. A user can now change his handle any number of times. But each new handle must not be equal to any handle that is already used or that was used at some point.
Misha has a list of handle change requests. After completing the requests he wants to understand the relation between the original and the new handles of the users. Help him to do that.
Input
The first line contains integer q (1 β€ q β€ 1000), the number of handle change requests.
Next q lines contain the descriptions of the requests, one per line.
Each query consists of two non-empty strings old and new, separated by a space. The strings consist of lowercase and uppercase Latin letters and digits. Strings old and new are distinct. The lengths of the strings do not exceed 20.
The requests are given chronologically. In other words, by the moment of a query there is a single person with handle old, and handle new is not used and has not been used by anyone.
Output
In the first line output the integer n β the number of users that changed their handles at least once.
In the next n lines print the mapping between the old and the new handles of the users. Each of them must contain two strings, old and new, separated by a space, meaning that before the user had handle old, and after all the requests are completed, his handle is new. You may output lines in any order.
Each user who changes the handle must occur exactly once in this description.
Examples
Input
5
Misha ILoveCodeforces
Vasya Petrov
Petrov VasyaPetrov123
ILoveCodeforces MikeMirzayanov
Petya Ivanov
Output
3
Petya Ivanov
Misha MikeMirzayanov
Vasya VasyaPetrov123
Submitted Solution:
```
users = dict()
nombres = list()
cant = input()
handles = []
class Handle:
def __init__(self, old):
self.old = old
self.new = None
def actualizar(self, new):
self.new = new
for i in range(int(cant)):
nombre = input().split(" ")
if nombre[0] not in nombres:
nombres.append(nombre[0])
u = Handle(nombre[0])
handles.append(u)
if nombre[1] not in nombres:
u.actualizar(nombre[1])
nombres.append(nombre[1])
else:
if nombre[1] not in nombres:
for i in handles:
if i.new == nombre[0]:
i.actualizar(nombre[1])
handles.remove(i)
handles.append(i)
break
print(len(handles))
a = len(handles) - 1
while a >= 0:
print("{} {}".format(handles[a].old, handles[a].new))
a -= 1
```
|
instruction
| 0
| 12,874
| 11
| 25,748
|
No
|
output
| 1
| 12,874
| 11
| 25,749
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
|
instruction
| 0
| 12,998
| 11
| 25,996
|
Tags: implementation, sortings
Correct Solution:
```
n = int(input())
r = 1
l = []
c = []
for i in range(n):
x, y = map(int, input().split())
if x != y:
r = 0
break
l.append(x)
c.append(x)
c.sort()
if r == 0:
print("rated")
else:
if c == l[::-1]:
print("maybe")
else:
print("unrated")
```
|
output
| 1
| 12,998
| 11
| 25,997
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
|
instruction
| 0
| 12,999
| 11
| 25,998
|
Tags: implementation, sortings
Correct Solution:
```
n=int(input())
d=list(range(n))
u=True
e=True
for i in d:
a,b=map(int,list(input().split()))
if (a!=b):
print("rated")
e=False
break
d[i]=a
if e:
for i in range(n):
if (i>0):
if d[i]>d[i-1]:
print("unrated")
u=False
break
if u:
print("maybe")
```
|
output
| 1
| 12,999
| 11
| 25,999
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
|
instruction
| 0
| 13,000
| 11
| 26,000
|
Tags: implementation, sortings
Correct Solution:
```
n=int(input())
m=4126
f=False
for i in range(n):
a,b=[int(i) for i in input().split()]
if a!=b:
print("rated")
exit(0)
if a>m:
f=True
m=min(a,m)
if f:
print("unrated")
exit(0)
print("maybe")
```
|
output
| 1
| 13,000
| 11
| 26,001
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
|
instruction
| 0
| 13,001
| 11
| 26,002
|
Tags: implementation, sortings
Correct Solution:
```
r = [tuple(map(int, input().split())) for _ in range(int(input()))]
if any([p[0] != p[1] for p in r]):
print('rated')
elif r != list(sorted(r, reverse=True)):
print('unrated')
else:
print('maybe')
```
|
output
| 1
| 13,001
| 11
| 26,003
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
|
instruction
| 0
| 13,002
| 11
| 26,004
|
Tags: implementation, sortings
Correct Solution:
```
n=int(input())
first_rate=[]
second_rate=[]
for i in range(n):
k=[int(i) for i in input().split()]
first_rate.append(k[0])
second_rate.append(k[1])
if first_rate == second_rate :
first_rate_sorted=sorted(first_rate,reverse=True)
if first_rate_sorted == first_rate:
print("maybe")
else :
print("unrated")
else :
print("rated")
```
|
output
| 1
| 13,002
| 11
| 26,005
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
|
instruction
| 0
| 13,003
| 11
| 26,006
|
Tags: implementation, sortings
Correct Solution:
```
# ===================================
# (c) MidAndFeed aka ASilentVoice
# ===================================
# import math, fractions, collections
# ===================================
n = int(input())
q = [[int(x) for x in input().split()] for i in range(n)]
if any(x[0] != x[1] for x in q):
print("rated")
else:
unrated = 0
for i in range(n-1):
piv = q[i]
for j in range(i+1, n):
temp = q[j]
if piv < temp:
unrated = 1
break
if unrated:
break
print("unrated" if unrated else "maybe")
```
|
output
| 1
| 13,003
| 11
| 26,007
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
|
instruction
| 0
| 13,004
| 11
| 26,008
|
Tags: implementation, sortings
Correct Solution:
```
n=int(input())
a=[]
for i in range(n):
r=list(map(int,input().split()))
a.append(r)
b=sorted(a,reverse=True)
for i in a:
if(i[1]-i[0]!=0):
print("rated")
exit(0)
for i in range(n):
if a[i]!=b[i]:
print("unrated")
exit(0)
c=0
for i in a:
if(i[1]-i[0]==0):
c+=1
if(c==n):
print("maybe")
```
|
output
| 1
| 13,004
| 11
| 26,009
|
Provide tags and a correct Python 3 solution for this coding contest problem.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
|
instruction
| 0
| 13,005
| 11
| 26,010
|
Tags: implementation, sortings
Correct Solution:
```
a=[]
for _ in range(int(input())):
a.append(list(map(int,input().split())))
a=list(zip(*a))
if a[0]!=a[1]:
print('rated')
else:
if list(a[0])==sorted(a[0],reverse=True):
print('maybe')
else:
print('unrated')
```
|
output
| 1
| 13,005
| 11
| 26,011
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
Submitted Solution:
```
R = lambda:map(int,input().split())
n, = R()
rate = []
for i in range(n):
x, y = R()
if x != y:
exit(print('rated'))
rate.append(y)
if rate == sorted(rate, reverse=True):
print("maybe")
else:
print("unrated")
```
|
instruction
| 0
| 13,006
| 11
| 26,012
|
Yes
|
output
| 1
| 13,006
| 11
| 26,013
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
Submitted Solution:
```
from math import inf
def solve():
global rates
for b, a in rates:
if b != a: return 'rated'
for i in range(1, len(rates)):
if rates[i-1] < rates[i]:
return 'unrated'
return 'maybe'
def main():
global rates
n = int(input())
rates = [list(map(int, input().split())) for _ in range(n)]
print(solve())
main()
```
|
instruction
| 0
| 13,007
| 11
| 26,014
|
Yes
|
output
| 1
| 13,007
| 11
| 26,015
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
Submitted Solution:
```
n = int(input())
check = []
for i in range(n):
x, y = map(int, input().split())
if x != y:
print('rated')
exit()
check.append(x)
comp = sorted(check, reverse=True)
if comp == check:
print('maybe')
else: print('unrated')
```
|
instruction
| 0
| 13,008
| 11
| 26,016
|
Yes
|
output
| 1
| 13,008
| 11
| 26,017
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
Submitted Solution:
```
k=[];n=0
for i in " "*int(input()):a,b=map(int,input().split());k+=[b];n+=a!=b
if n:print("rated")
elif all(i==j for i,j in zip(k,sorted(k,reverse=True))):print("maybe")
else:print("unrated")
```
|
instruction
| 0
| 13,009
| 11
| 26,018
|
Yes
|
output
| 1
| 13,009
| 11
| 26,019
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
Submitted Solution:
```
n = int(input())
count = 0
unrated = 0
may = 0
listX = []
listY = []
x , y = input().split()
listX.append(x)
listY.append(y)
for a in range(1 , n):
x , y = input().split()
listX.append(x)
listY.append(y)
if x != y:
count += 1
else:
if int(listX[a]) > int(listX[a-1]):
unrated += 1
elif int(listX[a]) <= int(listX[a-1]):
may += 1
if count > 0:
print("rated")
elif unrated > 0:
print("unrated")
elif may > 0:
print("maybe")
```
|
instruction
| 0
| 13,010
| 11
| 26,020
|
No
|
output
| 1
| 13,010
| 11
| 26,021
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
Submitted Solution:
```
#collaborated with Bhumi Patel
n = int(input())
finalarray=[]
finalarray1=[]
rating=False
temp=True
for i in range(n):
temp_var=input()
temp_var=temp_var.split()
finalarray.append(int(temp_var[0]))
finalarray1.append(int(temp_var[1]))
if finalarray[i]!=finalarray1[i]:
rating=True
for i in range(1,len(finalarray)):
if finalarray[i]>finalarray[i-1] or finalarray1[i]>finalarray1[i-1]:
temp=True
break
if rating==True:
print("rated")
elif temp==False and rating==False:
print("unrated")
else:
print("maybe")
```
|
instruction
| 0
| 13,011
| 11
| 26,022
|
No
|
output
| 1
| 13,011
| 11
| 26,023
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
Submitted Solution:
```
from bisect import bisect_right as br
import sys
from collections import *
from math import *
import re
def sieve(n):
prime=[True for i in range(n+1)]
p=2
while p*p<=n:
if prime[p]==True:
for i in range(p*p,n+1,p):
prime[i]=False
p+=1
c=0
for i in range(2,n):
if prime[i]:
#print(i)
c+=1
return c
def totient(n):
res,p=n,2
while p*p<=n:
if n%p==0:
while n%p==0:
n=n//p
res-=int(res/p)
p+=1
if n>1:res-=int(res/n)
return res
def iseven(n):return[False,True][0 if n%2 else 1]
def inp_matrix(n):return list([input().split()] for i in range(n))
def inp_arr():return list(map(int,input().split()))
def inp_integers():return map(int,input().split())
def inp_strings():return input().split()
def lcm(a,b):return (a*b)/gcd(a,b)
max_int = sys.maxsize
mod = 10**9+7
flag1=False
flag2=False
n=int(input())
a=[input().split() for i in range(n)]
for i in range(n):
if a[i][0]!=a[i][1]:flag1=True
for i in range(1,n):
if a[i][0]>a[i-1][0]:flag2=True
#print(flag1,flag2)
if flag1:print('rated')
elif not flag1 and flag2:print('unrated')
else:print('maybe')
```
|
instruction
| 0
| 13,012
| 11
| 26,024
|
No
|
output
| 1
| 13,012
| 11
| 26,025
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Is it rated?
Here it is. The Ultimate Question of Competitive Programming, Codeforces, and Everything. And you are here to answer it.
Another Codeforces round has been conducted. No two participants have the same number of points. For each participant, from the top to the bottom of the standings, their rating before and after the round is known.
It's known that if at least one participant's rating has changed, then the round was rated for sure.
It's also known that if the round was rated and a participant with lower rating took a better place in the standings than a participant with higher rating, then at least one round participant's rating has changed.
In this problem, you should not make any other assumptions about the rating system.
Determine if the current round is rated, unrated, or it's impossible to determine whether it is rated of not.
Input
The first line contains a single integer n (2 β€ n β€ 1000) β the number of round participants.
Each of the next n lines contains two integers ai and bi (1 β€ ai, bi β€ 4126) β the rating of the i-th participant before and after the round, respectively. The participants are listed in order from the top to the bottom of the standings.
Output
If the round is rated for sure, print "rated". If the round is unrated for sure, print "unrated". If it's impossible to determine whether the round is rated or not, print "maybe".
Examples
Input
6
3060 3060
2194 2194
2876 2903
2624 2624
3007 2991
2884 2884
Output
rated
Input
4
1500 1500
1300 1300
1200 1200
1400 1400
Output
unrated
Input
5
3123 3123
2777 2777
2246 2246
2246 2246
1699 1699
Output
maybe
Note
In the first example, the ratings of the participants in the third and fifth places have changed, therefore, the round was rated.
In the second example, no one's rating has changed, but the participant in the second place has lower rating than the participant in the fourth place. Therefore, if the round was rated, someone's rating would've changed for sure.
In the third example, no one's rating has changed, and the participants took places in non-increasing order of their rating. Therefore, it's impossible to determine whether the round is rated or not.
Submitted Solution:
```
def main():
ans = 'unrated'
n_old = None
quan = int(input())
while quan:
num = input().split()
if num[0] != num[1]:
ans = 'rated'
if n_old != None and n_old == num[0] and ans == 'unrated':
ans = 'maybe'
n_old = num[0]
quan -= 1
return ans
print(main())
```
|
instruction
| 0
| 13,013
| 11
| 26,026
|
No
|
output
| 1
| 13,013
| 11
| 26,027
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ringo Kingdom Congress is voting on a bill.
N members are present, and the i-th member (1 β€ i β€ N) has w_i white ballots and b_i blue ballots. Each member i will put all the w_i white ballots into the box if he/she is in favor of the bill, and put all the b_i blue ballots into the box if he/she is not in favor of the bill. No other action is allowed. For example, a member must not forfeit voting, or put only a part of his/her white ballots or a part of his/her blue ballots into the box.
After all the members vote, if at least P percent of the ballots in the box is white, the bill is passed; if less than P percent of the ballots is white, the bill is rejected.
In order for the bill to pass, at least how many members must be in favor of it?
Constraints
* 1 β€ N β€ 10^5
* 1 β€ P β€ 100
* 1 β€ w_i β€ 10^9
* 1 β€ b_i β€ 10^9
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N P
w_1 b_1
w_2 b_2
:
w_N b_N
Output
Print the minimum number of members in favor of the bill required for passage.
Examples
Input
4 75
1 1
1 1
1 1
1 1
Output
3
Input
4 75
1 1
1 1
1 1
100 1
Output
1
Input
5 60
6 3
5 9
3 4
7 8
4 7
Output
3
Submitted Solution:
```
from itertools import accumulate
def solve(l, r):
if l > r:
return l
m = (l + r) // 2
w, b = whites[m], blues[m]
if w * 100 // (w + b) < p:
l = m + 1
else:
r = m - 1
return solve(l, r)
n, p = map(int, input().split())
wbs = [tuple(map(int, input().split())) for _ in range(n)]
ass = [(sum(t), i) for i, t in enumerate(wbs)]
ass.sort(reverse=True)
whites = list(accumulate(wbs[i][0] for s, i in ass))
blues = [wbs[i][1] for s, i in ass[1:]]
blues = list(reversed(list(accumulate(reversed(blues))))) + [0]
# print(list(whites))
# print(list(blues))
print(solve(0, n - 1) + 1)
```
|
instruction
| 0
| 13,184
| 11
| 26,368
|
No
|
output
| 1
| 13,184
| 11
| 26,369
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ringo Kingdom Congress is voting on a bill.
N members are present, and the i-th member (1 β€ i β€ N) has w_i white ballots and b_i blue ballots. Each member i will put all the w_i white ballots into the box if he/she is in favor of the bill, and put all the b_i blue ballots into the box if he/she is not in favor of the bill. No other action is allowed. For example, a member must not forfeit voting, or put only a part of his/her white ballots or a part of his/her blue ballots into the box.
After all the members vote, if at least P percent of the ballots in the box is white, the bill is passed; if less than P percent of the ballots is white, the bill is rejected.
In order for the bill to pass, at least how many members must be in favor of it?
Constraints
* 1 β€ N β€ 10^5
* 1 β€ P β€ 100
* 1 β€ w_i β€ 10^9
* 1 β€ b_i β€ 10^9
* All input values are integers.
Input
Input is given from Standard Input in the following format:
N P
w_1 b_1
w_2 b_2
:
w_N b_N
Output
Print the minimum number of members in favor of the bill required for passage.
Examples
Input
4 75
1 1
1 1
1 1
1 1
Output
3
Input
4 75
1 1
1 1
1 1
100 1
Output
1
Input
5 60
6 3
5 9
3 4
7 8
4 7
Output
3
Submitted Solution:
```
from itertools import accumulate
def solve(l, r):
if l > r:
return l
m = (l + r) // 2
w, b = whites[m], blues[m]
if w * 100 // (w + b) < p:
l = m + 1
else:
r = m - 1
return solve(l, r)
n, p = map(int, input().split())
wbs = [tuple(map(int, input().split())) for _ in range(n)]
ass = [(sum(t), t[0], i) for i, t in enumerate(wbs)]
ass.sort(reverse=True)
whites = list(accumulate(wbs[i][0] for s, t, i in ass))
blues = [wbs[i][1] for s, t, i in ass[1:]]
blues = list(reversed(list(accumulate(reversed(blues))))) + [0]
# print(list(whites))
# print(list(blues))
print(solve(0, n - 1) + 1)
```
|
instruction
| 0
| 13,185
| 11
| 26,370
|
No
|
output
| 1
| 13,185
| 11
| 26,371
|
Provide a correct Python 3 solution for this coding contest problem.
G: Working
Kou decided to do the same number of jobs every day for the next $ N $.
$ A_i $ jobs are added on the $ i $ day of the $ N $ day.
Mr. Kou has no work to do now, and he doesn't have to finish all the work by the $ N $ day.
How many jobs can you do in a day?
However, Mr. Kou is excellent, so he can do as many jobs as he has.
input
$ N $ is given on the first line.
On the second line, $ N $ integers $ A_1, A_2, A_3, \ dots, A_N $ are given, separated by blanks.
output
Output the maximum number of jobs you can do in a day. Insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A_1, A_2, A_3, \ dots, A_N $ are integers between $ 1 $ and $ 100 $
Input example 1
Five
4 2 5 3 1
Output example 1
3
If you decide to work more than $ 4 $ a day, you'll run out of work on the second day.
Input example 2
Five
9 9 1 9 9
Output example 2
6
Example
Input
5
4 2 5 3 1
Output
3
|
instruction
| 0
| 13,260
| 11
| 26,520
|
"Correct Solution:
```
N = int(input())
A = [int(x) for x in input().split()]
sum = A[0]
ans = A[0]
for i in range(1, N):
sum += A[i]
ans = min(ans, sum // (i + 1))
print(ans)
```
|
output
| 1
| 13,260
| 11
| 26,521
|
Provide a correct Python 3 solution for this coding contest problem.
G: Working
Kou decided to do the same number of jobs every day for the next $ N $.
$ A_i $ jobs are added on the $ i $ day of the $ N $ day.
Mr. Kou has no work to do now, and he doesn't have to finish all the work by the $ N $ day.
How many jobs can you do in a day?
However, Mr. Kou is excellent, so he can do as many jobs as he has.
input
$ N $ is given on the first line.
On the second line, $ N $ integers $ A_1, A_2, A_3, \ dots, A_N $ are given, separated by blanks.
output
Output the maximum number of jobs you can do in a day. Insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A_1, A_2, A_3, \ dots, A_N $ are integers between $ 1 $ and $ 100 $
Input example 1
Five
4 2 5 3 1
Output example 1
3
If you decide to work more than $ 4 $ a day, you'll run out of work on the second day.
Input example 2
Five
9 9 1 9 9
Output example 2
6
Example
Input
5
4 2 5 3 1
Output
3
|
instruction
| 0
| 13,261
| 11
| 26,522
|
"Correct Solution:
```
n = map(int,input().split())
a = list(map(int,input().split()))
ok,ng = 1,10000
while ng-ok > 1:
mid = (ok+ng)//2
valid = True
now = 0
for i in a:
now += i
now -= mid
if now < 0:
valid = False
if valid:
ok = mid
else:
ng = mid
print(ok)
```
|
output
| 1
| 13,261
| 11
| 26,523
|
Provide a correct Python 3 solution for this coding contest problem.
G: Working
Kou decided to do the same number of jobs every day for the next $ N $.
$ A_i $ jobs are added on the $ i $ day of the $ N $ day.
Mr. Kou has no work to do now, and he doesn't have to finish all the work by the $ N $ day.
How many jobs can you do in a day?
However, Mr. Kou is excellent, so he can do as many jobs as he has.
input
$ N $ is given on the first line.
On the second line, $ N $ integers $ A_1, A_2, A_3, \ dots, A_N $ are given, separated by blanks.
output
Output the maximum number of jobs you can do in a day. Insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A_1, A_2, A_3, \ dots, A_N $ are integers between $ 1 $ and $ 100 $
Input example 1
Five
4 2 5 3 1
Output example 1
3
If you decide to work more than $ 4 $ a day, you'll run out of work on the second day.
Input example 2
Five
9 9 1 9 9
Output example 2
6
Example
Input
5
4 2 5 3 1
Output
3
|
instruction
| 0
| 13,263
| 11
| 26,526
|
"Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
r = 100
for i in range(n):
r = min(r, sum(a[:i+1])//(i+1))
print(r)
```
|
output
| 1
| 13,263
| 11
| 26,527
|
Provide a correct Python 3 solution for this coding contest problem.
G: Working
Kou decided to do the same number of jobs every day for the next $ N $.
$ A_i $ jobs are added on the $ i $ day of the $ N $ day.
Mr. Kou has no work to do now, and he doesn't have to finish all the work by the $ N $ day.
How many jobs can you do in a day?
However, Mr. Kou is excellent, so he can do as many jobs as he has.
input
$ N $ is given on the first line.
On the second line, $ N $ integers $ A_1, A_2, A_3, \ dots, A_N $ are given, separated by blanks.
output
Output the maximum number of jobs you can do in a day. Insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A_1, A_2, A_3, \ dots, A_N $ are integers between $ 1 $ and $ 100 $
Input example 1
Five
4 2 5 3 1
Output example 1
3
If you decide to work more than $ 4 $ a day, you'll run out of work on the second day.
Input example 2
Five
9 9 1 9 9
Output example 2
6
Example
Input
5
4 2 5 3 1
Output
3
|
instruction
| 0
| 13,265
| 11
| 26,530
|
"Correct Solution:
```
N = int(input())
A = [int(x) for x in input().split()]
def check(x):
work = 0
for a in A:
work += a
if work < x:
return 0
work -= x
return 1
ans = 0
for i in range(101):
if check(i):
ans = i
print(ans)
```
|
output
| 1
| 13,265
| 11
| 26,531
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
G: Working
Kou decided to do the same number of jobs every day for the next $ N $.
$ A_i $ jobs are added on the $ i $ day of the $ N $ day.
Mr. Kou has no work to do now, and he doesn't have to finish all the work by the $ N $ day.
How many jobs can you do in a day?
However, Mr. Kou is excellent, so he can do as many jobs as he has.
input
$ N $ is given on the first line.
On the second line, $ N $ integers $ A_1, A_2, A_3, \ dots, A_N $ are given, separated by blanks.
output
Output the maximum number of jobs you can do in a day. Insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A_1, A_2, A_3, \ dots, A_N $ are integers between $ 1 $ and $ 100 $
Input example 1
Five
4 2 5 3 1
Output example 1
3
If you decide to work more than $ 4 $ a day, you'll run out of work on the second day.
Input example 2
Five
9 9 1 9 9
Output example 2
6
Example
Input
5
4 2 5 3 1
Output
3
Submitted Solution:
```
from itertools import accumulate
n=int(input())
a=list(accumulate(map(int,input().split())))
for i in range(100,0,-1):
for j in range(n):
if i*(j+1)>a[j]:break
else:
print(i)
break
```
|
instruction
| 0
| 13,267
| 11
| 26,534
|
Yes
|
output
| 1
| 13,267
| 11
| 26,535
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
G: Working
Kou decided to do the same number of jobs every day for the next $ N $.
$ A_i $ jobs are added on the $ i $ day of the $ N $ day.
Mr. Kou has no work to do now, and he doesn't have to finish all the work by the $ N $ day.
How many jobs can you do in a day?
However, Mr. Kou is excellent, so he can do as many jobs as he has.
input
$ N $ is given on the first line.
On the second line, $ N $ integers $ A_1, A_2, A_3, \ dots, A_N $ are given, separated by blanks.
output
Output the maximum number of jobs you can do in a day. Insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A_1, A_2, A_3, \ dots, A_N $ are integers between $ 1 $ and $ 100 $
Input example 1
Five
4 2 5 3 1
Output example 1
3
If you decide to work more than $ 4 $ a day, you'll run out of work on the second day.
Input example 2
Five
9 9 1 9 9
Output example 2
6
Example
Input
5
4 2 5 3 1
Output
3
Submitted Solution:
```
N=int(input())
A=list(map(int,input().split()))
SUM=[A[0]]
for i in range(1,N):
SUM.append(SUM[-1]+A[i])
B=[SUM[i]//(i+1) for i in range(N)]
print(min(B))
```
|
instruction
| 0
| 13,268
| 11
| 26,536
|
Yes
|
output
| 1
| 13,268
| 11
| 26,537
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
G: Working
Kou decided to do the same number of jobs every day for the next $ N $.
$ A_i $ jobs are added on the $ i $ day of the $ N $ day.
Mr. Kou has no work to do now, and he doesn't have to finish all the work by the $ N $ day.
How many jobs can you do in a day?
However, Mr. Kou is excellent, so he can do as many jobs as he has.
input
$ N $ is given on the first line.
On the second line, $ N $ integers $ A_1, A_2, A_3, \ dots, A_N $ are given, separated by blanks.
output
Output the maximum number of jobs you can do in a day. Insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A_1, A_2, A_3, \ dots, A_N $ are integers between $ 1 $ and $ 100 $
Input example 1
Five
4 2 5 3 1
Output example 1
3
If you decide to work more than $ 4 $ a day, you'll run out of work on the second day.
Input example 2
Five
9 9 1 9 9
Output example 2
6
Example
Input
5
4 2 5 3 1
Output
3
Submitted Solution:
```
n = int(input())
a = list(map(int,input().split()))
ans = 10 ** 7
sum = 0
for i in range(n):
sum+=a[i]
ans = min(ans,sum // (i + 1))
print(ans)
```
|
instruction
| 0
| 13,269
| 11
| 26,538
|
Yes
|
output
| 1
| 13,269
| 11
| 26,539
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
G: Working
Kou decided to do the same number of jobs every day for the next $ N $.
$ A_i $ jobs are added on the $ i $ day of the $ N $ day.
Mr. Kou has no work to do now, and he doesn't have to finish all the work by the $ N $ day.
How many jobs can you do in a day?
However, Mr. Kou is excellent, so he can do as many jobs as he has.
input
$ N $ is given on the first line.
On the second line, $ N $ integers $ A_1, A_2, A_3, \ dots, A_N $ are given, separated by blanks.
output
Output the maximum number of jobs you can do in a day. Insert a line break at the end.
Constraint
* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A_1, A_2, A_3, \ dots, A_N $ are integers between $ 1 $ and $ 100 $
Input example 1
Five
4 2 5 3 1
Output example 1
3
If you decide to work more than $ 4 $ a day, you'll run out of work on the second day.
Input example 2
Five
9 9 1 9 9
Output example 2
6
Example
Input
5
4 2 5 3 1
Output
3
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
ans=float("inf")
s=0
for i in range(n):
s+=a[i]
ans=min(ans,int(s/(i+1)))
print(ans)
```
|
instruction
| 0
| 13,270
| 11
| 26,540
|
Yes
|
output
| 1
| 13,270
| 11
| 26,541
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sereja has got an array, consisting of n integers, a1, a2, ..., an. Sereja is an active boy, so he is now going to complete m operations. Each operation will have one of the three forms:
1. Make vi-th array element equal to xi. In other words, perform the assignment avi = xi.
2. Increase each array element by yi. In other words, perform n assignments ai = ai + yi (1 β€ i β€ n).
3. Take a piece of paper and write out the qi-th array element. That is, the element aqi.
Help Sereja, complete all his operations.
Input
The first line contains integers n, m (1 β€ n, m β€ 105). The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109) β the original array.
Next m lines describe operations, the i-th line describes the i-th operation. The first number in the i-th line is integer ti (1 β€ ti β€ 3) that represents the operation type. If ti = 1, then it is followed by two integers vi and xi, (1 β€ vi β€ n, 1 β€ xi β€ 109). If ti = 2, then it is followed by integer yi (1 β€ yi β€ 104). And if ti = 3, then it is followed by integer qi (1 β€ qi β€ n).
Output
For each third type operation print value aqi. Print the values in the order, in which the corresponding queries follow in the input.
Examples
Input
10 11
1 2 3 4 5 6 7 8 9 10
3 2
3 9
2 10
3 1
3 10
1 1 10
2 10
2 10
3 1
3 10
3 9
Output
2
9
11
20
30
40
39
Submitted Solution:
```
#python3
import sys, threading, os.path
import collections, heapq, math,bisect
import string
from platform import python_version
import itertools
sys.setrecursionlimit(10**6)
threading.stack_size(2**27)
def main():
if os.path.exists('input.txt'):
input = open('input.txt', 'r')
else:
input = sys.stdin
#--------------------------------INPUT---------------------------------
n,m = list(map(int, input.readline().split()))
lis = list(map(int, input.readline().split()))
sumall = 0
sol=[]
for i in range(m):
temlis = list(map(int, input.readline().split()))
#print(temlis)
if temlis[0]==1:
lis[temlis[1]-1]=temlis[2]-sumall
elif temlis[0]==2:
sumall+=temlis[1]
elif temlis[0]==3:
sol.append(lis[temlis[1]-1]+sumall)
#print(lis,sumall)
output = '\n'.join(map(str, sol))
#-------------------------------OUTPUT----------------------------------
if os.path.exists('output.txt'):
open('output.txt', 'w').writelines(str(output))
else:
sys.stdout.write(str(output))
if __name__ == '__main__':
main()
#threading.Thread(target=main).start()
```
|
instruction
| 0
| 13,608
| 11
| 27,216
|
Yes
|
output
| 1
| 13,608
| 11
| 27,217
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nathan O. Davis has been running an electronic bulletin board system named JAG-channel. He is now having hard time to add a new feature there --- threaded view.
Like many other bulletin board systems, JAG-channel is thread-based. Here a thread (also called a topic) refers to a single conversation with a collection of posts. Each post can be an opening post, which initiates a new thread, or a reply to a previous post in an existing thread.
Threaded view is a tree-like view that reflects the logical reply structure among the posts: each post forms a node of the tree and contains its replies as its subnodes in the chronological order (i.e. older replies precede newer ones). Note that a post along with its direct and indirect replies forms a subtree as a whole.
Let us take an example. Suppose: a user made an opening post with a message `hoge`; another user replied to it with `fuga`; yet another user also replied to the opening post with `piyo`; someone else replied to the second post (i.e. `fuga`β) with `foobar`; and the fifth user replied to the same post with `jagjag`. The tree of this thread would look like:
hoge
ββfuga
βγββfoobar
βγββjagjag
ββpiyo
For easier implementation, Nathan is thinking of a simpler format: the depth of each post from the opening post is represented by dots. Each reply gets one more dot than its parent post. The tree of the above thread would then look like:
hoge
.fuga
..foobar
..jagjag
.piyo
Your task in this problem is to help Nathan by writing a program that prints a tree in the Nathan's format for the given posts in a single thread.
Input
Input contains a single dataset in the following format:
n
k_1
M_1
k_2
M_2
:
:
k_n
M_n
The first line contains an integer n (1 β€ n β€ 1,000), which is the number of posts in the thread. Then 2n lines follow. Each post is represented by two lines: the first line contains an integer k_i (k_1 = 0, 1 β€ k_i < i for 2 β€ i β€ n) and indicates the i-th post is a reply to the k_i-th post; the second line contains a string M_i and represents the message of the i-th post. k_1 is always 0, which means the first post is not replying to any other post, i.e. it is an opening post.
Each message contains 1 to 50 characters, consisting of uppercase, lowercase, and numeric letters.
Output
Print the given n messages as specified in the problem statement.
Sample Input 1
1
0
icpc
Output for the Sample Input 1
icpc
Sample Input 2
5
0
hoge
1
fuga
1
piyo
2
foobar
2
jagjag
Output for the Sample Input 2
hoge
.fuga
..foobar
..jagjag
.piyo
Sample Input 3
8
0
jagjag
1
hogehoge
1
buhihi
2
fugafuga
4
ponyoponyo
5
evaeva
4
nowawa
5
pokemon
Output for the Sample Input 3
jagjag
.hogehoge
..fugafuga
...ponyoponyo
....evaeva
....pokemon
...nowawa
.buhihi
Sample Input 4
6
0
nakachan
1
fan
2
yamemasu
3
nennryou2
4
dannyaku4
5
kouzai11
Output for the Sample Input 4
nakachan
.fan
..yamemasu
...nennryou2
....dannyaku4
.....kouzai11
Sample Input 5
34
0
LoveLive
1
honoka
2
borarara
2
sunohare
2
mogyu
1
eri
6
kasikoi
7
kawaii
8
eriichika
1
kotori
10
WR
10
haetekurukotori
10
ichigo
1
umi
14
love
15
arrow
16
shoot
1
rin
18
nyanyanya
1
maki
20
6th
20
star
22
nishikino
1
nozomi
24
spiritual
25
power
1
hanayo
27
darekatasukete
28
chottomattete
1
niko
30
natsuiro
30
nikkonikkoni
30
sekaino
33
YAZAWA
Output for the Sample Input 5
LoveLive
.honoka
..borarara
..sunohare
..mogyu
.eri
..kasikoi
...kawaii
....eriichika
.kotori
..WR
..haetekurukotori
..ichigo
.umi
..love
...arrow
....shoot
.rin
..nyanyanya
.maki
..6th
..star
...nishikino
.nozomi
..spiritual
...power
.hanayo
..darekatasukete
...chottomattete
.niko
..natsuiro
..nikkonikkoni
..sekaino
...YAZAWA
Sample Input 6
6
0
2ch
1
1ostu
1
2get
1
1otsu
1
1ostu
3
pgr
Output for the Sample Input 6
2ch
.1ostu
.2get
..pgr
.1otsu
.1ostu
Example
Input
1
0
icpc
Output
icpc
Submitted Solution:
```
N = int(input())
src = []
for i in range(N):
k = int(input())
s = input()
src.append((s,[]))
if i == 0: continue
src[k-1][1].append(i)
def dfs(i,depth):
s,ch = src[i]
print('.'*depth + s)
for c in ch:
dfs(c,depth+1)
dfs(0,0)
```
|
instruction
| 0
| 14,101
| 11
| 28,202
|
No
|
output
| 1
| 14,101
| 11
| 28,203
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is choosing a laptop. The shop has n laptops to all tastes.
Vasya is interested in the following properties: processor speed, ram and hdd. Vasya is a programmer and not a gamer which is why he is not interested in all other properties.
If all three properties of a laptop are strictly less than those properties of some other laptop, then the first laptop is considered outdated by Vasya. Among all laptops Vasya does not consider outdated, he chooses the cheapest one.
There are very many laptops, which is why Vasya decided to write a program that chooses the suitable laptop. However, Vasya doesn't have his own laptop yet and he asks you to help him.
Input
The first line contains number n (1 β€ n β€ 100).
Then follow n lines. Each describes a laptop as speed ram hdd cost. Besides,
* speed, ram, hdd and cost are integers
* 1000 β€ speed β€ 4200 is the processor's speed in megahertz
* 256 β€ ram β€ 4096 the RAM volume in megabytes
* 1 β€ hdd β€ 500 is the HDD in gigabytes
* 100 β€ cost β€ 1000 is price in tugriks
All laptops have different prices.
Output
Print a single number β the number of a laptop Vasya will choose. The laptops are numbered with positive integers from 1 to n in the order in which they are given in the input data.
Examples
Input
5
2100 512 150 200
2000 2048 240 350
2300 1024 200 320
2500 2048 80 300
2000 512 180 150
Output
4
Note
In the third sample Vasya considers the first and fifth laptops outdated as all of their properties cannot match those of the third laptop. The fourth one is the cheapest among the laptops that are left. Thus, Vasya chooses the fourth laptop.
Submitted Solution:
```
n=int(input())
Z=[]
for i in range(n):
S,R,H,C=map(int,input().split())
Z.append((S,R,H,C))
Y=[]
for i in range(n):
for j in range(n):
if(Z[i][0]<Z[j][0] and Z[i][1]<Z[j][1] and Z[i][2]<Z[j][2]):
Y.append(Z[i])
Y=set(Y)
minn=100000
ans=0
for i in range(n):
item=Z[i]
if(item not in Y and item[3]<minn):
minn=item[3]
ans=i+1
print(ans)
```
|
instruction
| 0
| 14,162
| 11
| 28,324
|
Yes
|
output
| 1
| 14,162
| 11
| 28,325
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is choosing a laptop. The shop has n laptops to all tastes.
Vasya is interested in the following properties: processor speed, ram and hdd. Vasya is a programmer and not a gamer which is why he is not interested in all other properties.
If all three properties of a laptop are strictly less than those properties of some other laptop, then the first laptop is considered outdated by Vasya. Among all laptops Vasya does not consider outdated, he chooses the cheapest one.
There are very many laptops, which is why Vasya decided to write a program that chooses the suitable laptop. However, Vasya doesn't have his own laptop yet and he asks you to help him.
Input
The first line contains number n (1 β€ n β€ 100).
Then follow n lines. Each describes a laptop as speed ram hdd cost. Besides,
* speed, ram, hdd and cost are integers
* 1000 β€ speed β€ 4200 is the processor's speed in megahertz
* 256 β€ ram β€ 4096 the RAM volume in megabytes
* 1 β€ hdd β€ 500 is the HDD in gigabytes
* 100 β€ cost β€ 1000 is price in tugriks
All laptops have different prices.
Output
Print a single number β the number of a laptop Vasya will choose. The laptops are numbered with positive integers from 1 to n in the order in which they are given in the input data.
Examples
Input
5
2100 512 150 200
2000 2048 240 350
2300 1024 200 320
2500 2048 80 300
2000 512 180 150
Output
4
Note
In the third sample Vasya considers the first and fifth laptops outdated as all of their properties cannot match those of the third laptop. The fourth one is the cheapest among the laptops that are left. Thus, Vasya chooses the fourth laptop.
Submitted Solution:
```
n=int(input())
A=[]
for i in range(n):
A+=[list(map(int,input().split()))]
cost=10**18
I=0
for i in range(n):
ans=True
for j in range(n):
if(A[i][0]<A[j][0] and A[i][1]<A[j][1] and A[i][2]<A[j][2]):
ans=False
if(ans):
cost=min(cost,A[i][3])
if(cost==A[i][3]):
I=i+1
print(I)
```
|
instruction
| 0
| 14,163
| 11
| 28,326
|
Yes
|
output
| 1
| 14,163
| 11
| 28,327
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is choosing a laptop. The shop has n laptops to all tastes.
Vasya is interested in the following properties: processor speed, ram and hdd. Vasya is a programmer and not a gamer which is why he is not interested in all other properties.
If all three properties of a laptop are strictly less than those properties of some other laptop, then the first laptop is considered outdated by Vasya. Among all laptops Vasya does not consider outdated, he chooses the cheapest one.
There are very many laptops, which is why Vasya decided to write a program that chooses the suitable laptop. However, Vasya doesn't have his own laptop yet and he asks you to help him.
Input
The first line contains number n (1 β€ n β€ 100).
Then follow n lines. Each describes a laptop as speed ram hdd cost. Besides,
* speed, ram, hdd and cost are integers
* 1000 β€ speed β€ 4200 is the processor's speed in megahertz
* 256 β€ ram β€ 4096 the RAM volume in megabytes
* 1 β€ hdd β€ 500 is the HDD in gigabytes
* 100 β€ cost β€ 1000 is price in tugriks
All laptops have different prices.
Output
Print a single number β the number of a laptop Vasya will choose. The laptops are numbered with positive integers from 1 to n in the order in which they are given in the input data.
Examples
Input
5
2100 512 150 200
2000 2048 240 350
2300 1024 200 320
2500 2048 80 300
2000 512 180 150
Output
4
Note
In the third sample Vasya considers the first and fifth laptops outdated as all of their properties cannot match those of the third laptop. The fourth one is the cheapest among the laptops that are left. Thus, Vasya chooses the fourth laptop.
Submitted Solution:
```
def chck(m,l):
p,q,r,s=l
for i in m:
a,b,e,d=i
if a>p and b>q and e>r:return 0
else:return 1
n=int(input());l,c=[],[];m=10**4
for i in range(n):
t=list(map(int,input().split()))
l.append(t);c.append(t[-1])
for i in range(n):
if chck(l,l[i]):m=min(l[i][-1],m)
print(c.index(m)+1)
```
|
instruction
| 0
| 14,164
| 11
| 28,328
|
Yes
|
output
| 1
| 14,164
| 11
| 28,329
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is choosing a laptop. The shop has n laptops to all tastes.
Vasya is interested in the following properties: processor speed, ram and hdd. Vasya is a programmer and not a gamer which is why he is not interested in all other properties.
If all three properties of a laptop are strictly less than those properties of some other laptop, then the first laptop is considered outdated by Vasya. Among all laptops Vasya does not consider outdated, he chooses the cheapest one.
There are very many laptops, which is why Vasya decided to write a program that chooses the suitable laptop. However, Vasya doesn't have his own laptop yet and he asks you to help him.
Input
The first line contains number n (1 β€ n β€ 100).
Then follow n lines. Each describes a laptop as speed ram hdd cost. Besides,
* speed, ram, hdd and cost are integers
* 1000 β€ speed β€ 4200 is the processor's speed in megahertz
* 256 β€ ram β€ 4096 the RAM volume in megabytes
* 1 β€ hdd β€ 500 is the HDD in gigabytes
* 100 β€ cost β€ 1000 is price in tugriks
All laptops have different prices.
Output
Print a single number β the number of a laptop Vasya will choose. The laptops are numbered with positive integers from 1 to n in the order in which they are given in the input data.
Examples
Input
5
2100 512 150 200
2000 2048 240 350
2300 1024 200 320
2500 2048 80 300
2000 512 180 150
Output
4
Note
In the third sample Vasya considers the first and fifth laptops outdated as all of their properties cannot match those of the third laptop. The fourth one is the cheapest among the laptops that are left. Thus, Vasya chooses the fourth laptop.
Submitted Solution:
```
def readln(): return tuple(map(int, input().split()))
n, = readln()
ans = 0
best = (0, 0, 0, 0)
lst = [readln() + (_ + 1,) for _ in range(n)]
lst = [v for v in lst if not [1 for u in lst if v[0] < u[0] and v[1] < u[1] and v[2] < u[2]]]
lst.sort(key=lambda x: x[3])
print(lst[0][4])
```
|
instruction
| 0
| 14,165
| 11
| 28,330
|
Yes
|
output
| 1
| 14,165
| 11
| 28,331
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is choosing a laptop. The shop has n laptops to all tastes.
Vasya is interested in the following properties: processor speed, ram and hdd. Vasya is a programmer and not a gamer which is why he is not interested in all other properties.
If all three properties of a laptop are strictly less than those properties of some other laptop, then the first laptop is considered outdated by Vasya. Among all laptops Vasya does not consider outdated, he chooses the cheapest one.
There are very many laptops, which is why Vasya decided to write a program that chooses the suitable laptop. However, Vasya doesn't have his own laptop yet and he asks you to help him.
Input
The first line contains number n (1 β€ n β€ 100).
Then follow n lines. Each describes a laptop as speed ram hdd cost. Besides,
* speed, ram, hdd and cost are integers
* 1000 β€ speed β€ 4200 is the processor's speed in megahertz
* 256 β€ ram β€ 4096 the RAM volume in megabytes
* 1 β€ hdd β€ 500 is the HDD in gigabytes
* 100 β€ cost β€ 1000 is price in tugriks
All laptops have different prices.
Output
Print a single number β the number of a laptop Vasya will choose. The laptops are numbered with positive integers from 1 to n in the order in which they are given in the input data.
Examples
Input
5
2100 512 150 200
2000 2048 240 350
2300 1024 200 320
2500 2048 80 300
2000 512 180 150
Output
4
Note
In the third sample Vasya considers the first and fifth laptops outdated as all of their properties cannot match those of the third laptop. The fourth one is the cheapest among the laptops that are left. Thus, Vasya chooses the fourth laptop.
Submitted Solution:
```
n = int(input())
l = []
mspec = [0 for i in range(4)]
index = 0
price = 1000
for i in range(n):
spec = list(map(int, input().split()))
l.append(spec)
if spec[0] >= mspec[0] and spec[1] >= mspec[1] and spec[2] >= mspec[2]:
mspec = spec
for i in range(n):
if (l[i][0] >= mspec[0] or l[i][1] >= mspec[1] or l[i][2] >= mspec[2]) and l[i][
3
] <= price:
price = l[i][3]
index = i + 1
print(index)
```
|
instruction
| 0
| 14,166
| 11
| 28,332
|
No
|
output
| 1
| 14,166
| 11
| 28,333
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is choosing a laptop. The shop has n laptops to all tastes.
Vasya is interested in the following properties: processor speed, ram and hdd. Vasya is a programmer and not a gamer which is why he is not interested in all other properties.
If all three properties of a laptop are strictly less than those properties of some other laptop, then the first laptop is considered outdated by Vasya. Among all laptops Vasya does not consider outdated, he chooses the cheapest one.
There are very many laptops, which is why Vasya decided to write a program that chooses the suitable laptop. However, Vasya doesn't have his own laptop yet and he asks you to help him.
Input
The first line contains number n (1 β€ n β€ 100).
Then follow n lines. Each describes a laptop as speed ram hdd cost. Besides,
* speed, ram, hdd and cost are integers
* 1000 β€ speed β€ 4200 is the processor's speed in megahertz
* 256 β€ ram β€ 4096 the RAM volume in megabytes
* 1 β€ hdd β€ 500 is the HDD in gigabytes
* 100 β€ cost β€ 1000 is price in tugriks
All laptops have different prices.
Output
Print a single number β the number of a laptop Vasya will choose. The laptops are numbered with positive integers from 1 to n in the order in which they are given in the input data.
Examples
Input
5
2100 512 150 200
2000 2048 240 350
2300 1024 200 320
2500 2048 80 300
2000 512 180 150
Output
4
Note
In the third sample Vasya considers the first and fifth laptops outdated as all of their properties cannot match those of the third laptop. The fourth one is the cheapest among the laptops that are left. Thus, Vasya chooses the fourth laptop.
Submitted Solution:
```
# ========= /\ /| |====/|
# | / \ | | / |
# | /____\ | | / |
# | / \ | | / |
# ========= / \ ===== |/====|
# code
if __name__ == "__main__":
p = []
c = []
r = []
n = int(input())
x = range(n)
i = 0
for i in x:
a,b,e,d = map(int,input().split())
c.append(d)
p.append((a,b,e))
for i in x:
for j in x:
if p[j] < p[i]:
r.append(j)
for i in r:
c[i] = 1000000
print(c.index(min(c)) + 1)
```
|
instruction
| 0
| 14,167
| 11
| 28,334
|
No
|
output
| 1
| 14,167
| 11
| 28,335
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is choosing a laptop. The shop has n laptops to all tastes.
Vasya is interested in the following properties: processor speed, ram and hdd. Vasya is a programmer and not a gamer which is why he is not interested in all other properties.
If all three properties of a laptop are strictly less than those properties of some other laptop, then the first laptop is considered outdated by Vasya. Among all laptops Vasya does not consider outdated, he chooses the cheapest one.
There are very many laptops, which is why Vasya decided to write a program that chooses the suitable laptop. However, Vasya doesn't have his own laptop yet and he asks you to help him.
Input
The first line contains number n (1 β€ n β€ 100).
Then follow n lines. Each describes a laptop as speed ram hdd cost. Besides,
* speed, ram, hdd and cost are integers
* 1000 β€ speed β€ 4200 is the processor's speed in megahertz
* 256 β€ ram β€ 4096 the RAM volume in megabytes
* 1 β€ hdd β€ 500 is the HDD in gigabytes
* 100 β€ cost β€ 1000 is price in tugriks
All laptops have different prices.
Output
Print a single number β the number of a laptop Vasya will choose. The laptops are numbered with positive integers from 1 to n in the order in which they are given in the input data.
Examples
Input
5
2100 512 150 200
2000 2048 240 350
2300 1024 200 320
2500 2048 80 300
2000 512 180 150
Output
4
Note
In the third sample Vasya considers the first and fifth laptops outdated as all of their properties cannot match those of the third laptop. The fourth one is the cheapest among the laptops that are left. Thus, Vasya chooses the fourth laptop.
Submitted Solution:
```
from sys import*
input= stdin.readline
t=int(input())
speed=[]
ram=[]
hdd=[]
cost=[]
res=[]
for i in range(t):
l=[0]*3
res.append(l)
a,b,c,d=map(int,input().split())
speed.append([a,i])
ram.append([b,i])
hdd.append([c,i])
cost.append(d)
speed.sort()
ram.sort()
hdd.sort()
for i in range(t):
res[speed[i][1]][0]=i
res[ram[i][1]][1]=i
res[hdd[i][1]][2]=i
for i in range(t):
if(t-1 not in res[i]):
cost[i]=10000
print(cost.index(min(cost))+1)
```
|
instruction
| 0
| 14,168
| 11
| 28,336
|
No
|
output
| 1
| 14,168
| 11
| 28,337
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is choosing a laptop. The shop has n laptops to all tastes.
Vasya is interested in the following properties: processor speed, ram and hdd. Vasya is a programmer and not a gamer which is why he is not interested in all other properties.
If all three properties of a laptop are strictly less than those properties of some other laptop, then the first laptop is considered outdated by Vasya. Among all laptops Vasya does not consider outdated, he chooses the cheapest one.
There are very many laptops, which is why Vasya decided to write a program that chooses the suitable laptop. However, Vasya doesn't have his own laptop yet and he asks you to help him.
Input
The first line contains number n (1 β€ n β€ 100).
Then follow n lines. Each describes a laptop as speed ram hdd cost. Besides,
* speed, ram, hdd and cost are integers
* 1000 β€ speed β€ 4200 is the processor's speed in megahertz
* 256 β€ ram β€ 4096 the RAM volume in megabytes
* 1 β€ hdd β€ 500 is the HDD in gigabytes
* 100 β€ cost β€ 1000 is price in tugriks
All laptops have different prices.
Output
Print a single number β the number of a laptop Vasya will choose. The laptops are numbered with positive integers from 1 to n in the order in which they are given in the input data.
Examples
Input
5
2100 512 150 200
2000 2048 240 350
2300 1024 200 320
2500 2048 80 300
2000 512 180 150
Output
4
Note
In the third sample Vasya considers the first and fifth laptops outdated as all of their properties cannot match those of the third laptop. The fourth one is the cheapest among the laptops that are left. Thus, Vasya chooses the fourth laptop.
Submitted Solution:
```
n = int(input())
lap = [list(map(int,input().split())) for _ in range(n) ]
outdated = []
for i in range(n):
for j in range(n):
if lap[i][0] > lap[j][0] and lap[i][1] > lap[j][1] and lap[i][2] > lap[j][2]:
outdated.append(j)
q = len(outdated)
out = 0
min_c = float("inf")
min_i = 0
for i in range(n):
if out < q and outdated[out] == i:
out += 1
else:
if lap[i][3] < min_c:
min_c = lap[i][3]
min_i = i
print(i)
```
|
instruction
| 0
| 14,169
| 11
| 28,338
|
No
|
output
| 1
| 14,169
| 11
| 28,339
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To help those contestants who struggle a lot in contests, the headquarters of Codeforces are planning to introduce Division 5. In this new division, the tags of all problems will be announced prior to the round to help the contestants.
The contest consists of n problems, where the tag of the i-th problem is denoted by an integer a_i.
You want to AK (solve all problems). To do that, you must solve the problems in some order. To make the contest funnier, you created extra limitations on yourself. You do not want to solve two problems consecutively with the same tag since it is boring. Also, you are afraid of big jumps in difficulties while solving them, so you want to minimize the number of times that you solve two problems consecutively that are not adjacent in the contest order.
Formally, your solve order can be described by a permutation p of length n. The cost of a permutation is defined as the number of indices i (1β€ i<n) where |p_{i+1}-p_i|>1. You have the requirement that a_{p_i}β a_{p_{i+1}} for all 1β€ i< n.
You want to know the minimum possible cost of permutation that satisfies the requirement. If no permutations meet this requirement, you should report about it.
Input
The first line contains a single integer t (1β€ tβ€ 10^4) β the number of test cases.
The first line of the description of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of problems in the contest.
The next line contains n integers a_1,a_2,β¦ a_n (1 β€ a_i β€ n) β the tags of the problems.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, if there are no permutations that satisfy the required condition, print -1. Otherwise, print the minimum possible cost of a permutation that satisfies the required condition.
Example
Input
4
6
2 1 2 3 1 1
5
1 1 1 2 2
8
7 7 2 7 7 1 8 7
10
1 2 3 4 1 1 2 3 4 1
Output
1
3
-1
2
Note
In the first test case, let p=[5, 4, 3, 2, 1, 6]. The cost is 1 because we jump from p_5=1 to p_6=6, and |6-1|>1. This permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 1.
In the second test case, let p=[1,5,2,4,3]. The cost is 3 because |p_2-p_1|>1, |p_3-p_2|>1, and |p_4-p_3|>1. The permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 3.
In the third test case, for any order of solving the problems, we will solve two problems with the same tag consecutively, so the answer is -1.
Submitted Solution:
```
import sys
input = lambda: sys.stdin.readline().rstrip()
def calc(A):
N = len(A)
if N == 1:
return 0
X = [0] * N
for a in A:
X[a] += 1
if max(X) > (N + 1) // 2:
return -1
Y = [0] * N
Y[A[0]] += 1
Y[A[-1]] += 1
for a, b in zip(A, A[1:]):
if a == b:
Y[a] += 2
su, ma = sum(Y), max(Y)
cc = su - ma
return su // 2 - 1 + max(ma - cc - 2, 0) // 2
T = int(input())
for _ in range(T):
N = int(input())
A = [int(a) - 1 for a in input().split()]
print(calc(A))
```
|
instruction
| 0
| 14,377
| 11
| 28,754
|
Yes
|
output
| 1
| 14,377
| 11
| 28,755
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To help those contestants who struggle a lot in contests, the headquarters of Codeforces are planning to introduce Division 5. In this new division, the tags of all problems will be announced prior to the round to help the contestants.
The contest consists of n problems, where the tag of the i-th problem is denoted by an integer a_i.
You want to AK (solve all problems). To do that, you must solve the problems in some order. To make the contest funnier, you created extra limitations on yourself. You do not want to solve two problems consecutively with the same tag since it is boring. Also, you are afraid of big jumps in difficulties while solving them, so you want to minimize the number of times that you solve two problems consecutively that are not adjacent in the contest order.
Formally, your solve order can be described by a permutation p of length n. The cost of a permutation is defined as the number of indices i (1β€ i<n) where |p_{i+1}-p_i|>1. You have the requirement that a_{p_i}β a_{p_{i+1}} for all 1β€ i< n.
You want to know the minimum possible cost of permutation that satisfies the requirement. If no permutations meet this requirement, you should report about it.
Input
The first line contains a single integer t (1β€ tβ€ 10^4) β the number of test cases.
The first line of the description of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of problems in the contest.
The next line contains n integers a_1,a_2,β¦ a_n (1 β€ a_i β€ n) β the tags of the problems.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, if there are no permutations that satisfy the required condition, print -1. Otherwise, print the minimum possible cost of a permutation that satisfies the required condition.
Example
Input
4
6
2 1 2 3 1 1
5
1 1 1 2 2
8
7 7 2 7 7 1 8 7
10
1 2 3 4 1 1 2 3 4 1
Output
1
3
-1
2
Note
In the first test case, let p=[5, 4, 3, 2, 1, 6]. The cost is 1 because we jump from p_5=1 to p_6=6, and |6-1|>1. This permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 1.
In the second test case, let p=[1,5,2,4,3]. The cost is 3 because |p_2-p_1|>1, |p_3-p_2|>1, and |p_4-p_3|>1. The permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 3.
In the third test case, for any order of solving the problems, we will solve two problems with the same tag consecutively, so the answer is -1.
Submitted Solution:
```
def calc(A):
N = len(A)
if N == 1:
return 0
X = [0] * N
for a in A:
X[a] += 1
if max(X) > (N + 1) // 2:
return -1
Y = [0] * N
Y[A[0]] += 1
Y[A[-1]] += 1
for a, b in zip(A, A[1:]):
if a == b:
Y[a] += 2
su, ma = sum(Y), max(Y)
cc = su - ma
return su // 2 - 1 + max(ma - cc - 2, 0) // 2
T = int(input())
for _ in range(T):
N = int(input())
A = [int(a) - 1 for a in input().split()]
print(calc(A))
```
|
instruction
| 0
| 14,378
| 11
| 28,756
|
Yes
|
output
| 1
| 14,378
| 11
| 28,757
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To help those contestants who struggle a lot in contests, the headquarters of Codeforces are planning to introduce Division 5. In this new division, the tags of all problems will be announced prior to the round to help the contestants.
The contest consists of n problems, where the tag of the i-th problem is denoted by an integer a_i.
You want to AK (solve all problems). To do that, you must solve the problems in some order. To make the contest funnier, you created extra limitations on yourself. You do not want to solve two problems consecutively with the same tag since it is boring. Also, you are afraid of big jumps in difficulties while solving them, so you want to minimize the number of times that you solve two problems consecutively that are not adjacent in the contest order.
Formally, your solve order can be described by a permutation p of length n. The cost of a permutation is defined as the number of indices i (1β€ i<n) where |p_{i+1}-p_i|>1. You have the requirement that a_{p_i}β a_{p_{i+1}} for all 1β€ i< n.
You want to know the minimum possible cost of permutation that satisfies the requirement. If no permutations meet this requirement, you should report about it.
Input
The first line contains a single integer t (1β€ tβ€ 10^4) β the number of test cases.
The first line of the description of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of problems in the contest.
The next line contains n integers a_1,a_2,β¦ a_n (1 β€ a_i β€ n) β the tags of the problems.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, if there are no permutations that satisfy the required condition, print -1. Otherwise, print the minimum possible cost of a permutation that satisfies the required condition.
Example
Input
4
6
2 1 2 3 1 1
5
1 1 1 2 2
8
7 7 2 7 7 1 8 7
10
1 2 3 4 1 1 2 3 4 1
Output
1
3
-1
2
Note
In the first test case, let p=[5, 4, 3, 2, 1, 6]. The cost is 1 because we jump from p_5=1 to p_6=6, and |6-1|>1. This permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 1.
In the second test case, let p=[1,5,2,4,3]. The cost is 3 because |p_2-p_1|>1, |p_3-p_2|>1, and |p_4-p_3|>1. The permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 3.
In the third test case, for any order of solving the problems, we will solve two problems with the same tag consecutively, so the answer is -1.
Submitted Solution:
```
import sys
input = sys.stdin.readline
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
c = [0]*(n+1)
for e in a:
c[e]+=1
if max(c) >= (n+1)//2 +1:
print(-1)
continue
count = 1
s=0
c = [0]*(n+1)
for i in range(n-1):
if a[i]==a[i+1]:
count +=1
c[a[s]]+=1
c[a[i]]+=1
s = i+1
c[a[s]]+=1
c[a[n-1]]+=1
mx = max(c)
ss = sum(c)
if mx-2 <= ss-mx :
print(count-1)
else:
count += (mx-2-(ss-mx))//2
print(count-1)
```
|
instruction
| 0
| 14,379
| 11
| 28,758
|
Yes
|
output
| 1
| 14,379
| 11
| 28,759
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To help those contestants who struggle a lot in contests, the headquarters of Codeforces are planning to introduce Division 5. In this new division, the tags of all problems will be announced prior to the round to help the contestants.
The contest consists of n problems, where the tag of the i-th problem is denoted by an integer a_i.
You want to AK (solve all problems). To do that, you must solve the problems in some order. To make the contest funnier, you created extra limitations on yourself. You do not want to solve two problems consecutively with the same tag since it is boring. Also, you are afraid of big jumps in difficulties while solving them, so you want to minimize the number of times that you solve two problems consecutively that are not adjacent in the contest order.
Formally, your solve order can be described by a permutation p of length n. The cost of a permutation is defined as the number of indices i (1β€ i<n) where |p_{i+1}-p_i|>1. You have the requirement that a_{p_i}β a_{p_{i+1}} for all 1β€ i< n.
You want to know the minimum possible cost of permutation that satisfies the requirement. If no permutations meet this requirement, you should report about it.
Input
The first line contains a single integer t (1β€ tβ€ 10^4) β the number of test cases.
The first line of the description of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of problems in the contest.
The next line contains n integers a_1,a_2,β¦ a_n (1 β€ a_i β€ n) β the tags of the problems.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, if there are no permutations that satisfy the required condition, print -1. Otherwise, print the minimum possible cost of a permutation that satisfies the required condition.
Example
Input
4
6
2 1 2 3 1 1
5
1 1 1 2 2
8
7 7 2 7 7 1 8 7
10
1 2 3 4 1 1 2 3 4 1
Output
1
3
-1
2
Note
In the first test case, let p=[5, 4, 3, 2, 1, 6]. The cost is 1 because we jump from p_5=1 to p_6=6, and |6-1|>1. This permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 1.
In the second test case, let p=[1,5,2,4,3]. The cost is 3 because |p_2-p_1|>1, |p_3-p_2|>1, and |p_4-p_3|>1. The permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 3.
In the third test case, for any order of solving the problems, we will solve two problems with the same tag consecutively, so the answer is -1.
Submitted Solution:
```
import sys,io,os;Z=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
o=[]
for _ in range(int(Z())):
n=int(Z());a=[*map(int,Z().split())]
cn=p=a[0];pn=d=0;e=[0]*n;b=[0]*n;c=[0]*n
for i in range(n):
if a[i]==p:
if pn:
if pn!=p:b[pn-1]+=1
else:e[p-1]+=1
b[p-1]+=1;d+=1
pn=p
p=a[i];c[p-1]+=1
if pn!=p:b[pn-1]+=1
else:e[p-1]+=1
b[p-1]+=1
if 2*max(c)-1>n:o.append('-1');continue
m=0;s=sum(b)
for i in range(n):
dl=max(0,e[i]-2-d+b[i]);m=max(dl,m)
o.append(str(d+m))
print('\n'.join(o))
```
|
instruction
| 0
| 14,380
| 11
| 28,760
|
Yes
|
output
| 1
| 14,380
| 11
| 28,761
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To help those contestants who struggle a lot in contests, the headquarters of Codeforces are planning to introduce Division 5. In this new division, the tags of all problems will be announced prior to the round to help the contestants.
The contest consists of n problems, where the tag of the i-th problem is denoted by an integer a_i.
You want to AK (solve all problems). To do that, you must solve the problems in some order. To make the contest funnier, you created extra limitations on yourself. You do not want to solve two problems consecutively with the same tag since it is boring. Also, you are afraid of big jumps in difficulties while solving them, so you want to minimize the number of times that you solve two problems consecutively that are not adjacent in the contest order.
Formally, your solve order can be described by a permutation p of length n. The cost of a permutation is defined as the number of indices i (1β€ i<n) where |p_{i+1}-p_i|>1. You have the requirement that a_{p_i}β a_{p_{i+1}} for all 1β€ i< n.
You want to know the minimum possible cost of permutation that satisfies the requirement. If no permutations meet this requirement, you should report about it.
Input
The first line contains a single integer t (1β€ tβ€ 10^4) β the number of test cases.
The first line of the description of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of problems in the contest.
The next line contains n integers a_1,a_2,β¦ a_n (1 β€ a_i β€ n) β the tags of the problems.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, if there are no permutations that satisfy the required condition, print -1. Otherwise, print the minimum possible cost of a permutation that satisfies the required condition.
Example
Input
4
6
2 1 2 3 1 1
5
1 1 1 2 2
8
7 7 2 7 7 1 8 7
10
1 2 3 4 1 1 2 3 4 1
Output
1
3
-1
2
Note
In the first test case, let p=[5, 4, 3, 2, 1, 6]. The cost is 1 because we jump from p_5=1 to p_6=6, and |6-1|>1. This permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 1.
In the second test case, let p=[1,5,2,4,3]. The cost is 3 because |p_2-p_1|>1, |p_3-p_2|>1, and |p_4-p_3|>1. The permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 3.
In the third test case, for any order of solving the problems, we will solve two problems with the same tag consecutively, so the answer is -1.
Submitted Solution:
```
import sys
input = sys.stdin.readline
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
c = [0]*(n+1)
for e in a:
c[e]+=1
if max(c) >= (n+1)//2 +1:
print(-1)
continue
count = 1
s=0
c = [0]*(n+1)
for i in range(n-1):
if a[i]==a[i+1]:
count +=1
if a[s]!=a[i]:
c[a[s]]+=1
c[a[i]]+=1
else:
c[a[i]]+=1
s = i+1
if a[s]!=a[n-1]:
c[a[s]]+=1
c[a[n-1]]+=1
else:
c[a[s]]+=1
mx = max(c)
ss = sum(c)
if mx < (ss+1)//2 +1:
print(count-1)
else:
while mx >= (ss+1)//2 +1:
count +=1
ss += 2
print(count-1)
```
|
instruction
| 0
| 14,381
| 11
| 28,762
|
No
|
output
| 1
| 14,381
| 11
| 28,763
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To help those contestants who struggle a lot in contests, the headquarters of Codeforces are planning to introduce Division 5. In this new division, the tags of all problems will be announced prior to the round to help the contestants.
The contest consists of n problems, where the tag of the i-th problem is denoted by an integer a_i.
You want to AK (solve all problems). To do that, you must solve the problems in some order. To make the contest funnier, you created extra limitations on yourself. You do not want to solve two problems consecutively with the same tag since it is boring. Also, you are afraid of big jumps in difficulties while solving them, so you want to minimize the number of times that you solve two problems consecutively that are not adjacent in the contest order.
Formally, your solve order can be described by a permutation p of length n. The cost of a permutation is defined as the number of indices i (1β€ i<n) where |p_{i+1}-p_i|>1. You have the requirement that a_{p_i}β a_{p_{i+1}} for all 1β€ i< n.
You want to know the minimum possible cost of permutation that satisfies the requirement. If no permutations meet this requirement, you should report about it.
Input
The first line contains a single integer t (1β€ tβ€ 10^4) β the number of test cases.
The first line of the description of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of problems in the contest.
The next line contains n integers a_1,a_2,β¦ a_n (1 β€ a_i β€ n) β the tags of the problems.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, if there are no permutations that satisfy the required condition, print -1. Otherwise, print the minimum possible cost of a permutation that satisfies the required condition.
Example
Input
4
6
2 1 2 3 1 1
5
1 1 1 2 2
8
7 7 2 7 7 1 8 7
10
1 2 3 4 1 1 2 3 4 1
Output
1
3
-1
2
Note
In the first test case, let p=[5, 4, 3, 2, 1, 6]. The cost is 1 because we jump from p_5=1 to p_6=6, and |6-1|>1. This permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 1.
In the second test case, let p=[1,5,2,4,3]. The cost is 3 because |p_2-p_1|>1, |p_3-p_2|>1, and |p_4-p_3|>1. The permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 3.
In the third test case, for any order of solving the problems, we will solve two problems with the same tag consecutively, so the answer is -1.
Submitted Solution:
```
from sys import stdin
input = stdin.readline
q = int(input())
for _ in range(q):
n = int(input())
l = list(map(int,input().split()))
ile = [0] * (n+1)
for i in l:
ile[i] += 1
if max(ile) >= (n+1)//2 + 1:
print(-1)
else:
if n == 1:
print(0)
else:
cyk = []
count = 0
cur = l[0]
for i in range(n):
if l[i] != cur:
cyk.append([cur,count])
count = 1
cur = l[i]
else:
count += 1
if count != 1:
cyk.append([cur, count])
if l[-1] != l[-2]:
cyk.append([cur,count])
wyn = 0
for i in cyk:
wyn += max(0, i[1]-1)
if cyk[0][1] == 1 and cyk[-1][1] == 1 and cyk[0][0] == cyk[-1][0]:
for j in range(1, len(cyk)-1):
if cyk[j][0] == cyk[0][0] and cyk[j][1] > 1:
wyn += 1
break
print(wyn)
```
|
instruction
| 0
| 14,382
| 11
| 28,764
|
No
|
output
| 1
| 14,382
| 11
| 28,765
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To help those contestants who struggle a lot in contests, the headquarters of Codeforces are planning to introduce Division 5. In this new division, the tags of all problems will be announced prior to the round to help the contestants.
The contest consists of n problems, where the tag of the i-th problem is denoted by an integer a_i.
You want to AK (solve all problems). To do that, you must solve the problems in some order. To make the contest funnier, you created extra limitations on yourself. You do not want to solve two problems consecutively with the same tag since it is boring. Also, you are afraid of big jumps in difficulties while solving them, so you want to minimize the number of times that you solve two problems consecutively that are not adjacent in the contest order.
Formally, your solve order can be described by a permutation p of length n. The cost of a permutation is defined as the number of indices i (1β€ i<n) where |p_{i+1}-p_i|>1. You have the requirement that a_{p_i}β a_{p_{i+1}} for all 1β€ i< n.
You want to know the minimum possible cost of permutation that satisfies the requirement. If no permutations meet this requirement, you should report about it.
Input
The first line contains a single integer t (1β€ tβ€ 10^4) β the number of test cases.
The first line of the description of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of problems in the contest.
The next line contains n integers a_1,a_2,β¦ a_n (1 β€ a_i β€ n) β the tags of the problems.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, if there are no permutations that satisfy the required condition, print -1. Otherwise, print the minimum possible cost of a permutation that satisfies the required condition.
Example
Input
4
6
2 1 2 3 1 1
5
1 1 1 2 2
8
7 7 2 7 7 1 8 7
10
1 2 3 4 1 1 2 3 4 1
Output
1
3
-1
2
Note
In the first test case, let p=[5, 4, 3, 2, 1, 6]. The cost is 1 because we jump from p_5=1 to p_6=6, and |6-1|>1. This permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 1.
In the second test case, let p=[1,5,2,4,3]. The cost is 3 because |p_2-p_1|>1, |p_3-p_2|>1, and |p_4-p_3|>1. The permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 3.
In the third test case, for any order of solving the problems, we will solve two problems with the same tag consecutively, so the answer is -1.
Submitted Solution:
```
from sys import stdin
input = stdin.readline
q = int(input())
for _ in range(q):
n = int(input())
l = list(map(int,input().split()))
ile = [0] * (n+1)
for i in l:
ile[i] += 1
if max(ile) >= (n+1)//2 + 1:
print(-1)
else:
if n == 1:
print(0)
else:
frag = []
now = [l[0]]
for i in range(1,n):
if l[i] == l[i-1]:
frag.append(now)
now = [l[i]]
else:
now.append(l[i])
frag.append(now)
konce = []
for i in frag:
if i[0] == i[-1]:
konce.append(i[0])
k = len(frag)
d = {}
for i in konce:
d[i] = 0
for i in konce:
d[i] += 1
m = 0
for i in d:
m = max(m, d[i])
dif = max(0, 2*m-k-1)
print(k + dif-1)
```
|
instruction
| 0
| 14,383
| 11
| 28,766
|
No
|
output
| 1
| 14,383
| 11
| 28,767
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
To help those contestants who struggle a lot in contests, the headquarters of Codeforces are planning to introduce Division 5. In this new division, the tags of all problems will be announced prior to the round to help the contestants.
The contest consists of n problems, where the tag of the i-th problem is denoted by an integer a_i.
You want to AK (solve all problems). To do that, you must solve the problems in some order. To make the contest funnier, you created extra limitations on yourself. You do not want to solve two problems consecutively with the same tag since it is boring. Also, you are afraid of big jumps in difficulties while solving them, so you want to minimize the number of times that you solve two problems consecutively that are not adjacent in the contest order.
Formally, your solve order can be described by a permutation p of length n. The cost of a permutation is defined as the number of indices i (1β€ i<n) where |p_{i+1}-p_i|>1. You have the requirement that a_{p_i}β a_{p_{i+1}} for all 1β€ i< n.
You want to know the minimum possible cost of permutation that satisfies the requirement. If no permutations meet this requirement, you should report about it.
Input
The first line contains a single integer t (1β€ tβ€ 10^4) β the number of test cases.
The first line of the description of each test case contains a single integer n (1 β€ n β€ 10^5) β the number of problems in the contest.
The next line contains n integers a_1,a_2,β¦ a_n (1 β€ a_i β€ n) β the tags of the problems.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case, if there are no permutations that satisfy the required condition, print -1. Otherwise, print the minimum possible cost of a permutation that satisfies the required condition.
Example
Input
4
6
2 1 2 3 1 1
5
1 1 1 2 2
8
7 7 2 7 7 1 8 7
10
1 2 3 4 1 1 2 3 4 1
Output
1
3
-1
2
Note
In the first test case, let p=[5, 4, 3, 2, 1, 6]. The cost is 1 because we jump from p_5=1 to p_6=6, and |6-1|>1. This permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 1.
In the second test case, let p=[1,5,2,4,3]. The cost is 3 because |p_2-p_1|>1, |p_3-p_2|>1, and |p_4-p_3|>1. The permutation is valid because we don't solve problems with the same tag twice in a row. We cannot find a permutation with a cost smaller than 3.
In the third test case, for any order of solving the problems, we will solve two problems with the same tag consecutively, so the answer is -1.
Submitted Solution:
```
import sys
input = sys.stdin.readline
outL = []
t = int(input())
for _ in range(t):
n = int(input())
l = list(map(lambda x: int(x) - 1, input().split()))
intervals = []
start = 0
for i in range(1,n):
if l[i] == l[i-1]:
intervals.append((start, i-1))
start = i
intervals.append((start,n-1))
count = [0] * n
for le,ri in intervals:
count[l[le]] += 1
count[l[ri]] += 1
want = len(intervals)-1
if max(count) <= want+2:
outL.append(want)
continue
count2 = [0] * n
for v in l:
count2[v] += 1
for issue in range(n):
if count2[issue] == max(count2):
break
num_issue = max(count2)
rest = n - issue
best = max(count) - 2
if num_issue > rest + 1:
outL.append(-1)
else:
outL.append(best)
print('\n'.join(map(str,outL)))
```
|
instruction
| 0
| 14,384
| 11
| 28,768
|
No
|
output
| 1
| 14,384
| 11
| 28,769
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
People do many crazy things to stand out in a crowd. Some of them dance, some learn by heart rules of Russian language, some try to become an outstanding competitive programmers, while others collect funny math objects.
Alis is among these collectors. Right now she wants to get one of k-special tables. In case you forget, the table n Γ n is called k-special if the following three conditions are satisfied:
* every integer from 1 to n2 appears in the table exactly once;
* in each row numbers are situated in increasing order;
* the sum of numbers in the k-th column is maximum possible.
Your goal is to help Alice and find at least one k-special table of size n Γ n. Both rows and columns are numbered from 1 to n, with rows numbered from top to bottom and columns numbered from left to right.
Input
The first line of the input contains two integers n and k (1 β€ n β€ 500, 1 β€ k β€ n) β the size of the table Alice is looking for and the column that should have maximum possible sum.
Output
First print the sum of the integers in the k-th column of the required table.
Next n lines should contain the description of the table itself: first line should contains n elements of the first row, second line should contain n elements of the second row and so on.
If there are multiple suitable table, you are allowed to print any.
Examples
Input
4 1
Output
28
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Input
5 3
Output
85
5 6 17 18 19
9 10 23 24 25
7 8 20 21 22
3 4 14 15 16
1 2 11 12 13
Submitted Solution:
```
def main():
(n, k) = (int(x) for x in input().split())
(total, table) = solver(n, k)
print(total)
for row in range(n):
for col in range(n):
print(table[row][col], end = " ")
print()
def solver(n, k):
col = k - 1
smallest = col * n + 1
table = [list(range(col * row + 1, col * (row + 1) + 1)) +
list(range((n - col) * row + smallest, (n - col) * (row + 1) + smallest))
for row in range(n)]
total = n * smallest + n * (n - 1) * (n - col) // 2
return (total, table)
#print(solver(5, 2))
main()
```
|
instruction
| 0
| 14,648
| 11
| 29,296
|
Yes
|
output
| 1
| 14,648
| 11
| 29,297
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
People do many crazy things to stand out in a crowd. Some of them dance, some learn by heart rules of Russian language, some try to become an outstanding competitive programmers, while others collect funny math objects.
Alis is among these collectors. Right now she wants to get one of k-special tables. In case you forget, the table n Γ n is called k-special if the following three conditions are satisfied:
* every integer from 1 to n2 appears in the table exactly once;
* in each row numbers are situated in increasing order;
* the sum of numbers in the k-th column is maximum possible.
Your goal is to help Alice and find at least one k-special table of size n Γ n. Both rows and columns are numbered from 1 to n, with rows numbered from top to bottom and columns numbered from left to right.
Input
The first line of the input contains two integers n and k (1 β€ n β€ 500, 1 β€ k β€ n) β the size of the table Alice is looking for and the column that should have maximum possible sum.
Output
First print the sum of the integers in the k-th column of the required table.
Next n lines should contain the description of the table itself: first line should contains n elements of the first row, second line should contain n elements of the second row and so on.
If there are multiple suitable table, you are allowed to print any.
Examples
Input
4 1
Output
28
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Input
5 3
Output
85
5 6 17 18 19
9 10 23 24 25
7 8 20 21 22
3 4 14 15 16
1 2 11 12 13
Submitted Solution:
```
def main():
n, k = [int(t) for t in input().split()]
k -= 1
lst = [[0 for t in range(n)] for t in range(n)]
for col in range(n-1, k-1, -1):
for row in range(n):
lst[row][col] = n*n - (n-1-col) - row*(n-k)
for col in range(k-1, -1, -1):
for row in range(n):
lst[row][col] = n*k - (k-1-col) - row*k
print(sum([lst[i][k] for i in range(n)]))
out = "\n".join(" ".join(str(v) for v in lst[i]) for i in range(n))
print(out)
main()
# PYTHON3!!
```
|
instruction
| 0
| 14,649
| 11
| 29,298
|
Yes
|
output
| 1
| 14,649
| 11
| 29,299
|
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