text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a given sequence $A = \\{a_0, a_1, ..., a_{n-1}\\}$ which is sorted by ascending order, find a specific value $k$ given as a query.
Constraints
* $1 \leq n \leq 100,000$
* $1 \leq q \leq 200,000$
* $0 \leq a_0 \leq a_1 \leq ... \leq a_{n-1} \leq 1,000,000,000$
* $0 \leq k_i \leq 1,000,000,000$
Input
The input is given in the following format.
$n$
$a_0 \; a_1 \; ,..., \; a_{n-1}$
$q$
$k_1$
$k_2$
:
$k_q$
The number of elements $n$ and each element $a_i$ are given in the first line and the second line respectively. In the third line, the number of queries $q$ is given and the following $q$ lines, $q$ integers $k_i$ are given as queries.
Output
For each query, print 1 if any element in $A$ is equivalent to $k$, and 0 otherwise.
Example
Input
4
1 2 2 4
3
2
3
5
Output
1
0
0
Submitted Solution:
```
# http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ITP2_6_A&lang=jp
# Binary Search : python3
# 2018.11.26 yonezawa
import sys
input = sys.stdin.readline
def main():
n = int(input())
l = list(map(int,input().split()))
for i in range(int(input())):
max = n -1
min = 0
mid = max // 2
d = int(input())
while True:
if (l[mid] > d ):
max = mid
elif (l[mid] == d ):
print ("1")
break
else:
min = mid
mid = (max + min ) // 2
if ( mid == max or mid == min ):
if ( l[mid] == d or l[max] == d):
print ("1")
break
#print ("debug:",min,mid,max,l[mid])
print ("0")
break
if __name__ == '__main__':
main()
```
Yes
| 13,300 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a given sequence $A = \\{a_0, a_1, ..., a_{n-1}\\}$ which is sorted by ascending order, find a specific value $k$ given as a query.
Constraints
* $1 \leq n \leq 100,000$
* $1 \leq q \leq 200,000$
* $0 \leq a_0 \leq a_1 \leq ... \leq a_{n-1} \leq 1,000,000,000$
* $0 \leq k_i \leq 1,000,000,000$
Input
The input is given in the following format.
$n$
$a_0 \; a_1 \; ,..., \; a_{n-1}$
$q$
$k_1$
$k_2$
:
$k_q$
The number of elements $n$ and each element $a_i$ are given in the first line and the second line respectively. In the third line, the number of queries $q$ is given and the following $q$ lines, $q$ integers $k_i$ are given as queries.
Output
For each query, print 1 if any element in $A$ is equivalent to $k$, and 0 otherwise.
Example
Input
4
1 2 2 4
3
2
3
5
Output
1
0
0
Submitted Solution:
```
import math
n=int(input())
a=list(map(int,input().split()))
q=int(input())
for i in range(q):
k=int(input())
t=0
max=n-1
min=0
while max-min>1:
nn=math.floor((max+min)/2)
if a[nn]==k:
t=1
print("1")
break
elif a[nn]<=k:
min=nn
elif a[nn]>=k:
max=nn
if t==0:
if a[max]==k or a[min]==k:
t=1
print("1")
else:
print("0")
```
Yes
| 13,301 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a given sequence $A = \\{a_0, a_1, ..., a_{n-1}\\}$ which is sorted by ascending order, find a specific value $k$ given as a query.
Constraints
* $1 \leq n \leq 100,000$
* $1 \leq q \leq 200,000$
* $0 \leq a_0 \leq a_1 \leq ... \leq a_{n-1} \leq 1,000,000,000$
* $0 \leq k_i \leq 1,000,000,000$
Input
The input is given in the following format.
$n$
$a_0 \; a_1 \; ,..., \; a_{n-1}$
$q$
$k_1$
$k_2$
:
$k_q$
The number of elements $n$ and each element $a_i$ are given in the first line and the second line respectively. In the third line, the number of queries $q$ is given and the following $q$ lines, $q$ integers $k_i$ are given as queries.
Output
For each query, print 1 if any element in $A$ is equivalent to $k$, and 0 otherwise.
Example
Input
4
1 2 2 4
3
2
3
5
Output
1
0
0
Submitted Solution:
```
def solve():
from sys import stdin
f_i = stdin
n = int(f_i.readline())
A = list(map(int, f_i.readline().split()))
A.append(1000000001)
q = int(f_i.readline())
from bisect import bisect_left
ans = []
for i in range(q):
k = int(f_i.readline())
if k == A[bisect_left(A, k)]:
ans.append('1')
else:
ans.append('0')
print('\n'.join(ans))
solve()
```
Yes
| 13,302 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a given sequence $A = \\{a_0, a_1, ..., a_{n-1}\\}$ which is sorted by ascending order, find a specific value $k$ given as a query.
Constraints
* $1 \leq n \leq 100,000$
* $1 \leq q \leq 200,000$
* $0 \leq a_0 \leq a_1 \leq ... \leq a_{n-1} \leq 1,000,000,000$
* $0 \leq k_i \leq 1,000,000,000$
Input
The input is given in the following format.
$n$
$a_0 \; a_1 \; ,..., \; a_{n-1}$
$q$
$k_1$
$k_2$
:
$k_q$
The number of elements $n$ and each element $a_i$ are given in the first line and the second line respectively. In the third line, the number of queries $q$ is given and the following $q$ lines, $q$ integers $k_i$ are given as queries.
Output
For each query, print 1 if any element in $A$ is equivalent to $k$, and 0 otherwise.
Example
Input
4
1 2 2 4
3
2
3
5
Output
1
0
0
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split(' ')))
def binary_search(val):
left, right = 0, n-1
while right - left > 1:
mid = (right+left)//2
if a[mid] == val: return 1
if a[mid] < val:
left = mid
elif val < a[mid]:
right = mid
return 0
q = int(input())
for i in range(q):
k = int(input())
print(binary_search(k))
```
No
| 13,303 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a given sequence $A = \\{a_0, a_1, ..., a_{n-1}\\}$ which is sorted by ascending order, find a specific value $k$ given as a query.
Constraints
* $1 \leq n \leq 100,000$
* $1 \leq q \leq 200,000$
* $0 \leq a_0 \leq a_1 \leq ... \leq a_{n-1} \leq 1,000,000,000$
* $0 \leq k_i \leq 1,000,000,000$
Input
The input is given in the following format.
$n$
$a_0 \; a_1 \; ,..., \; a_{n-1}$
$q$
$k_1$
$k_2$
:
$k_q$
The number of elements $n$ and each element $a_i$ are given in the first line and the second line respectively. In the third line, the number of queries $q$ is given and the following $q$ lines, $q$ integers $k_i$ are given as queries.
Output
For each query, print 1 if any element in $A$ is equivalent to $k$, and 0 otherwise.
Example
Input
4
1 2 2 4
3
2
3
5
Output
1
0
0
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split(' ')))
def binary_search(val):
left, right = 0, n-1
while right - left >= 1:
mid = (right+left)//2
if a[mid] == val: return 1
if a[mid] < val:
left = mid
elif val < a[mid]:
right = mid
return 0
q = int(input())
for i in range(q):
k = int(input())
print(binary_search(k))
```
No
| 13,304 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a given sequence $A = \\{a_0, a_1, ..., a_{n-1}\\}$ which is sorted by ascending order, find a specific value $k$ given as a query.
Constraints
* $1 \leq n \leq 100,000$
* $1 \leq q \leq 200,000$
* $0 \leq a_0 \leq a_1 \leq ... \leq a_{n-1} \leq 1,000,000,000$
* $0 \leq k_i \leq 1,000,000,000$
Input
The input is given in the following format.
$n$
$a_0 \; a_1 \; ,..., \; a_{n-1}$
$q$
$k_1$
$k_2$
:
$k_q$
The number of elements $n$ and each element $a_i$ are given in the first line and the second line respectively. In the third line, the number of queries $q$ is given and the following $q$ lines, $q$ integers $k_i$ are given as queries.
Output
For each query, print 1 if any element in $A$ is equivalent to $k$, and 0 otherwise.
Example
Input
4
1 2 2 4
3
2
3
5
Output
1
0
0
Submitted Solution:
```
input()
b = list(map(int,input().split()))
def nibu(low,high,i):
middle = (low+high)//2
if b[middle] <= i:
low = middle
else:
high = middle
if low == high:
if i == b[low]:
return 1
else:
return 0
a = int(input())
for i in range(a):
k = int(input())
print(nibu(0,len(b)-1,k))
```
No
| 13,305 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a given sequence $A = \\{a_0, a_1, ..., a_{n-1}\\}$ which is sorted by ascending order, find a specific value $k$ given as a query.
Constraints
* $1 \leq n \leq 100,000$
* $1 \leq q \leq 200,000$
* $0 \leq a_0 \leq a_1 \leq ... \leq a_{n-1} \leq 1,000,000,000$
* $0 \leq k_i \leq 1,000,000,000$
Input
The input is given in the following format.
$n$
$a_0 \; a_1 \; ,..., \; a_{n-1}$
$q$
$k_1$
$k_2$
:
$k_q$
The number of elements $n$ and each element $a_i$ are given in the first line and the second line respectively. In the third line, the number of queries $q$ is given and the following $q$ lines, $q$ integers $k_i$ are given as queries.
Output
For each query, print 1 if any element in $A$ is equivalent to $k$, and 0 otherwise.
Example
Input
4
1 2 2 4
3
2
3
5
Output
1
0
0
Submitted Solution:
```
a = int(input())
b = list(map(int,input().split()))
def nibu(low,high,i):
middle = (low+high)//2
if b[middle] <= i:
low = middle
else:
high = middle
if low == high:
if i == b[low]:
return 1
else:
return 0
for i in range(a):
k = int(input())
print(nibu(0,len(b),k))
```
No
| 13,306 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party.
Tags: brute force, greedy
Correct Solution:
```
n, m = map(int, input().split())
parties = [0 for _ in range(m)]
pairs = []
for i in range(n):
p, c = map(int, input().split())
parties[p - 1] += 1
pairs.append((c, p - 1))
pairs.sort()
min_sum = 10 ** 31
for i in range(n, parties[0] - 1, -1):
cur_sum = 0
cur_amount = parties[0]
colours = [0 for _ in range(n)]
need_to_buy = [max(parties[j] + 1 - i, 0) for j in range(m)]
need_to_buy[0] = 0
tmp = sum(need_to_buy)
if tmp + parties[0] > i:
print(min_sum)
exit()
for j in range(n):
if need_to_buy[pairs[j][1]] > 0:
need_to_buy[pairs[j][1]] -= 1
colours[j] = 1
cur_amount += 1
cur_sum += pairs[j][0]
j = 0
while cur_amount < i:
if colours[j] == 0 and pairs[j][1] != 0:
cur_amount += 1
cur_sum += pairs[j][0]
j += 1
min_sum = min(min_sum, cur_sum)
print(min_sum)
```
| 13,307 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party.
Tags: brute force, greedy
Correct Solution:
```
#!/usr/bin/python3
def solve(N, M, A):
pv = [0] * M
pm = [[] for _ in range(M)]
allm = {}
for (p, c) in A:
p -= 1
pv[p] += 1
if p == 0:
continue
pm[p].append(c)
if c not in allm:
allm[c] = 0
allm[c] += 1
for m in pm:
m.sort()
allml = list(allm)
allml.sort()
maxv = max(pv)
best = 10 ** 9 * 3000 + 1
for iv in range(1, min(maxv + 2, N + 1)):
allmrest = dict(allm)
cost = 0
reqd = 0
if pv[0] < iv:
reqd = iv - pv[0]
got = 0
for i in range(1, M):
if pv[i] >= iv:
ad = pv[i] - iv + 1
for j in range(ad):
cost += pm[i][j]
allmrest[pm[i][j]] -= 1
got += ad
if reqd > got:
for m in allml:
c = allmrest[m]
toget = min(reqd - got, c)
cost += toget * m
got += toget
if reqd == got:
break
if reqd > got:
continue
best = min(best, cost)
return best
def main():
N, M = [int(e) for e in input().split(' ')]
A = [[int(e) for e in input().split(' ')] for _ in range(N)]
print(solve(N, M, A))
if __name__ == '__main__':
main()
```
| 13,308 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party.
Tags: brute force, greedy
Correct Solution:
```
import sys
n,m=map(int,input().split())
V=[None]*n
for i in range(n):
V[i]=tuple(map(int,input().split()))
V.sort(key=lambda x:x[1])
if n==1:
if V[0][0]==1:
print(0)
else:
print(V[0][1])
sys.exit()
Voterlist=[[] for i in range(m+1)]
Partylist=[0 for i in range(m+1)]
Vnot1=[]
for i in range(n):
Voterlist[V[i][0]]+=[V[i][1]]
Partylist[V[i][0]]+=1
if V[i][0]!=1:
Vnot1.append(V[i])
Vnot1.sort(key=lambda x:x[1])
maxvote=max(Partylist)
Votenumberlist=[[] for i in range(maxvote+1)]
for i in range(1,m+1):
Votenumberlist[Partylist[i]].append(i)
if Votenumberlist[maxvote]==[1]:
print(0)
sys.exit()
#print(Votenumberlist,maxvote,Votenumberlist[maxvote])
ANS=0
for i in range(maxvote+1-Partylist[1]):
ANS+=Vnot1[i][1]
maxvoteminus=1
while True:
checkparty=[0 for i in range(m+1)]
money=0
for i in range(maxvoteminus):
for j in Votenumberlist[maxvote-i]:
checkparty[j]+=maxvoteminus-i
#print(checkparty)
for i in range(2,m+1):
money+=sum(Voterlist[i][:checkparty[i]])
if money>ANS:
break
neednumber=maxvote+1-maxvoteminus-Partylist[1]
for i in range(maxvoteminus):
neednumber-=len(Votenumberlist[maxvote-i])*(maxvoteminus-i)
#print(checkparty,neednumber,money)
for pm in Vnot1:
if checkparty[pm[0]]!=0:
checkparty[pm[0]]-=1
else:
money+=pm[1]
neednumber-=1
if neednumber==0:
break
#print(money,ANS)
if money<ANS:
ANS=money
maxvoteminus+=1
print(ANS)
```
| 13,309 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party.
Tags: brute force, greedy
Correct Solution:
```
import sys
#sys.stdin=open("data.txt")
input=sys.stdin.readline
n,m=map(int,input().split())
party=[[] for _ in range(m+5)]
pc=sorted([list(map(int,input().split())) for _ in range(n)],key=lambda x:x[1])
choose=[0]*n
for i in range(n):
party[pc[i][0]].append(i)
want=10**18
for i in range(1,n+1):
p1=len(party[1])
# want all other parties to have <i voters
for j in range(2,m+5):
if len(party[j])<i: continue
for k in range(len(party[j])-i+1):
p1+=1
choose[party[j][k]]=1
# want party 1 to have >=i voters
want2=0
for j in range(n):
if p1<i and choose[j]==0 and pc[j][0]!=1:
choose[j]=1
p1+=1
if choose[j]==1:
want2+=pc[j][1]
if want>want2:
want=want2
#print(i,want2)
# reset
choose=[0]*n
print(want)
```
| 13,310 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party.
Tags: brute force, greedy
Correct Solution:
```
import sys
import os
def solve(m, candidates):
n = len(candidates)
candidates.sort(key=lambda x: x[1])
party = dict()
granted = 0
for i in range(len(candidates)):
p = candidates[i][0]
c = candidates[i][1]
if p == 1:
granted += 1
continue
if p in party:
party[p].append((i, c))
else:
party[p] = [(i, c)]
result = None
for t in range(1, n // 2 + 2):
total = 0
chosen = set()
for k, v in party.items():
if len(v) >= t:
for i in range(len(v) + 1 - t):
chosen.add(v[i][0])
total += v[i][1]
for i in range(n):
if len(chosen) + granted >= t:
break
if i not in chosen and candidates[i][0] != 1:
chosen.add(i)
total += candidates[i][1]
if result is None:
result = total
else:
result = min(result, total)
return result
def main():
n, m = map(int, input().split())
candidates = []
for i in range(n):
p, c = map(int, input().split())
candidates.append((p, c))
print(solve(m, candidates))
if __name__ == '__main__':
main()
```
| 13,311 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party.
Tags: brute force, greedy
Correct Solution:
```
class Solver:
def solve(self):
self.num_voters, self.num_parties = (int(x) for x in input().split())
self.votes_per_party = [[] for _ in range(self.num_parties)]
for _ in range(self.num_voters):
party, price = (int(x) for x in input().split())
party -= 1
self.votes_per_party[party].append(price)
for party in range(self.num_parties):
self.votes_per_party[party].sort()
max_necessary_votes = self.num_voters//2 + 1
cost = lambda min_votes : self.conversion_price_with_fixed_votes(min_votes)
return self.ternary_search(cost, 0, max_necessary_votes + 2)
def ternary_search(self, func, begin, end):
while begin + 1 < end:
mid = (begin + end - 1) // 2
if func(mid) <= func(mid + 1):
end = mid + 1
else:
begin = mid + 1
return func(begin)
def conversion_price_with_fixed_votes(self, num_votes):
current_votes = len(self.votes_per_party[0])
total_cost = 0
for votes in self.votes_per_party[1:]:
if len(votes) >= num_votes:
num_bought_votes = len(votes) - num_votes + 1
total_cost += sum(votes[:num_bought_votes])
current_votes += num_bought_votes
if current_votes >= num_votes:
return total_cost
num_votes_to_buy = num_votes - current_votes
votes_left = []
for party in range(1, self.num_parties):
votes_left += self.votes_per_party[party][-(num_votes-1):]
votes_left.sort()
return total_cost + sum(votes_left[:num_votes_to_buy])
solver = Solver()
min_price = solver.solve()
print(min_price)
```
| 13,312 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party.
Tags: brute force, greedy
Correct Solution:
```
from sys import stdin
from collections import deque
n,m = map(int, stdin.readline().split())
pc=dict()
ph=dict()
costs=[]
class Cost:
def __init__(self, id, cost, party) -> None:
self.id=id
self.cost=cost
self.party=party
self.removed=False
for i in range(n):
p, c = map(int,stdin.readline().split())
if p != 1:
cost = Cost(i, c, p)
costs.append(cost)
pc.setdefault(p,[]).append(cost)
ph[p] = ph.setdefault(p, 0) + 1
max_h=0
for p, pa in pc.items():
pa.sort(reverse=True, key=lambda x: x.cost)
max_h = max(max_h, len(pa))
hp = [[] for _ in range(max_h+2)]
for p, h in ph.items():
if p != 1:
hp[h].append(p)
ans=[0]*(max_h+2)
costs.sort(key=lambda x:x.cost)
dq = deque(costs)
p_set = set()
height_p1 = ph[1] if 1 in ph else 0
top_sum = 0
top_count = 0
for i in range(max_h+1, -1, -1):
if len(costs) < i:
ans[i] = float('inf')
continue
for p in hp[i]:
p_set.add(p)
for p in p_set:
if len(pc[p]) > 0:
min_cost = pc[p].pop()
min_cost.removed=True
top_sum += min_cost.cost
top_count += 1
ans[i] += top_sum
if height_p1+top_count < i:
for j in range(i-height_p1-top_count):
while dq[0].removed: dq.popleft()
ans[i] += dq[0].cost
dq.rotate(-1)
dq.rotate(i-height_p1-top_count)
print(min(ans))
```
| 13,313 |
Provide tags and a correct Python 3 solution for this coding contest problem.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party.
Tags: brute force, greedy
Correct Solution:
```
n,m=map(int,input().split())
men=[]
for i in range(n):
x,y=map(int,input().split())
men.append((y,x-1))
def Calc(lim):
cnt=[0]*m
vis=[False]*n
for i in range(n):
cnt[men[i][1]]+=1
cost=0
for i in range(n):
if men[i][1]!=0 and cnt[men[i][1]]>=lim:
cnt[men[i][1]]-=1
cost+=men[i][0]
vis[i]=True
cnt[0]+=1
for i in range(n):
if cnt[0]<lim and vis[i] == False and men[i][1]!=0:
cnt[0]+=1
cost+=men[i][0]
return cost
men.sort()
ans = 10**18
for i in range(n):
ans=min(ans,Calc(i))
print(ans)
```
| 13,314 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party.
Submitted Solution:
```
import sys
from collections import defaultdict
n, m = map(int, sys.stdin.readline().rstrip('\n').split(' '))
p = defaultdict(list)
for _ in range(n):
x, y = map(int, sys.stdin.readline().rstrip('\n').split(' '))
p[x].append(y)
for key in p:
p[key] = sorted(p[key])
ans = 10**100
for k in range(1, n + 1):
cur = 0
r = []
for key in p:
if key == 1:
continue
a = p[key]
cnt = max(0, len(a) - (k - 1))
cur += sum(a[:cnt])
r += a[cnt:]
r = sorted(r)
cnt = max(0, len(r) - (n - k))
cur += sum(r[:cnt])
ans = min(ans, cur)
print(ans)
```
Yes
| 13,315 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party.
Submitted Solution:
```
import sys
def pro():
return sys.stdin.readline().strip()
def rop():
return map(int, pro().split())
a, s = rop()
q = [0] * a
w = [0] * s
for i in range(a):
n, p = rop()
q[i] = (p, n-1)
w[n - 1] += 1
q.sort()
t = 1e100
for k in range(1, a + 1):
p = w[::]
r = [True] * a
m = 0
for i in range(a):
po, rty = q[i]
if rty != 0 and p[rty] >= k:
p[0] += 1
p[rty] -= 1
m += po
r[i] = False
for i in range(a):
po, rty = q[i]
if rty != 0 and r[i] and p[0] < k:
p[0] += 1
p[rty] -= 1
m += po
r[i] = False
t = min(t, m)
print(t)
```
Yes
| 13,316 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party.
Submitted Solution:
```
import sys
input=sys.stdin.readline
n,m=map(int,input().split())
men=[]
for i in range(n):
x,y=map(int,input().split())
men.append((y,x-1))
def solve(lim):
cnt=[0]*m
vis=[0]*n
cost=0
for i in range(n):
cnt[men[i][1]]+=1
for i in range(n):
if(men[i][1]!=0 and cnt[men[i][1]]>=lim):
cnt[men[i][1]]-=1
cnt[0]+=1
cost+=men[i][0]
vis[i]=1
for i in range(n):
if cnt[0]<lim and vis[i] == False and men[i][1]!=0:
cnt[0]+=1
cost+=men[i][0]
return cost
men.sort()
ans=10**18
for i in range(n):
ans=min(ans,solve(i))
print(ans)
```
Yes
| 13,317 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party.
Submitted Solution:
```
# -*- coding:utf-8 -*-
"""
created by shuangquan.huang at 1/29/19
Let's iterate over final number of votes for The United Party of Berland.
We can see that all opponents should get less votes than our party,
and our party should get at least our chosen number of votes.
We can sort all voters by their costs, and solve the problem in two passes.
First, if we need to get π₯ votes, we should definitely buy all cheap votes for parties that have at least π₯ votes.
Second, if we don't have π₯ votes yet, we should by the cheapest votes to get π₯ votes.
We can see that this solution is optimal: consider the optimal answer, and see how many votes The United Party got.
We tried such number of votes, and we tried to achieve this number of votes by cheapest way, so we couldn't miss the
optimal answer. This can be implemented in π(π2logπ) or even π(πlogπ).
"""
import collections
import time
import os
import sys
import bisect
import heapq
def solve(N, M, A):
votes = [[] for _ in range(M+1)]
for p, v in A:
votes[p].append(v)
for p in range(1, M+1):
votes[p].sort()
own = len(votes[1])
votes = votes[2:]
votes.sort(reverse=True, key=len)
size = [len(v) for v in votes]
if not size or own > size[0]:
return 0
nvotes = len(votes)
ans = float('inf')
for buy in range((size[0]-own)//2+1, min(N, (N+1) // 2 + 1) + 1):
cost = 0
target = own + buy
done = 0
for p in range(nvotes):
if size[p] >= target:
t = size[p] - target + 1
cost += sum(votes[p][: t] or [0])
done += t
else:
break
if done >= buy:
ans = min(ans, cost)
else:
more = buy - done
q = []
for p in range(nvotes):
t = max(size[p] - target + 1, 0)
q.extend(votes[p][t: t+more])
q.sort()
cost += sum(q[:more])
ans = min(ans, cost)
return ans
N, M = map(int, input().split())
A = []
for i in range(N):
p, v = map(int, input().split())
A.append((p, v))
print(solve(N, M, A))
```
Yes
| 13,318 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party.
Submitted Solution:
```
# https://codeforces.com/contest/1019/problem/A
n, m = map(int, input().split())
d = {}
count = 0
for _ in range(n):
p, c = map(int, input().split())
if p == 1:
count += 1
continue
if p not in d:
d[p] = []
d[p].append(c)
for k in d:
d[k] = sorted(d[k])
Min = float('inf')
for i in range(n-count+1):
cur = count + i
remain = []
c = 0
s = 0
for k in d:
if len(d[k]) >= cur:
for j, x in enumerate(d[k]):
if j <= len(d[k]) - cur:
s += x
c += 1
else:
remain.append(x)
else:
remain.extend(d[k])
if c < cur:
remain = sorted(remain)
for x in remain[:cur-c]:
s += x
Min = min(Min, s)
print(Min)
#5 5
#2 100
#3 200
#4 300
#5 800
#5 900
```
No
| 13,319 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party.
Submitted Solution:
```
import math
n,m = map(int,(input().split()))
arr = []
for _ in range(n):
arr.append(list(map(int,input().split())))
arr.sort(key = lambda x : x[0])
arrC= []
flag = -1
for i in range(n):
if (arr[i][0] != 1):
flag = i
break
if(flag == -1):
print(0)
else:
ans = 0
arrC= arr[flag:]
arrC.sort(key = lambda x:x[1])
count = [0]*m
count[0] = flag
for num in arr[flag:]:
count[num[0]-1] += 1
j = 0
while(count[0] <= max(count) ):
maxV = max(count)
valV = 0
for i in range(j,n-flag):
if count[arrC[i][0] -1] == maxV :
valV = arrC[i][1]
break
if(maxV == count[0]):
if(valV == 0 ):
break
else:
ans += arrC[j][1]
break
if(valV > arrC[j][1] + arrC[j+1][1]):
ans += arrC[j][1]
count[arrC[j][0] -1] -= 1
j += 1
count[0] += 1
else:
ans += valV
j += 1
count[arrC[i][0] -1] -= 1
count[0] += 1
print(ans)
```
No
| 13,320 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party.
Submitted Solution:
```
import math
n,m = map(int,(input().split()))
arr = []
for _ in range(n):
arr.append(list(map(int,input().split())))
arr.sort(key = lambda x : x[0])
arrC= []
flag = -1
for i in range(n):
if (arr[i][0] != 1):
flag = i
break
if(flag == -1):
print(0)
else:
ans = 0
arrC= arr[flag:]
arrC.sort(key = lambda x:x[1])
count = [0]*m
count[0] = flag
for num in arr[flag:]:
count[num[0]-1] += 1
j = 0
while(count[0] < max(count) - 1):
ans += arrC[j][1]
count[arrC[j][0] -1] -= 1
count[0] += 1
j += 1
val = 0
maxV = max(count)
pos = -1
valV = 0
for i in range(1,m):
if maxV == count[i]:
pos = i
val += 1
if(maxV != count[0]):
if(val == 1):
for i in range(j,n):
if arrC[i][0] == pos+1 :
valV = arr[i][1]
break
if(valV < arrC[j][1] + arr[j+1][1]):
ans += valV
else:
ans += arrC[j][1] + arr[j+1][1]
else:
ans += arrC[j][1]+arr[j+1][1]
print(ans)
else:
if(val == 0):
print(0)
else:
print(arrC[j][1])
```
No
| 13,321 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year. Since corruption there is quite widespread, the following story is not uncommon.
Elections are coming. You know the number of voters and the number of parties β n and m respectively. For each voter you know the party he is going to vote for. However, he can easily change his vote given a certain amount of money. In particular, if you give i-th voter c_i bytecoins you can ask him to vote for any other party you choose.
The United Party of Berland has decided to perform a statistical study β you need to calculate the minimum number of bytecoins the Party needs to spend to ensure its victory. In order for a party to win the elections, it needs to receive strictly more votes than any other party.
Input
The first line of input contains two integers n and m (1 β€ n, m β€ 3000) β the number of voters and the number of parties respectively.
Each of the following n lines contains two integers p_i and c_i (1 β€ p_i β€ m, 1 β€ c_i β€ 10^9) β the index of this voter's preferred party and the number of bytecoins needed for him to reconsider his decision.
The United Party of Berland has the index 1.
Output
Print a single number β the minimum number of bytecoins needed for The United Party of Berland to win the elections.
Examples
Input
1 2
1 100
Output
0
Input
5 5
2 100
3 200
4 300
5 400
5 900
Output
500
Input
5 5
2 100
3 200
4 300
5 800
5 900
Output
600
Note
In the first sample, The United Party wins the elections even without buying extra votes.
In the second sample, The United Party can buy the votes of the first and the fourth voter. This way The Party gets two votes, while parties 3, 4 and 5 get one vote and party number 2 gets no votes.
In the third sample, The United Party can buy the votes of the first three voters and win, getting three votes against two votes of the fifth party.
Submitted Solution:
```
n, m = map(int, input().split())
cnt = {}
un = 0
arr = [[] for _ in range(m+1)]
for i in range(n):
p, c = map(int, input().split())
if p != 1:
if p not in cnt:
cnt[p] = 1
else:
cnt[p] += 1
arr[p].append(c)
else:
un += 1
for i in range(m+1):
arr[i].sort(reverse=True)
total = 0
while cnt:
if un > max(cnt.values()):
break
mn1 = 1000000001
mn1_index = -1
for i in range(1, m+1):
if arr[i] and arr[i][-1] < mn1:
mn1 = arr[i][-1]
mn1_index = i
mn2 = 1000000001
mn2_index = -1
for i in range(1, m+1):
if arr[i] and i != mn1_index and arr[i][-1] < mn2:
mn2 = arr[i][-1]
mn2_index = i
if len(arr[mn1_index]) > 1 and arr[mn1_index][-2] < mn2:
mn2 = arr[mn1_index][-2]
mn2_index = mn1_index
mx = 0
mx_index = -1
for x in cnt:
if cnt[x] > mx:
mx = cnt[x]
mx_index = x
# print(mn1,mn1_index,mn2,mx_index,mn2_index,arr[mx_index][-1])
if cnt[mx_index] > un:
if mn1+mn2 < arr[mx_index][-1]:
total += mn1
arr[mn1_index].pop()
cnt[mn1_index] -= 1
else:
total += arr[mx_index][-1]
arr[mx_index].pop()
cnt[mx_index] -= 1
else:
if mn1 < arr[mx_index][-1]:
total += mn1
arr[mn1_index].pop()
cnt[mn1_index] -= 1
else:
total += arr[mx_index][-1]
arr[mx_index].pop()
cnt[mx_index] -= 1
un += 1
print(total)
```
No
| 13,322 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a tube which is reflective inside represented as two non-coinciding, but parallel to Ox lines. Each line has some special integer points β positions of sensors on sides of the tube.
You are going to emit a laser ray in the tube. To do so, you have to choose two integer points A and B on the first and the second line respectively (coordinates can be negative): the point A is responsible for the position of the laser, and the point B β for the direction of the laser ray. The laser ray is a ray starting at A and directed at B which will reflect from the sides of the tube (it doesn't matter if there are any sensors at a reflection point or not). A sensor will only register the ray if the ray hits exactly at the position of the sensor.
<image> Examples of laser rays. Note that image contains two examples. The 3 sensors (denoted by black bold points on the tube sides) will register the blue ray but only 2 will register the red.
Calculate the maximum number of sensors which can register your ray if you choose points A and B on the first and the second lines respectively.
Input
The first line contains two integers n and y_1 (1 β€ n β€ 10^5, 0 β€ y_1 β€ 10^9) β number of sensors on the first line and its y coordinate.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9) β x coordinates of the sensors on the first line in the ascending order.
The third line contains two integers m and y_2 (1 β€ m β€ 10^5, y_1 < y_2 β€ 10^9) β number of sensors on the second line and its y coordinate.
The fourth line contains m integers b_1, b_2, β¦, b_m (0 β€ b_i β€ 10^9) β x coordinates of the sensors on the second line in the ascending order.
Output
Print the only integer β the maximum number of sensors which can register the ray.
Example
Input
3 1
1 5 6
1 3
3
Output
3
Note
One of the solutions illustrated on the image by pair A_2 and B_2.
Tags: data structures, divide and conquer, dp, math
Correct Solution:
```
n, y1 = map(int, input().split())
a = list(map(int, input().split()))
m, y2 = map(int, input().split())
b = list(map(int, input().split()))
a_st, b_st = dict(), dict()
osn = 2 ** 30
k_a, k_b = set(), set()
for el in a:
try:
a_st[el % osn] += 1
except KeyError:
a_st[el % osn] = 1
finally:
k_a.add(el % osn)
for el in b:
try:
b_st[el % osn] += 1
except KeyError:
b_st[el % osn] = 1
finally:
k_b.add(el % osn)
ans = 2
for i in range(1, 31)[::-1]:
temp = 0
for el in k_a:
try:
temp = max(temp, a_st[el] + b_st[(el + osn // 2) % osn])
except KeyError:
temp = max(temp, a_st[el])
for el in k_b:
try:
temp = max(temp, b_st[el] + a_st[(el + osn // 2) % osn])
except KeyError:
temp = max(temp, b_st[el])
ans = max(ans, temp)
osn //= 2
k_ = set()
k_add = set()
for el in k_a:
if el >= osn:
try:
a_st[el - osn] += a_st[el]
except KeyError:
a_st[el - osn] = a_st[el]
finally:
del a_st[el]
k_add.add(el - osn)
k_.add(el)
for el in k_:
k_a.remove(el)
for el in k_add:
k_a.add(el)
k_ = set()
k_add = set()
for el in k_b:
if el >= osn:
try:
b_st[el - osn] += b_st[el]
except KeyError:
b_st[el - osn] = b_st[el]
finally:
del b_st[el]
k_add.add(el - osn)
k_.add(el)
for el in k_:
k_b.remove(el)
for el in k_add:
k_b.add(el)
print(ans)
```
| 13,323 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a tube which is reflective inside represented as two non-coinciding, but parallel to Ox lines. Each line has some special integer points β positions of sensors on sides of the tube.
You are going to emit a laser ray in the tube. To do so, you have to choose two integer points A and B on the first and the second line respectively (coordinates can be negative): the point A is responsible for the position of the laser, and the point B β for the direction of the laser ray. The laser ray is a ray starting at A and directed at B which will reflect from the sides of the tube (it doesn't matter if there are any sensors at a reflection point or not). A sensor will only register the ray if the ray hits exactly at the position of the sensor.
<image> Examples of laser rays. Note that image contains two examples. The 3 sensors (denoted by black bold points on the tube sides) will register the blue ray but only 2 will register the red.
Calculate the maximum number of sensors which can register your ray if you choose points A and B on the first and the second lines respectively.
Input
The first line contains two integers n and y_1 (1 β€ n β€ 10^5, 0 β€ y_1 β€ 10^9) β number of sensors on the first line and its y coordinate.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9) β x coordinates of the sensors on the first line in the ascending order.
The third line contains two integers m and y_2 (1 β€ m β€ 10^5, y_1 < y_2 β€ 10^9) β number of sensors on the second line and its y coordinate.
The fourth line contains m integers b_1, b_2, β¦, b_m (0 β€ b_i β€ 10^9) β x coordinates of the sensors on the second line in the ascending order.
Output
Print the only integer β the maximum number of sensors which can register the ray.
Example
Input
3 1
1 5 6
1 3
3
Output
3
Note
One of the solutions illustrated on the image by pair A_2 and B_2.
Tags: data structures, divide and conquer, dp, math
Correct Solution:
```
#coding:utf-8
n, _ = map (int, input().strip ().split ())
a = list (map (int, input().strip ().split ()))
m, _ = map (int, input().strip ().split ())
b = list (map (int, input().strip ().split ()))
def solve (T):
global a, b
d = dict()
for i in b:
pos = i%T
if pos in d:
d[pos] += 1
else:
d[pos] = 1
for i in a:
pos = (i+T/2)%T
if pos in d:
d[pos] += 1
else:
d[pos] = 1
return max (d.values())
T = 2
ans = 0
while True:
ans = max (ans, solve (T))
T *= 2
if T > 1e9:
break
if len (set (a) and set (b)) > 0:
ans = max (ans, 2)
print (ans)
```
| 13,324 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a tube which is reflective inside represented as two non-coinciding, but parallel to Ox lines. Each line has some special integer points β positions of sensors on sides of the tube.
You are going to emit a laser ray in the tube. To do so, you have to choose two integer points A and B on the first and the second line respectively (coordinates can be negative): the point A is responsible for the position of the laser, and the point B β for the direction of the laser ray. The laser ray is a ray starting at A and directed at B which will reflect from the sides of the tube (it doesn't matter if there are any sensors at a reflection point or not). A sensor will only register the ray if the ray hits exactly at the position of the sensor.
<image> Examples of laser rays. Note that image contains two examples. The 3 sensors (denoted by black bold points on the tube sides) will register the blue ray but only 2 will register the red.
Calculate the maximum number of sensors which can register your ray if you choose points A and B on the first and the second lines respectively.
Input
The first line contains two integers n and y_1 (1 β€ n β€ 10^5, 0 β€ y_1 β€ 10^9) β number of sensors on the first line and its y coordinate.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9) β x coordinates of the sensors on the first line in the ascending order.
The third line contains two integers m and y_2 (1 β€ m β€ 10^5, y_1 < y_2 β€ 10^9) β number of sensors on the second line and its y coordinate.
The fourth line contains m integers b_1, b_2, β¦, b_m (0 β€ b_i β€ 10^9) β x coordinates of the sensors on the second line in the ascending order.
Output
Print the only integer β the maximum number of sensors which can register the ray.
Example
Input
3 1
1 5 6
1 3
3
Output
3
Note
One of the solutions illustrated on the image by pair A_2 and B_2.
Tags: data structures, divide and conquer, dp, math
Correct Solution:
```
n, y1 = map(int, input().split())
a = list(map(int, input().split()))
m, y2 = map(int, input().split())
b = list(map(int, input().split()))
a_st, b_st = dict(), dict()
osn = 2 ** 32
k_a, k_b = set(), set()
for el in a:
try:
a_st[el % osn] += 1
except KeyError:
a_st[el % osn] = 1
finally:
k_a.add(el % osn)
for el in b:
try:
b_st[el % osn] += 1
except KeyError:
b_st[el % osn] = 1
finally:
k_b.add(el % osn)
ans = 2
for i in range(1, 33)[::-1]:
temp = 0
for el in k_a:
try:
temp = max(temp, a_st[el] + b_st[(el + osn // 2) % osn])
except KeyError:
temp = max(temp, a_st[el])
for el in k_b:
try:
temp = max(temp, b_st[el] + a_st[(el + osn // 2) % osn])
except KeyError:
temp = max(temp, b_st[el])
ans = max(ans, temp)
osn //= 2
k_ = set()
k_add = set()
for el in k_a:
if el >= osn:
try:
a_st[el - osn] += a_st[el]
except KeyError:
a_st[el - osn] = a_st[el]
finally:
del a_st[el]
k_add.add(el - osn)
k_.add(el)
for el in k_:
k_a.remove(el)
for el in k_add:
k_a.add(el)
k_ = set()
k_add = set()
for el in k_b:
if el >= osn:
try:
b_st[el - osn] += b_st[el]
except KeyError:
b_st[el - osn] = b_st[el]
finally:
del b_st[el]
k_add.add(el - osn)
k_.add(el)
for el in k_:
k_b.remove(el)
for el in k_add:
k_b.add(el)
print(ans)
```
| 13,325 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a tube which is reflective inside represented as two non-coinciding, but parallel to Ox lines. Each line has some special integer points β positions of sensors on sides of the tube.
You are going to emit a laser ray in the tube. To do so, you have to choose two integer points A and B on the first and the second line respectively (coordinates can be negative): the point A is responsible for the position of the laser, and the point B β for the direction of the laser ray. The laser ray is a ray starting at A and directed at B which will reflect from the sides of the tube (it doesn't matter if there are any sensors at a reflection point or not). A sensor will only register the ray if the ray hits exactly at the position of the sensor.
<image> Examples of laser rays. Note that image contains two examples. The 3 sensors (denoted by black bold points on the tube sides) will register the blue ray but only 2 will register the red.
Calculate the maximum number of sensors which can register your ray if you choose points A and B on the first and the second lines respectively.
Input
The first line contains two integers n and y_1 (1 β€ n β€ 10^5, 0 β€ y_1 β€ 10^9) β number of sensors on the first line and its y coordinate.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9) β x coordinates of the sensors on the first line in the ascending order.
The third line contains two integers m and y_2 (1 β€ m β€ 10^5, y_1 < y_2 β€ 10^9) β number of sensors on the second line and its y coordinate.
The fourth line contains m integers b_1, b_2, β¦, b_m (0 β€ b_i β€ 10^9) β x coordinates of the sensors on the second line in the ascending order.
Output
Print the only integer β the maximum number of sensors which can register the ray.
Example
Input
3 1
1 5 6
1 3
3
Output
3
Note
One of the solutions illustrated on the image by pair A_2 and B_2.
Tags: data structures, divide and conquer, dp, math
Correct Solution:
```
n, y1 = map(int, input().split())
a = list(map(int, input().split()))
m, y2 = map(int, input().split())
b = list(map(int, input().split()))
ans = 2
for i in range(30):
step = (2 << i)
half = (step >> 1)
counts = dict()
for x in a:
rem = x % step
count = counts.get(rem, 0)
counts[rem] = count + 1
for x in b:
rem = (x + half) % step
count = counts.get(rem, 0)
counts[rem] = count + 1
res = max(counts.values())
ans = max(ans, res)
print(ans)
```
| 13,326 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a tube which is reflective inside represented as two non-coinciding, but parallel to Ox lines. Each line has some special integer points β positions of sensors on sides of the tube.
You are going to emit a laser ray in the tube. To do so, you have to choose two integer points A and B on the first and the second line respectively (coordinates can be negative): the point A is responsible for the position of the laser, and the point B β for the direction of the laser ray. The laser ray is a ray starting at A and directed at B which will reflect from the sides of the tube (it doesn't matter if there are any sensors at a reflection point or not). A sensor will only register the ray if the ray hits exactly at the position of the sensor.
<image> Examples of laser rays. Note that image contains two examples. The 3 sensors (denoted by black bold points on the tube sides) will register the blue ray but only 2 will register the red.
Calculate the maximum number of sensors which can register your ray if you choose points A and B on the first and the second lines respectively.
Input
The first line contains two integers n and y_1 (1 β€ n β€ 10^5, 0 β€ y_1 β€ 10^9) β number of sensors on the first line and its y coordinate.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9) β x coordinates of the sensors on the first line in the ascending order.
The third line contains two integers m and y_2 (1 β€ m β€ 10^5, y_1 < y_2 β€ 10^9) β number of sensors on the second line and its y coordinate.
The fourth line contains m integers b_1, b_2, β¦, b_m (0 β€ b_i β€ 10^9) β x coordinates of the sensors on the second line in the ascending order.
Output
Print the only integer β the maximum number of sensors which can register the ray.
Example
Input
3 1
1 5 6
1 3
3
Output
3
Note
One of the solutions illustrated on the image by pair A_2 and B_2.
Tags: data structures, divide and conquer, dp, math
Correct Solution:
```
n, y1 = map(int, input().split())
a = list(map(int, input().split()))
m, y2 = map(int, input().split())
b = list(map(int, input().split()))
a_st, b_st = dict(), dict()
osn = 2 ** 29
k_a, k_b = set(), set()
for el in a:
try:
a_st[el % osn] += 1
except KeyError:
a_st[el % osn] = 1
finally:
k_a.add(el % osn)
for el in b:
try:
b_st[el % osn] += 1
except KeyError:
b_st[el % osn] = 1
finally:
k_b.add(el % osn)
ans = 2
for i in range(1, 30)[::-1]:
temp = 0
for el in k_a:
try:
temp = max(temp, a_st[el] + b_st[(el + osn // 2) % osn])
except KeyError:
temp = max(temp, a_st[el])
for el in k_b:
try:
temp = max(temp, b_st[el] + a_st[(el + osn // 2) % osn])
except KeyError:
temp = max(temp, b_st[el])
ans = max(ans, temp)
osn //= 2
k_ = set()
k_add = set()
for el in k_a:
if el >= osn:
try:
a_st[el - osn] += a_st[el]
except KeyError:
a_st[el - osn] = a_st[el]
finally:
del a_st[el]
k_add.add(el - osn)
k_.add(el)
for el in k_:
k_a.remove(el)
for el in k_add:
k_a.add(el)
k_ = set()
k_add = set()
for el in k_b:
if el >= osn:
try:
b_st[el - osn] += b_st[el]
except KeyError:
b_st[el - osn] = b_st[el]
finally:
del b_st[el]
k_add.add(el - osn)
k_.add(el)
for el in k_:
k_b.remove(el)
for el in k_add:
k_b.add(el)
print(ans)
```
| 13,327 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a tube which is reflective inside represented as two non-coinciding, but parallel to Ox lines. Each line has some special integer points β positions of sensors on sides of the tube.
You are going to emit a laser ray in the tube. To do so, you have to choose two integer points A and B on the first and the second line respectively (coordinates can be negative): the point A is responsible for the position of the laser, and the point B β for the direction of the laser ray. The laser ray is a ray starting at A and directed at B which will reflect from the sides of the tube (it doesn't matter if there are any sensors at a reflection point or not). A sensor will only register the ray if the ray hits exactly at the position of the sensor.
<image> Examples of laser rays. Note that image contains two examples. The 3 sensors (denoted by black bold points on the tube sides) will register the blue ray but only 2 will register the red.
Calculate the maximum number of sensors which can register your ray if you choose points A and B on the first and the second lines respectively.
Input
The first line contains two integers n and y_1 (1 β€ n β€ 10^5, 0 β€ y_1 β€ 10^9) β number of sensors on the first line and its y coordinate.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9) β x coordinates of the sensors on the first line in the ascending order.
The third line contains two integers m and y_2 (1 β€ m β€ 10^5, y_1 < y_2 β€ 10^9) β number of sensors on the second line and its y coordinate.
The fourth line contains m integers b_1, b_2, β¦, b_m (0 β€ b_i β€ 10^9) β x coordinates of the sensors on the second line in the ascending order.
Output
Print the only integer β the maximum number of sensors which can register the ray.
Example
Input
3 1
1 5 6
1 3
3
Output
3
Note
One of the solutions illustrated on the image by pair A_2 and B_2.
Tags: data structures, divide and conquer, dp, math
Correct Solution:
```
n,w56=map(int,input().split())
if(n==1):
x=[int(input())]
else:
x=[int(i) for i in input().split()]
k,w90=map(int,input().split())
if(k==1):
y=[int(input())]
else:
y=[int(i) for i in input().split()]
t=0
import collections
for i in range(1,35):
m=collections.Counter()
for j in range(n):
m[((x[j]-(2**(i-1)))%(2**i))]+=1
for j in range(k):
m[(y[j])%(2**i)]+=1
t=max(t,max([m[o] for o in m.keys()]))
if(t>=2):
print(t)
else:
p=0
for i in range(len(x)):
for j in range(len(y)):
if(x[i]==y[j]):
print(2)
p=1
break
if(p==1):
break
if(p==0):
print(1)
```
| 13,328 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a tube which is reflective inside represented as two non-coinciding, but parallel to Ox lines. Each line has some special integer points β positions of sensors on sides of the tube.
You are going to emit a laser ray in the tube. To do so, you have to choose two integer points A and B on the first and the second line respectively (coordinates can be negative): the point A is responsible for the position of the laser, and the point B β for the direction of the laser ray. The laser ray is a ray starting at A and directed at B which will reflect from the sides of the tube (it doesn't matter if there are any sensors at a reflection point or not). A sensor will only register the ray if the ray hits exactly at the position of the sensor.
<image> Examples of laser rays. Note that image contains two examples. The 3 sensors (denoted by black bold points on the tube sides) will register the blue ray but only 2 will register the red.
Calculate the maximum number of sensors which can register your ray if you choose points A and B on the first and the second lines respectively.
Input
The first line contains two integers n and y_1 (1 β€ n β€ 10^5, 0 β€ y_1 β€ 10^9) β number of sensors on the first line and its y coordinate.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9) β x coordinates of the sensors on the first line in the ascending order.
The third line contains two integers m and y_2 (1 β€ m β€ 10^5, y_1 < y_2 β€ 10^9) β number of sensors on the second line and its y coordinate.
The fourth line contains m integers b_1, b_2, β¦, b_m (0 β€ b_i β€ 10^9) β x coordinates of the sensors on the second line in the ascending order.
Output
Print the only integer β the maximum number of sensors which can register the ray.
Example
Input
3 1
1 5 6
1 3
3
Output
3
Note
One of the solutions illustrated on the image by pair A_2 and B_2.
Tags: data structures, divide and conquer, dp, math
Correct Solution:
```
x,y = map(int,input().split())
a = list(map(int,input().split()))
o,p = map(int,input().split())
b = list(map(int,input().split()))
ans = 0
from collections import defaultdict
for i in range(34):
fac = 2**i
fac2 = fac*2
if i == 33:
fac = fac2
map = defaultdict(int)
for j in a:
map[j%fac2] += 1
for j in b:
map[(j+fac)%fac2] += 1
for v in map.values():
ans = max(ans,v)
print(ans)
```
| 13,329 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a tube which is reflective inside represented as two non-coinciding, but parallel to Ox lines. Each line has some special integer points β positions of sensors on sides of the tube.
You are going to emit a laser ray in the tube. To do so, you have to choose two integer points A and B on the first and the second line respectively (coordinates can be negative): the point A is responsible for the position of the laser, and the point B β for the direction of the laser ray. The laser ray is a ray starting at A and directed at B which will reflect from the sides of the tube (it doesn't matter if there are any sensors at a reflection point or not). A sensor will only register the ray if the ray hits exactly at the position of the sensor.
<image> Examples of laser rays. Note that image contains two examples. The 3 sensors (denoted by black bold points on the tube sides) will register the blue ray but only 2 will register the red.
Calculate the maximum number of sensors which can register your ray if you choose points A and B on the first and the second lines respectively.
Input
The first line contains two integers n and y_1 (1 β€ n β€ 10^5, 0 β€ y_1 β€ 10^9) β number of sensors on the first line and its y coordinate.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9) β x coordinates of the sensors on the first line in the ascending order.
The third line contains two integers m and y_2 (1 β€ m β€ 10^5, y_1 < y_2 β€ 10^9) β number of sensors on the second line and its y coordinate.
The fourth line contains m integers b_1, b_2, β¦, b_m (0 β€ b_i β€ 10^9) β x coordinates of the sensors on the second line in the ascending order.
Output
Print the only integer β the maximum number of sensors which can register the ray.
Example
Input
3 1
1 5 6
1 3
3
Output
3
Note
One of the solutions illustrated on the image by pair A_2 and B_2.
Tags: data structures, divide and conquer, dp, math
Correct Solution:
```
n, y1 = map(int, input().split())
a = list(map(int, input().split()))
m, y2 = map(int, input().split())
b = list(map(int, input().split()))
ans = 2
for i in range(31):
step = (2 << i)
half = (step >> 1)
counts = dict()
for x in a:
rem = x % step
count = counts.get(rem, 0)
counts[rem] = count + 1
for x in b:
rem = (x + half) % step
count = counts.get(rem, 0)
counts[rem] = count + 1
res = max(counts.values())
ans = max(ans, res)
print(ans)
```
| 13,330 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a tube which is reflective inside represented as two non-coinciding, but parallel to Ox lines. Each line has some special integer points β positions of sensors on sides of the tube.
You are going to emit a laser ray in the tube. To do so, you have to choose two integer points A and B on the first and the second line respectively (coordinates can be negative): the point A is responsible for the position of the laser, and the point B β for the direction of the laser ray. The laser ray is a ray starting at A and directed at B which will reflect from the sides of the tube (it doesn't matter if there are any sensors at a reflection point or not). A sensor will only register the ray if the ray hits exactly at the position of the sensor.
<image> Examples of laser rays. Note that image contains two examples. The 3 sensors (denoted by black bold points on the tube sides) will register the blue ray but only 2 will register the red.
Calculate the maximum number of sensors which can register your ray if you choose points A and B on the first and the second lines respectively.
Input
The first line contains two integers n and y_1 (1 β€ n β€ 10^5, 0 β€ y_1 β€ 10^9) β number of sensors on the first line and its y coordinate.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9) β x coordinates of the sensors on the first line in the ascending order.
The third line contains two integers m and y_2 (1 β€ m β€ 10^5, y_1 < y_2 β€ 10^9) β number of sensors on the second line and its y coordinate.
The fourth line contains m integers b_1, b_2, β¦, b_m (0 β€ b_i β€ 10^9) β x coordinates of the sensors on the second line in the ascending order.
Output
Print the only integer β the maximum number of sensors which can register the ray.
Example
Input
3 1
1 5 6
1 3
3
Output
3
Note
One of the solutions illustrated on the image by pair A_2 and B_2.
Submitted Solution:
```
import math
nsensors1, ycoord1 = [int(x) for x in input().split()] # number of sensors in the first line, y coordinate of the line
sensors_x_coord1 = [int(x) for x in input().split()] # x coordinates of the sensors in the first line
nsensors2, ycoord2 = [int(x) for x in input().split()] # number of sensors in the second line, y coordinate of the line
sensors_x_coord2 = [int(x) for x in input().split()] # x coordinates of the sensors in the second line
# sensor coordinates
sensors1 = []
sensors2 = []
for i in range(nsensors1):
sensors1.append([sensors_x_coord1[i], ycoord1])
for i in range(nsensors2):
sensors2.append([sensors_x_coord2[i], ycoord2])
pointAx, pointAy = [-3, ycoord1]
pointBx, pointBy = [-1, ycoord2]
distanceAB = math.sqrt((pointAx - pointBx) ** 2 + (pointAy - pointBy) ** 2)
angle = math.atan((pointBy - pointAy) / (pointBx - pointAx))
reflection_distance = 2 * distanceAB * math.cos(angle)
sensor_count = 0
for sensor in sensors1:
result = round((sensor[0] - pointAx) / reflection_distance, 2)
if result.is_integer() and ((pointAx < pointBx and result > 0) or (pointAx > pointBx and result < 0)):
sensor_count += 1
for sensor in sensors2:
result = round((sensor[0] - pointBx) / reflection_distance, 2)
if result.is_integer() and ((pointAx < pointBx and result > 0) or (pointAx > pointBx and result < 0)):
sensor_count += 1
print(3)
```
No
| 13,331 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a tube which is reflective inside represented as two non-coinciding, but parallel to Ox lines. Each line has some special integer points β positions of sensors on sides of the tube.
You are going to emit a laser ray in the tube. To do so, you have to choose two integer points A and B on the first and the second line respectively (coordinates can be negative): the point A is responsible for the position of the laser, and the point B β for the direction of the laser ray. The laser ray is a ray starting at A and directed at B which will reflect from the sides of the tube (it doesn't matter if there are any sensors at a reflection point or not). A sensor will only register the ray if the ray hits exactly at the position of the sensor.
<image> Examples of laser rays. Note that image contains two examples. The 3 sensors (denoted by black bold points on the tube sides) will register the blue ray but only 2 will register the red.
Calculate the maximum number of sensors which can register your ray if you choose points A and B on the first and the second lines respectively.
Input
The first line contains two integers n and y_1 (1 β€ n β€ 10^5, 0 β€ y_1 β€ 10^9) β number of sensors on the first line and its y coordinate.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9) β x coordinates of the sensors on the first line in the ascending order.
The third line contains two integers m and y_2 (1 β€ m β€ 10^5, y_1 < y_2 β€ 10^9) β number of sensors on the second line and its y coordinate.
The fourth line contains m integers b_1, b_2, β¦, b_m (0 β€ b_i β€ 10^9) β x coordinates of the sensors on the second line in the ascending order.
Output
Print the only integer β the maximum number of sensors which can register the ray.
Example
Input
3 1
1 5 6
1 3
3
Output
3
Note
One of the solutions illustrated on the image by pair A_2 and B_2.
Submitted Solution:
```
n, y1 = map(int, input().split())
a = list(map(int, input().split()))
m, y2 = map(int, input().split())
b = list(map(int, input().split()))
ans = 0
for i in range(30):
step = (2 << i)
half = (step >> 1)
counts = dict()
for x in a:
rem = x % step
count = counts.get(rem, 0)
counts[rem] = count + 1
for x in b:
rem = (x + half) % step
count = counts.get(rem, 0)
counts[rem] = count + 1
res = max(counts.values())
ans = max(ans, res)
print(ans)
```
No
| 13,332 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a tube which is reflective inside represented as two non-coinciding, but parallel to Ox lines. Each line has some special integer points β positions of sensors on sides of the tube.
You are going to emit a laser ray in the tube. To do so, you have to choose two integer points A and B on the first and the second line respectively (coordinates can be negative): the point A is responsible for the position of the laser, and the point B β for the direction of the laser ray. The laser ray is a ray starting at A and directed at B which will reflect from the sides of the tube (it doesn't matter if there are any sensors at a reflection point or not). A sensor will only register the ray if the ray hits exactly at the position of the sensor.
<image> Examples of laser rays. Note that image contains two examples. The 3 sensors (denoted by black bold points on the tube sides) will register the blue ray but only 2 will register the red.
Calculate the maximum number of sensors which can register your ray if you choose points A and B on the first and the second lines respectively.
Input
The first line contains two integers n and y_1 (1 β€ n β€ 10^5, 0 β€ y_1 β€ 10^9) β number of sensors on the first line and its y coordinate.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9) β x coordinates of the sensors on the first line in the ascending order.
The third line contains two integers m and y_2 (1 β€ m β€ 10^5, y_1 < y_2 β€ 10^9) β number of sensors on the second line and its y coordinate.
The fourth line contains m integers b_1, b_2, β¦, b_m (0 β€ b_i β€ 10^9) β x coordinates of the sensors on the second line in the ascending order.
Output
Print the only integer β the maximum number of sensors which can register the ray.
Example
Input
3 1
1 5 6
1 3
3
Output
3
Note
One of the solutions illustrated on the image by pair A_2 and B_2.
Submitted Solution:
```
import math
# return the distance between two consecutive points
def get_distance(point_a, point_b):
return math.sqrt((point_a[0] - point_b[0]) ** 2 + (point_a[1] - point_b[1]) ** 2)
# return true if the distance between any two consecutive points is constant
def distance_const(sensor_list):
for sensor in range(len(sensor_list) - 2):
if get_distance(sensor[i], sensor[i + 1]) != get_distance(sensor[i + 1], sensor[i + 2]):
return False
return [True, get_distance(sensor_list[0], sensor_list[1])]
def get_slope(point_a, point_b):
return (point_b[1] - point_a[1]) / (point_b[0] - point_a[0])
nsensors1, ycoord1 = [int(x) for x in input().split()] # number of sensors in the first line, y coordinate of the line
sensors_x_coord1 = [int(x) for x in input().split()] # x coordinates of the sensors in the first line
nsensors2, ycoord2 = [int(x) for x in input().split()] # number of sensors in the second line, y coordinate of the line
sensors_x_coord2 = [int(x) for x in input().split()] # x coordinates of the sensors in the second line
# sensor coordinates
sensors1 = []
sensors2 = []
for i in range(nsensors1):
sensors1.append([sensors_x_coord1[i], ycoord1])
for i in range(nsensors2):
sensors2.append([sensors_x_coord2[i], ycoord2])
# distance from point n to point n + 1 on both horizontal lines
distance1 = []
distance2 = []
for i in range(len(sensors1) - 1):
distance1.append(get_distance(sensors1[i], sensors1[i + 1]))
# if there are no sensors on a given line and the distance between sensors on the opposite line is constant and greater
# than 1, the maximum number of hit sensors is equal to the number of sensors in the line that has them
if len(sensors_x_coord1) == 0 and distance_const(sensors2)[0] and distance_const(sensors2)[1] > 1:
print(len(sensors2))
exit(0)
elif len(sensors_x_coord2) == 0 and distance_const(sensors1)[0] and distance_const(sensors1)[1] > 1:
print(len(sensors1))
exit(0)
counter = []
i = 0
# First pass: check distance between consecutive sensors on line1
while i < len(sensors1) - 1:
# if len(sensors2) > 1 and len(sensors1) > 1:
# if get_distance(sensors1[i], sensors1[i + 1]) == 1 or get_distance(sensors2[i], [i + 1]) == 1:
# continue
middlePoint = [(sensors1[i][0] + sensors1[i + 1][0]) / 2, ycoord2]
distance_ab = get_distance(sensors1[i], middlePoint)
angle = math.atan(get_slope(sensors1[i], middlePoint))
distance_consecutive = 2 * distance_ab * math.cos(angle)
point_a = [sensors1[i][0] - distance_consecutive, ycoord1]
point_b = [middlePoint[0] - distance_consecutive, ycoord2]
moving_a = sensors1[0]
moving_b = sensors2[0]
j = 0
while moving_a[0] <= sensors1[-1][0] or moving_b[0] <= sensors2[-1][0]:
moving_a = [round(point_a[0] + abs(distance_consecutive * j)), point_a[1]]
moving_b = [round(point_b[0] + abs(distance_consecutive * j)), point_b[1]]
if moving_a in sensors1:
try:
counter[i] += 1
except IndexError:
counter.append(1)
if moving_b in sensors2:
try:
counter[i] += 1
except IndexError:
counter.append(1)
j += 1
i += 1
if i > 0:
i += 1
for j in range(len(sensors2) - 1):
# if len(sensors2) > 1 and len(sensors1) > 1 and (
# get_distance(sensors1[i], sensors1[i + 1]) == 1 or get_distance(sensors2[i], [i + 1]) == 1):
# continue
middlePoint = [(sensors2[j][0] + sensors2[j + 1][0]) / 2, ycoord1]
distance_ab = get_distance(sensors2[j], middlePoint)
angle = math.atan(get_slope(sensors2[j], middlePoint))
distance_consecutive = 2 * distance_ab * math.cos(angle)
point_a = [sensors2[j][0] - distance_consecutive, ycoord2]
point_b = [middlePoint[0] - distance_consecutive, ycoord1]
moving_a = sensors1[0]
moving_b = sensors2[0]
k = 0
while moving_a[0] <= sensors1[-1][0] or moving_b[0] <= sensors2[-1][0]:
moving_a = [round(point_a[0] + abs(distance_consecutive * k)), point_a[1]]
moving_b = [round(point_b[0] + abs(distance_consecutive * k)), point_b[1]]
if moving_a in sensors2:
try:
counter[i] += 1
except IndexError:
counter.append(1)
if moving_b in sensors1:
try:
counter[i] += 1
except IndexError:
counter.append(1)
k += 1
i += 1
if len(counter) == 0:
print(0)
exit(0)
_max = counter[0]
for i in range(len(counter)):
if counter[i] > _max:
_max = counter[i]
print(_max)
```
No
| 13,333 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a tube which is reflective inside represented as two non-coinciding, but parallel to Ox lines. Each line has some special integer points β positions of sensors on sides of the tube.
You are going to emit a laser ray in the tube. To do so, you have to choose two integer points A and B on the first and the second line respectively (coordinates can be negative): the point A is responsible for the position of the laser, and the point B β for the direction of the laser ray. The laser ray is a ray starting at A and directed at B which will reflect from the sides of the tube (it doesn't matter if there are any sensors at a reflection point or not). A sensor will only register the ray if the ray hits exactly at the position of the sensor.
<image> Examples of laser rays. Note that image contains two examples. The 3 sensors (denoted by black bold points on the tube sides) will register the blue ray but only 2 will register the red.
Calculate the maximum number of sensors which can register your ray if you choose points A and B on the first and the second lines respectively.
Input
The first line contains two integers n and y_1 (1 β€ n β€ 10^5, 0 β€ y_1 β€ 10^9) β number of sensors on the first line and its y coordinate.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^9) β x coordinates of the sensors on the first line in the ascending order.
The third line contains two integers m and y_2 (1 β€ m β€ 10^5, y_1 < y_2 β€ 10^9) β number of sensors on the second line and its y coordinate.
The fourth line contains m integers b_1, b_2, β¦, b_m (0 β€ b_i β€ 10^9) β x coordinates of the sensors on the second line in the ascending order.
Output
Print the only integer β the maximum number of sensors which can register the ray.
Example
Input
3 1
1 5 6
1 3
3
Output
3
Note
One of the solutions illustrated on the image by pair A_2 and B_2.
Submitted Solution:
```
import math
# return the distance between two consecutive points
def get_distance(point_a, point_b):
return math.sqrt((point_a[0] - point_b[0]) ** 2 + (point_a[1] - point_b[1]) ** 2)
def get_slope(point_a, point_b):
return (point_b[1] - point_a[1]) / (point_b[0] - point_a[0])
def generate_coordinates(nsensors, sensor_x, sensor_y, sensor_list):
for i in range(nsensors):
sensor_list.append([sensor_x[i], sensor_y])
def compute_distances(sensors, distances):
for i in range(len(sensors) - 1):
distances.append(get_distance(sensors[i], sensors[i + 1]))
def on_line(sensor, ylower, yupper):
if sensor[1] == ylower:
return 0
elif sensor[1] == yupper:
return 1
# number of sensors in the lower line and y coordinate of the line
nsensors_lower, ycoord_lower = [int(x) for x in input().split()]
# x coordinates of the sensors in the lower line
sensors_x_coord_lower = [int(x) for x in input().split()]
# number of sensors in the upper line, y coordinate of the line
nsensors_upper, ycoord_upper = [int(x) for x in input().split()]
# x coordinates of the sensors in the upper line
sensors_x_coord_upper = [int(x) for x in input().split()]
# sensor coordinates on the upper and lower lines
sensors_upper = []
sensors_lower = []
# distance between two consecutive sensors on the upper and lower lines
distance_sensors_upper = []
distance_sensors_lower = []
# generate coordinates for the sensors in each line
generate_coordinates(nsensors_lower, sensors_x_coord_lower, ycoord_lower, sensors_lower)
generate_coordinates(nsensors_upper, sensors_x_coord_upper, ycoord_upper, sensors_upper)
sensors_sorted = sorted(sensors_upper + sensors_lower)
compute_distances(sensors_upper, distance_sensors_upper)
compute_distances(sensors_lower, distance_sensors_lower)
leftmost_sensor = list(sensors_sorted[0])
start_sensor = list(leftmost_sensor)
counter_lower = []
max_distance_lower = get_distance(sensors_lower[0], sensors_lower[-1])
max_distance_upper = get_distance(sensors_upper[0], sensors_upper[-1])
# if the leftmost sensor is on line 0
if on_line(leftmost_sensor, ycoord_lower, ycoord_upper) == 0:
middlePoint = [round((start_sensor[0] + distance_sensors_lower[0] + 1) / 2), ycoord_upper]
for i in range(len(distance_sensors_lower)):
added_distance = i
while added_distance <= max_distance_lower:
if distance_sensors_lower[i] == 1:
break
if middlePoint in sensors_upper:
try:
counter_lower[i] += 1
except IndexError:
counter_lower.append(1)
if start_sensor in sensors_lower:
try:
counter_lower[i] += 1
except IndexError:
counter_lower.append(1)
added_distance += distance_sensors_lower[i]
middlePoint[0] += distance_sensors_lower[i]
start_sensor[0] += distance_sensors_lower[i]
start_sensor = list(leftmost_sensor)
middlePoint = [round((start_sensor[0] + distance_sensors_lower[i] + 1) / 2), ycoord_upper]
elif on_line(leftmost_sensor, ycoord_lower, ycoord_upper) == 1:
middlePoint = [round(start_sensor[0] + distance_sensors_upper[0] + 1) / 2, ycoord_lower]
for i in range(len(distance_sensors_upper)):
added_distance = i
while added_distance <= max_distance_upper:
if middlePoint in sensors_lower:
try:
counter_lower[i] += 1
except IndexError:
counter_lower.append(1)
if start_sensor in sensors_upper:
try:
counter_lower[i] += 1
except IndexError:
counter_lower.append(1)
added_distance += distance_sensors_upper[i]
middlePoint[0] += distance_sensors_upper[i]
start_sensor[0] += distance_sensors_upper[i]
start_sensor = list(leftmost_sensor)
middlePoint = [round((start_sensor[0] + distance_sensors_upper[i] + 1) / 2), ycoord_lower]
_max = counter_lower[0]
for i in range(len(counter_lower)):
if counter_lower[i] > _max:
_max = counter_lower[i]
print(_max)
```
No
| 13,334 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
In good old times dwarves tried to develop extrasensory abilities:
* Exactly n dwarves entered completely dark cave.
* Each dwarf received a hat β white or black. While in cave, none of the dwarves was able to see either his own hat or hats of other Dwarves.
* Dwarves went out of the cave to the meadow and sat at an arbitrary place one after the other. When a dwarf leaves the cave, he sees the colors of all hats of all dwarves that are seating on the meadow (i.e. left the cave before him). However, he is not able to see the color of his own hat and none of the dwarves can give him this information.
* The task for dwarves was to got diverged into two parts β one with dwarves with white hats and one with black hats.
After many centuries, dwarves finally managed to select the right place on the meadow without error. Will you be able to repeat their success?
You are asked to successively name n different integer points on the plane. After naming each new point you will be given its color β black or white. Your task is to ensure that the named points can be split by a line in such a way that all points of one color lie on the same side from the line and points of different colors lie on different sides. Moreover, no points can belong to the line. Also, you need to report any such line at the end of the process.
In this problem, the interactor is adaptive β the colors of the points in the tests are not fixed beforehand and the jury program can select them arbitrarily, in particular, depending on your program output.
Interaction
The first line of the standard input stream contains an integer n (1 β€ n β€ 30) β the number of points your program should name.
Then n times your program must print two integer coordinates x and y (0 β€ x β€ 109, 0 β€ y β€ 109). All points you print must be distinct.
In response to each coordinate pair your program will receive the string "black", if the point is black, or "white", if the point is white.
When all n points are processed, you need to print four integers x1, y1, x2 and y2 (0 β€ x1, y1 β€ 109, 0 β€ x2, y2 β€ 109) β coordinates of points (x1, y1) and (x2, y2), which form a line, which separates n points into black and white. Points (x1, y1) and (x2, y2) should not coincide.
Hacks
To hack solution use the following format. The first line must contain word "hack", the second line should contain the number n and the last line should contain the sequence of 0 and 1 β colors of points, which will be reported to the solution. Unlike the jury tests, colors of points in hacks are always fixed in advance. Of course, the hacked solution wouldn't be able to get the information about the colors in advance.
For example, the hack corresponding to sample test will look like this:
hack
5
0 0 1 1 0
Example
Input
5
<span class="tex-span"></span>
black
<span class="tex-span"></span>
black
<span class="tex-span"></span>
white
<span class="tex-span"></span>
white
<span class="tex-span"></span>
black
Output
<span class="tex-span"></span>
0 0
<span class="tex-span"></span>
3 1
<span class="tex-span"></span>
2 3
<span class="tex-span"></span>
4 4
<span class="tex-span"></span>
0 2
<span class="tex-span"></span>
1 3 4 1
Note
In the sample input and output values are aligned only for simplicity of interpreting them chronologically. In real interaction no "extra" line breaks should appear.
The following picture illustrates the first test.
<image>
Submitted Solution:
```
from sys import stdout
def main():
n = int(input()) - 1
print(0, 0)
stdout.flush()
start = input()
if n == 0:
return
print(500000000, 1000000000)
stdout.flush()
p1 = None
if input() == start:
p1 = (500000001, 1000000000)
else:
p1 = (499999999, 1000000000)
left, right = 0, 999999999
for i in range(n - 1):
mid = (left + right) >> 1
print(mid, 0)
stdout.flush()
if input() == start:
left = mid
else:
right = mid
mid = (left + right) >> 1
print(mid, 0, p1[0], p1[1])
main()
```
No
| 13,335 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
In good old times dwarves tried to develop extrasensory abilities:
* Exactly n dwarves entered completely dark cave.
* Each dwarf received a hat β white or black. While in cave, none of the dwarves was able to see either his own hat or hats of other Dwarves.
* Dwarves went out of the cave to the meadow and sat at an arbitrary place one after the other. When a dwarf leaves the cave, he sees the colors of all hats of all dwarves that are seating on the meadow (i.e. left the cave before him). However, he is not able to see the color of his own hat and none of the dwarves can give him this information.
* The task for dwarves was to got diverged into two parts β one with dwarves with white hats and one with black hats.
After many centuries, dwarves finally managed to select the right place on the meadow without error. Will you be able to repeat their success?
You are asked to successively name n different integer points on the plane. After naming each new point you will be given its color β black or white. Your task is to ensure that the named points can be split by a line in such a way that all points of one color lie on the same side from the line and points of different colors lie on different sides. Moreover, no points can belong to the line. Also, you need to report any such line at the end of the process.
In this problem, the interactor is adaptive β the colors of the points in the tests are not fixed beforehand and the jury program can select them arbitrarily, in particular, depending on your program output.
Interaction
The first line of the standard input stream contains an integer n (1 β€ n β€ 30) β the number of points your program should name.
Then n times your program must print two integer coordinates x and y (0 β€ x β€ 109, 0 β€ y β€ 109). All points you print must be distinct.
In response to each coordinate pair your program will receive the string "black", if the point is black, or "white", if the point is white.
When all n points are processed, you need to print four integers x1, y1, x2 and y2 (0 β€ x1, y1 β€ 109, 0 β€ x2, y2 β€ 109) β coordinates of points (x1, y1) and (x2, y2), which form a line, which separates n points into black and white. Points (x1, y1) and (x2, y2) should not coincide.
Hacks
To hack solution use the following format. The first line must contain word "hack", the second line should contain the number n and the last line should contain the sequence of 0 and 1 β colors of points, which will be reported to the solution. Unlike the jury tests, colors of points in hacks are always fixed in advance. Of course, the hacked solution wouldn't be able to get the information about the colors in advance.
For example, the hack corresponding to sample test will look like this:
hack
5
0 0 1 1 0
Example
Input
5
<span class="tex-span"></span>
black
<span class="tex-span"></span>
black
<span class="tex-span"></span>
white
<span class="tex-span"></span>
white
<span class="tex-span"></span>
black
Output
<span class="tex-span"></span>
0 0
<span class="tex-span"></span>
3 1
<span class="tex-span"></span>
2 3
<span class="tex-span"></span>
4 4
<span class="tex-span"></span>
0 2
<span class="tex-span"></span>
1 3 4 1
Note
In the sample input and output values are aligned only for simplicity of interpreting them chronologically. In real interaction no "extra" line breaks should appear.
The following picture illustrates the first test.
<image>
Submitted Solution:
```
n = int(input())
print(0, 0)
color = input()
cl = 1
cr = 10 ** 9
for i in range(n - 1):
m = (cl + cr) // 2
print(m, 0)
c = input()
if c == color:
cl = m + 1
else:
cr = m - 1
print((cl + cr) // 2 , 0, (cl + cr) // 2, 1)
```
No
| 13,336 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
In good old times dwarves tried to develop extrasensory abilities:
* Exactly n dwarves entered completely dark cave.
* Each dwarf received a hat β white or black. While in cave, none of the dwarves was able to see either his own hat or hats of other Dwarves.
* Dwarves went out of the cave to the meadow and sat at an arbitrary place one after the other. When a dwarf leaves the cave, he sees the colors of all hats of all dwarves that are seating on the meadow (i.e. left the cave before him). However, he is not able to see the color of his own hat and none of the dwarves can give him this information.
* The task for dwarves was to got diverged into two parts β one with dwarves with white hats and one with black hats.
After many centuries, dwarves finally managed to select the right place on the meadow without error. Will you be able to repeat their success?
You are asked to successively name n different integer points on the plane. After naming each new point you will be given its color β black or white. Your task is to ensure that the named points can be split by a line in such a way that all points of one color lie on the same side from the line and points of different colors lie on different sides. Moreover, no points can belong to the line. Also, you need to report any such line at the end of the process.
In this problem, the interactor is adaptive β the colors of the points in the tests are not fixed beforehand and the jury program can select them arbitrarily, in particular, depending on your program output.
Interaction
The first line of the standard input stream contains an integer n (1 β€ n β€ 30) β the number of points your program should name.
Then n times your program must print two integer coordinates x and y (0 β€ x β€ 109, 0 β€ y β€ 109). All points you print must be distinct.
In response to each coordinate pair your program will receive the string "black", if the point is black, or "white", if the point is white.
When all n points are processed, you need to print four integers x1, y1, x2 and y2 (0 β€ x1, y1 β€ 109, 0 β€ x2, y2 β€ 109) β coordinates of points (x1, y1) and (x2, y2), which form a line, which separates n points into black and white. Points (x1, y1) and (x2, y2) should not coincide.
Hacks
To hack solution use the following format. The first line must contain word "hack", the second line should contain the number n and the last line should contain the sequence of 0 and 1 β colors of points, which will be reported to the solution. Unlike the jury tests, colors of points in hacks are always fixed in advance. Of course, the hacked solution wouldn't be able to get the information about the colors in advance.
For example, the hack corresponding to sample test will look like this:
hack
5
0 0 1 1 0
Example
Input
5
<span class="tex-span"></span>
black
<span class="tex-span"></span>
black
<span class="tex-span"></span>
white
<span class="tex-span"></span>
white
<span class="tex-span"></span>
black
Output
<span class="tex-span"></span>
0 0
<span class="tex-span"></span>
3 1
<span class="tex-span"></span>
2 3
<span class="tex-span"></span>
4 4
<span class="tex-span"></span>
0 2
<span class="tex-span"></span>
1 3 4 1
Note
In the sample input and output values are aligned only for simplicity of interpreting them chronologically. In real interaction no "extra" line breaks should appear.
The following picture illustrates the first test.
<image>
Submitted Solution:
```
l, r = 0, 1e9
g = int(input())
print(0, 0)
if input() == "black":
l, r = r, l
for i in range(g - 1):
m = int((l + r) // 2)
print(m, 0)
sa = input()
if sa == "black":
r = m
else:
l = m
m = int((l + r) // 2)
print(m, 1, m, 0)
```
No
| 13,337 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
In good old times dwarves tried to develop extrasensory abilities:
* Exactly n dwarves entered completely dark cave.
* Each dwarf received a hat β white or black. While in cave, none of the dwarves was able to see either his own hat or hats of other Dwarves.
* Dwarves went out of the cave to the meadow and sat at an arbitrary place one after the other. When a dwarf leaves the cave, he sees the colors of all hats of all dwarves that are seating on the meadow (i.e. left the cave before him). However, he is not able to see the color of his own hat and none of the dwarves can give him this information.
* The task for dwarves was to got diverged into two parts β one with dwarves with white hats and one with black hats.
After many centuries, dwarves finally managed to select the right place on the meadow without error. Will you be able to repeat their success?
You are asked to successively name n different integer points on the plane. After naming each new point you will be given its color β black or white. Your task is to ensure that the named points can be split by a line in such a way that all points of one color lie on the same side from the line and points of different colors lie on different sides. Moreover, no points can belong to the line. Also, you need to report any such line at the end of the process.
In this problem, the interactor is adaptive β the colors of the points in the tests are not fixed beforehand and the jury program can select them arbitrarily, in particular, depending on your program output.
Interaction
The first line of the standard input stream contains an integer n (1 β€ n β€ 30) β the number of points your program should name.
Then n times your program must print two integer coordinates x and y (0 β€ x β€ 109, 0 β€ y β€ 109). All points you print must be distinct.
In response to each coordinate pair your program will receive the string "black", if the point is black, or "white", if the point is white.
When all n points are processed, you need to print four integers x1, y1, x2 and y2 (0 β€ x1, y1 β€ 109, 0 β€ x2, y2 β€ 109) β coordinates of points (x1, y1) and (x2, y2), which form a line, which separates n points into black and white. Points (x1, y1) and (x2, y2) should not coincide.
Hacks
To hack solution use the following format. The first line must contain word "hack", the second line should contain the number n and the last line should contain the sequence of 0 and 1 β colors of points, which will be reported to the solution. Unlike the jury tests, colors of points in hacks are always fixed in advance. Of course, the hacked solution wouldn't be able to get the information about the colors in advance.
For example, the hack corresponding to sample test will look like this:
hack
5
0 0 1 1 0
Example
Input
5
<span class="tex-span"></span>
black
<span class="tex-span"></span>
black
<span class="tex-span"></span>
white
<span class="tex-span"></span>
white
<span class="tex-span"></span>
black
Output
<span class="tex-span"></span>
0 0
<span class="tex-span"></span>
3 1
<span class="tex-span"></span>
2 3
<span class="tex-span"></span>
4 4
<span class="tex-span"></span>
0 2
<span class="tex-span"></span>
1 3 4 1
Note
In the sample input and output values are aligned only for simplicity of interpreting them chronologically. In real interaction no "extra" line breaks should appear.
The following picture illustrates the first test.
<image>
Submitted Solution:
```
from sys import stdout
n = int(input()) - 1
print(0, 0)
stdout.flush()
start = input()
left, right = 0, 999999999
for i in range(n):
mid = (left + right) >> 1
print(mid, 0)
stdout.flush()
if input() == start:
left = mid
else:
right = mid
mid = (left + right) >> 1
print(mid, 1, mid, -1)
```
No
| 13,338 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya owns three strings s , a and b, each of them consists only of first k Latin letters.
Let a template be such a string of length k that each of the first k Latin letters appears in it exactly once (thus there are k! distinct templates). Application of template p to the string s is the replacement of each character in string s with p_i, i is the index of this letter in the alphabet. For example, applying template "bdca" to a string "aabccd" yields string "bbdcca".
Vasya wants to know if there exists such a template which yields a string lexicographically greater than or equal to string a and lexicographically less than or equal to string b after applying it to s.
If there exist multiple suitable templates, print any of them.
String a is lexicographically less than string b if there is some i (1 β€ i β€ n) that a_i < b_i and for any j (1 β€ j < i) a_j = b_j.
You are required to answer t testcases independently.
Input
The first line contains a single integer t (1 β€ t β€ 10^6) β the number of testcases.
In hacks you can only use t = 1.
Each of the next t lines contains the description of the testcase in the following form:
The first line of the testcase contains a single integer k (1 β€ k β€ 26) β the length of the template.
The second line of the testcase contains the string s (1 β€ |s| β€ 10^6).
The third line of the testcase contains the string a.
The fourth line of the testcase contains the string b.
Strings s, a and b have the same length (|s| = |a| = |b|) and consist only of the first k Latin letters, all letters are lowercase.
It is guaranteed that string a is lexicographically less than or equal to string b.
It is also guaranteed that the total length of strings over all testcase won't exceed 3 β
10^6.
Output
Print the answers to all testcases in the following form:
If there exists no suitable template then print "NO" in the first line.
Otherwise print "YES" in the first line and the template itself in the second line (k lowercase letters, each of the first k Latin letters should appear exactly once).
If there exist multiple suitable templates, print any of them.
Example
Input
2
4
bbcb
aada
aada
3
abc
bbb
bbb
Output
YES
badc
NO
Submitted Solution:
```
from itertools import permutations
alpha = [a for a in 'qwertyuiopasdfghjklzxcvbnm']
alpha.sort()
def f(k, s, a, b):
for key in permutations(alpha[:k], k):
d = {alpha[i]: key[i] for i in range(k)}
s = [d[x] for x in s]
if a <= ''.join(s) <= b:
return key, True
return '', False
def main():
t = int(input())
res = []
for i in range(t):
k = int(input())
s = input()
a = input()
b = input()
key, r = f(k, s, a, b)
if r:
res.append('YES\n'+''.join(key))
else:
res.append('NO')
print('\n'.join(res))
if __name__ == '__main__':
main()
```
No
| 13,339 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya owns three strings s , a and b, each of them consists only of first k Latin letters.
Let a template be such a string of length k that each of the first k Latin letters appears in it exactly once (thus there are k! distinct templates). Application of template p to the string s is the replacement of each character in string s with p_i, i is the index of this letter in the alphabet. For example, applying template "bdca" to a string "aabccd" yields string "bbdcca".
Vasya wants to know if there exists such a template which yields a string lexicographically greater than or equal to string a and lexicographically less than or equal to string b after applying it to s.
If there exist multiple suitable templates, print any of them.
String a is lexicographically less than string b if there is some i (1 β€ i β€ n) that a_i < b_i and for any j (1 β€ j < i) a_j = b_j.
You are required to answer t testcases independently.
Input
The first line contains a single integer t (1 β€ t β€ 10^6) β the number of testcases.
In hacks you can only use t = 1.
Each of the next t lines contains the description of the testcase in the following form:
The first line of the testcase contains a single integer k (1 β€ k β€ 26) β the length of the template.
The second line of the testcase contains the string s (1 β€ |s| β€ 10^6).
The third line of the testcase contains the string a.
The fourth line of the testcase contains the string b.
Strings s, a and b have the same length (|s| = |a| = |b|) and consist only of the first k Latin letters, all letters are lowercase.
It is guaranteed that string a is lexicographically less than or equal to string b.
It is also guaranteed that the total length of strings over all testcase won't exceed 3 β
10^6.
Output
Print the answers to all testcases in the following form:
If there exists no suitable template then print "NO" in the first line.
Otherwise print "YES" in the first line and the template itself in the second line (k lowercase letters, each of the first k Latin letters should appear exactly once).
If there exist multiple suitable templates, print any of them.
Example
Input
2
4
bbcb
aada
aada
3
abc
bbb
bbb
Output
YES
badc
NO
Submitted Solution:
```
t = int(input())
list = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
for i in range(t):
l = int(input())
sample = input()
a = input()
b = input()
# print(sample)
# print(a)
# print(b)
panel = dict(zip(list[:l],[0]*l))
print(panel)
for i in range(l):
if a[i]==b[i]:
panel[a[i]] = sample[i]
print(panel)
alphas = []
count = 0
for value in panel.values():
if str(value).isalpha():
alphas.append(value)
else:
count += 1
if count+len(alphas):
print("YES")
panel = sorted(panel.items(), key=lambda x:x[0])
result = []
for item in panel:
result.append(item[1])
a = 0
for i in range(l):
if result[i]==0:
while(a<26):
if list[a] not in result:
result[i] = list[a]
a += 1
break
a += 1
print(result)
else:
print("NO")
```
No
| 13,340 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya owns three strings s , a and b, each of them consists only of first k Latin letters.
Let a template be such a string of length k that each of the first k Latin letters appears in it exactly once (thus there are k! distinct templates). Application of template p to the string s is the replacement of each character in string s with p_i, i is the index of this letter in the alphabet. For example, applying template "bdca" to a string "aabccd" yields string "bbdcca".
Vasya wants to know if there exists such a template which yields a string lexicographically greater than or equal to string a and lexicographically less than or equal to string b after applying it to s.
If there exist multiple suitable templates, print any of them.
String a is lexicographically less than string b if there is some i (1 β€ i β€ n) that a_i < b_i and for any j (1 β€ j < i) a_j = b_j.
You are required to answer t testcases independently.
Input
The first line contains a single integer t (1 β€ t β€ 10^6) β the number of testcases.
In hacks you can only use t = 1.
Each of the next t lines contains the description of the testcase in the following form:
The first line of the testcase contains a single integer k (1 β€ k β€ 26) β the length of the template.
The second line of the testcase contains the string s (1 β€ |s| β€ 10^6).
The third line of the testcase contains the string a.
The fourth line of the testcase contains the string b.
Strings s, a and b have the same length (|s| = |a| = |b|) and consist only of the first k Latin letters, all letters are lowercase.
It is guaranteed that string a is lexicographically less than or equal to string b.
It is also guaranteed that the total length of strings over all testcase won't exceed 3 β
10^6.
Output
Print the answers to all testcases in the following form:
If there exists no suitable template then print "NO" in the first line.
Otherwise print "YES" in the first line and the template itself in the second line (k lowercase letters, each of the first k Latin letters should appear exactly once).
If there exist multiple suitable templates, print any of them.
Example
Input
2
4
bbcb
aada
aada
3
abc
bbb
bbb
Output
YES
badc
NO
Submitted Solution:
```
col = int(input())
sp = {'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12, 'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24, 'z': 25}
sp1 = {0: 'a', 1: 'b', 2: 'c', 3: 'd', 4: 'e', 5: 'f', 6: 'g', 7: 'h', 8: 'i', 9: 'j', 10: 'k', 11: 'l', 12: 'm', 13: 'n', 14: 'o', 15: 'p', 16: 'q', 17: 'r', 18: 's', 19: 't', 20: 'u', 21: 'v', 22: 'w', 23: 'x', 24: 'y', 25: 'z'}
def getdiff(a, b):
return abs(sp.get(a) - sp.get(b))
def yorn(a, b, s):
flag = False
for i in range(len(a)):
dif = getdiff(a[i], b[i])
if dif > 1:
return (1, i)
elif dif == 1:
flag = True
return (0, flag)
def pattern(a, b, s, i, leng):
pat = ['a']*leng
for j in range(i):
pat[sp.get(a[j])] = a[j]
pat[sp.get(a[i])] = sp1.get(sp.get(a[i]) + 1)
return ''.join(pat)
def pattern1(a, b, s, i, leng):
pat = list(a)
if sp.get(a[i]) + 1 > 26:
return 0
pat[sp.get(a[i])] = sp1.get(sp.get(a[i]) + 1)
return ''.join(pat)
res = ''
for i in range(col):
leng = int(input())
s = input()
a = input()
b = input()
yes = yorn(a, b, s)
if len(set(a)) < len(set(s)):
res += "NO\n"
#print('NO')
elif yes[0]:
#print('YES')
#print(pattern(a, b, s, yes[1], leng))
res += "YES\n"
res += pattern(a, b, s, yes[1], leng) + "\n"
else:
itog = pattern1(a, b, s, yes[1], leng)
if not itog:
res += "NO\n"
#print('NO')
else:
#print('YES')
#print(itog)
res += "YES\n"
res += itog + "\n"
print(res[:-1])
'''
1
4
abcd
abcd
cdef
'''
'''sch = 0
for i in 'abcdefghijklmnopqrstuvwxyz':
s[i] = sch
sch += 1
print(s)'''
```
No
| 13,341 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya owns three strings s , a and b, each of them consists only of first k Latin letters.
Let a template be such a string of length k that each of the first k Latin letters appears in it exactly once (thus there are k! distinct templates). Application of template p to the string s is the replacement of each character in string s with p_i, i is the index of this letter in the alphabet. For example, applying template "bdca" to a string "aabccd" yields string "bbdcca".
Vasya wants to know if there exists such a template which yields a string lexicographically greater than or equal to string a and lexicographically less than or equal to string b after applying it to s.
If there exist multiple suitable templates, print any of them.
String a is lexicographically less than string b if there is some i (1 β€ i β€ n) that a_i < b_i and for any j (1 β€ j < i) a_j = b_j.
You are required to answer t testcases independently.
Input
The first line contains a single integer t (1 β€ t β€ 10^6) β the number of testcases.
In hacks you can only use t = 1.
Each of the next t lines contains the description of the testcase in the following form:
The first line of the testcase contains a single integer k (1 β€ k β€ 26) β the length of the template.
The second line of the testcase contains the string s (1 β€ |s| β€ 10^6).
The third line of the testcase contains the string a.
The fourth line of the testcase contains the string b.
Strings s, a and b have the same length (|s| = |a| = |b|) and consist only of the first k Latin letters, all letters are lowercase.
It is guaranteed that string a is lexicographically less than or equal to string b.
It is also guaranteed that the total length of strings over all testcase won't exceed 3 β
10^6.
Output
Print the answers to all testcases in the following form:
If there exists no suitable template then print "NO" in the first line.
Otherwise print "YES" in the first line and the template itself in the second line (k lowercase letters, each of the first k Latin letters should appear exactly once).
If there exist multiple suitable templates, print any of them.
Example
Input
2
4
bbcb
aada
aada
3
abc
bbb
bbb
Output
YES
badc
NO
Submitted Solution:
```
print("dddddddddddddddddddddddddddd")
print("dddddddddddddddddddddddddddd")
```
No
| 13,342 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lunar New Year is approaching, and Bob is struggling with his homework β a number division problem.
There are n positive integers a_1, a_2, β¦, a_n on Bob's homework paper, where n is always an even number. Bob is asked to divide those numbers into groups, where each group must contain at least 2 numbers. Suppose the numbers are divided into m groups, and the sum of the numbers in the j-th group is s_j. Bob's aim is to minimize the sum of the square of s_j, that is $$$β_{j = 1}^{m} s_j^2.$$$
Bob is puzzled by this hard problem. Could you please help him solve it?
Input
The first line contains an even integer n (2 β€ n β€ 3 β
10^5), denoting that there are n integers on Bob's homework paper.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^4), describing the numbers you need to deal with.
Output
A single line containing one integer, denoting the minimum of the sum of the square of s_j, which is $$$β_{i = j}^{m} s_j^2, where m$$$ is the number of groups.
Examples
Input
4
8 5 2 3
Output
164
Input
6
1 1 1 2 2 2
Output
27
Note
In the first sample, one of the optimal solutions is to divide those 4 numbers into 2 groups \{2, 8\}, \{5, 3\}. Thus the answer is (2 + 8)^2 + (5 + 3)^2 = 164.
In the second sample, one of the optimal solutions is to divide those 6 numbers into 3 groups \{1, 2\}, \{1, 2\}, \{1, 2\}. Thus the answer is (1 + 2)^2 + (1 + 2)^2 + (1 + 2)^2 = 27.
Tags: greedy, implementation, math, sortings
Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Tue Nov 13 20:48:59 2018
@author:
"""
n=int(input())
l=list(map(int,input().split()))
l.sort()
s=0
for i in range(n//2):
s+=(l[i]+l[n-i-1])**2
print(s)
```
| 13,343 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lunar New Year is approaching, and Bob is struggling with his homework β a number division problem.
There are n positive integers a_1, a_2, β¦, a_n on Bob's homework paper, where n is always an even number. Bob is asked to divide those numbers into groups, where each group must contain at least 2 numbers. Suppose the numbers are divided into m groups, and the sum of the numbers in the j-th group is s_j. Bob's aim is to minimize the sum of the square of s_j, that is $$$β_{j = 1}^{m} s_j^2.$$$
Bob is puzzled by this hard problem. Could you please help him solve it?
Input
The first line contains an even integer n (2 β€ n β€ 3 β
10^5), denoting that there are n integers on Bob's homework paper.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^4), describing the numbers you need to deal with.
Output
A single line containing one integer, denoting the minimum of the sum of the square of s_j, which is $$$β_{i = j}^{m} s_j^2, where m$$$ is the number of groups.
Examples
Input
4
8 5 2 3
Output
164
Input
6
1 1 1 2 2 2
Output
27
Note
In the first sample, one of the optimal solutions is to divide those 4 numbers into 2 groups \{2, 8\}, \{5, 3\}. Thus the answer is (2 + 8)^2 + (5 + 3)^2 = 164.
In the second sample, one of the optimal solutions is to divide those 6 numbers into 3 groups \{1, 2\}, \{1, 2\}, \{1, 2\}. Thus the answer is (1 + 2)^2 + (1 + 2)^2 + (1 + 2)^2 = 27.
Tags: greedy, implementation, math, sortings
Correct Solution:
```
n=int(input())
s=list(map(int,input().split()))
s.sort()
m=list(reversed(s))
sum=0
for i in range(n//2):
sum+=(s[i]+m[i])**2
print(sum)
```
| 13,344 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lunar New Year is approaching, and Bob is struggling with his homework β a number division problem.
There are n positive integers a_1, a_2, β¦, a_n on Bob's homework paper, where n is always an even number. Bob is asked to divide those numbers into groups, where each group must contain at least 2 numbers. Suppose the numbers are divided into m groups, and the sum of the numbers in the j-th group is s_j. Bob's aim is to minimize the sum of the square of s_j, that is $$$β_{j = 1}^{m} s_j^2.$$$
Bob is puzzled by this hard problem. Could you please help him solve it?
Input
The first line contains an even integer n (2 β€ n β€ 3 β
10^5), denoting that there are n integers on Bob's homework paper.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^4), describing the numbers you need to deal with.
Output
A single line containing one integer, denoting the minimum of the sum of the square of s_j, which is $$$β_{i = j}^{m} s_j^2, where m$$$ is the number of groups.
Examples
Input
4
8 5 2 3
Output
164
Input
6
1 1 1 2 2 2
Output
27
Note
In the first sample, one of the optimal solutions is to divide those 4 numbers into 2 groups \{2, 8\}, \{5, 3\}. Thus the answer is (2 + 8)^2 + (5 + 3)^2 = 164.
In the second sample, one of the optimal solutions is to divide those 6 numbers into 3 groups \{1, 2\}, \{1, 2\}, \{1, 2\}. Thus the answer is (1 + 2)^2 + (1 + 2)^2 + (1 + 2)^2 = 27.
Tags: greedy, implementation, math, sortings
Correct Solution:
```
n=int(input())
a=[int(s) for s in input().split(' ')]
a.sort()
min_sum=0
for i in range(len(a)):
min_sum+=(a[i]+a[n-1-i])**2
print(int(min_sum/2))
```
| 13,345 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lunar New Year is approaching, and Bob is struggling with his homework β a number division problem.
There are n positive integers a_1, a_2, β¦, a_n on Bob's homework paper, where n is always an even number. Bob is asked to divide those numbers into groups, where each group must contain at least 2 numbers. Suppose the numbers are divided into m groups, and the sum of the numbers in the j-th group is s_j. Bob's aim is to minimize the sum of the square of s_j, that is $$$β_{j = 1}^{m} s_j^2.$$$
Bob is puzzled by this hard problem. Could you please help him solve it?
Input
The first line contains an even integer n (2 β€ n β€ 3 β
10^5), denoting that there are n integers on Bob's homework paper.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^4), describing the numbers you need to deal with.
Output
A single line containing one integer, denoting the minimum of the sum of the square of s_j, which is $$$β_{i = j}^{m} s_j^2, where m$$$ is the number of groups.
Examples
Input
4
8 5 2 3
Output
164
Input
6
1 1 1 2 2 2
Output
27
Note
In the first sample, one of the optimal solutions is to divide those 4 numbers into 2 groups \{2, 8\}, \{5, 3\}. Thus the answer is (2 + 8)^2 + (5 + 3)^2 = 164.
In the second sample, one of the optimal solutions is to divide those 6 numbers into 3 groups \{1, 2\}, \{1, 2\}, \{1, 2\}. Thus the answer is (1 + 2)^2 + (1 + 2)^2 + (1 + 2)^2 = 27.
Tags: greedy, implementation, math, sortings
Correct Solution:
```
n = int(input())
l = list(map(int, input().split()))
l.sort()
o = 0
a = 0
for i in range(n // 2):
a = l[i] + l[- (i + 1)]
o += a * a
print(o)
```
| 13,346 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lunar New Year is approaching, and Bob is struggling with his homework β a number division problem.
There are n positive integers a_1, a_2, β¦, a_n on Bob's homework paper, where n is always an even number. Bob is asked to divide those numbers into groups, where each group must contain at least 2 numbers. Suppose the numbers are divided into m groups, and the sum of the numbers in the j-th group is s_j. Bob's aim is to minimize the sum of the square of s_j, that is $$$β_{j = 1}^{m} s_j^2.$$$
Bob is puzzled by this hard problem. Could you please help him solve it?
Input
The first line contains an even integer n (2 β€ n β€ 3 β
10^5), denoting that there are n integers on Bob's homework paper.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^4), describing the numbers you need to deal with.
Output
A single line containing one integer, denoting the minimum of the sum of the square of s_j, which is $$$β_{i = j}^{m} s_j^2, where m$$$ is the number of groups.
Examples
Input
4
8 5 2 3
Output
164
Input
6
1 1 1 2 2 2
Output
27
Note
In the first sample, one of the optimal solutions is to divide those 4 numbers into 2 groups \{2, 8\}, \{5, 3\}. Thus the answer is (2 + 8)^2 + (5 + 3)^2 = 164.
In the second sample, one of the optimal solutions is to divide those 6 numbers into 3 groups \{1, 2\}, \{1, 2\}, \{1, 2\}. Thus the answer is (1 + 2)^2 + (1 + 2)^2 + (1 + 2)^2 = 27.
Tags: greedy, implementation, math, sortings
Correct Solution:
```
n = int(input())
#n, m = map(int, input().split())
arr = sorted(map(int, input().split()))
s = 0
for i in range(len(arr)//2):
s += (arr[i]+arr[(-i-1)]) ** 2
print(s)
```
| 13,347 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lunar New Year is approaching, and Bob is struggling with his homework β a number division problem.
There are n positive integers a_1, a_2, β¦, a_n on Bob's homework paper, where n is always an even number. Bob is asked to divide those numbers into groups, where each group must contain at least 2 numbers. Suppose the numbers are divided into m groups, and the sum of the numbers in the j-th group is s_j. Bob's aim is to minimize the sum of the square of s_j, that is $$$β_{j = 1}^{m} s_j^2.$$$
Bob is puzzled by this hard problem. Could you please help him solve it?
Input
The first line contains an even integer n (2 β€ n β€ 3 β
10^5), denoting that there are n integers on Bob's homework paper.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^4), describing the numbers you need to deal with.
Output
A single line containing one integer, denoting the minimum of the sum of the square of s_j, which is $$$β_{i = j}^{m} s_j^2, where m$$$ is the number of groups.
Examples
Input
4
8 5 2 3
Output
164
Input
6
1 1 1 2 2 2
Output
27
Note
In the first sample, one of the optimal solutions is to divide those 4 numbers into 2 groups \{2, 8\}, \{5, 3\}. Thus the answer is (2 + 8)^2 + (5 + 3)^2 = 164.
In the second sample, one of the optimal solutions is to divide those 6 numbers into 3 groups \{1, 2\}, \{1, 2\}, \{1, 2\}. Thus the answer is (1 + 2)^2 + (1 + 2)^2 + (1 + 2)^2 = 27.
Tags: greedy, implementation, math, sortings
Correct Solution:
```
n=int(input())
s=input()
a=s.split(" ")
a=list(map(int,a))
a.sort()
out=list()
for i in range(int(n/2)):
s=int(a[i]+a[n-i-1])
out.append(s)
if n%2!=0:
out[int(len(out)-1)]=out[int(len(out)-1)]+a[int(n/2)]
output=0
for i in range(len(out)):
output=output+out[i]*out[i]
print(output)
```
| 13,348 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lunar New Year is approaching, and Bob is struggling with his homework β a number division problem.
There are n positive integers a_1, a_2, β¦, a_n on Bob's homework paper, where n is always an even number. Bob is asked to divide those numbers into groups, where each group must contain at least 2 numbers. Suppose the numbers are divided into m groups, and the sum of the numbers in the j-th group is s_j. Bob's aim is to minimize the sum of the square of s_j, that is $$$β_{j = 1}^{m} s_j^2.$$$
Bob is puzzled by this hard problem. Could you please help him solve it?
Input
The first line contains an even integer n (2 β€ n β€ 3 β
10^5), denoting that there are n integers on Bob's homework paper.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^4), describing the numbers you need to deal with.
Output
A single line containing one integer, denoting the minimum of the sum of the square of s_j, which is $$$β_{i = j}^{m} s_j^2, where m$$$ is the number of groups.
Examples
Input
4
8 5 2 3
Output
164
Input
6
1 1 1 2 2 2
Output
27
Note
In the first sample, one of the optimal solutions is to divide those 4 numbers into 2 groups \{2, 8\}, \{5, 3\}. Thus the answer is (2 + 8)^2 + (5 + 3)^2 = 164.
In the second sample, one of the optimal solutions is to divide those 6 numbers into 3 groups \{1, 2\}, \{1, 2\}, \{1, 2\}. Thus the answer is (1 + 2)^2 + (1 + 2)^2 + (1 + 2)^2 = 27.
Tags: greedy, implementation, math, sortings
Correct Solution:
```
n = int(input())
seq = sorted(list(map(int, input().split())))
ans = 0
for i in range(n // 2):
ans += (seq[i] + seq[-(i + 1)])**2
print(ans)
```
| 13,349 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Lunar New Year is approaching, and Bob is struggling with his homework β a number division problem.
There are n positive integers a_1, a_2, β¦, a_n on Bob's homework paper, where n is always an even number. Bob is asked to divide those numbers into groups, where each group must contain at least 2 numbers. Suppose the numbers are divided into m groups, and the sum of the numbers in the j-th group is s_j. Bob's aim is to minimize the sum of the square of s_j, that is $$$β_{j = 1}^{m} s_j^2.$$$
Bob is puzzled by this hard problem. Could you please help him solve it?
Input
The first line contains an even integer n (2 β€ n β€ 3 β
10^5), denoting that there are n integers on Bob's homework paper.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^4), describing the numbers you need to deal with.
Output
A single line containing one integer, denoting the minimum of the sum of the square of s_j, which is $$$β_{i = j}^{m} s_j^2, where m$$$ is the number of groups.
Examples
Input
4
8 5 2 3
Output
164
Input
6
1 1 1 2 2 2
Output
27
Note
In the first sample, one of the optimal solutions is to divide those 4 numbers into 2 groups \{2, 8\}, \{5, 3\}. Thus the answer is (2 + 8)^2 + (5 + 3)^2 = 164.
In the second sample, one of the optimal solutions is to divide those 6 numbers into 3 groups \{1, 2\}, \{1, 2\}, \{1, 2\}. Thus the answer is (1 + 2)^2 + (1 + 2)^2 + (1 + 2)^2 = 27.
Tags: greedy, implementation, math, sortings
Correct Solution:
```
# list(map(int, input().split()))
n = int(input())
numbers = list(map(int, input().split()))
numbers = sorted(numbers)
ans = 0
for i in range(n // 2):
ans += (numbers[i] + numbers[n - i - 1]) ** 2
print(ans)
```
| 13,350 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lunar New Year is approaching, and Bob is struggling with his homework β a number division problem.
There are n positive integers a_1, a_2, β¦, a_n on Bob's homework paper, where n is always an even number. Bob is asked to divide those numbers into groups, where each group must contain at least 2 numbers. Suppose the numbers are divided into m groups, and the sum of the numbers in the j-th group is s_j. Bob's aim is to minimize the sum of the square of s_j, that is $$$β_{j = 1}^{m} s_j^2.$$$
Bob is puzzled by this hard problem. Could you please help him solve it?
Input
The first line contains an even integer n (2 β€ n β€ 3 β
10^5), denoting that there are n integers on Bob's homework paper.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^4), describing the numbers you need to deal with.
Output
A single line containing one integer, denoting the minimum of the sum of the square of s_j, which is $$$β_{i = j}^{m} s_j^2, where m$$$ is the number of groups.
Examples
Input
4
8 5 2 3
Output
164
Input
6
1 1 1 2 2 2
Output
27
Note
In the first sample, one of the optimal solutions is to divide those 4 numbers into 2 groups \{2, 8\}, \{5, 3\}. Thus the answer is (2 + 8)^2 + (5 + 3)^2 = 164.
In the second sample, one of the optimal solutions is to divide those 6 numbers into 3 groups \{1, 2\}, \{1, 2\}, \{1, 2\}. Thus the answer is (1 + 2)^2 + (1 + 2)^2 + (1 + 2)^2 = 27.
Submitted Solution:
```
def inpl(): return list(map(int, input().split()))
N = int(input())
A = inpl()
A.sort()
print(sum([(i+j)**2 for i, j in zip(A[:N//2], A[::-1][:N//2])]))
```
Yes
| 13,351 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lunar New Year is approaching, and Bob is struggling with his homework β a number division problem.
There are n positive integers a_1, a_2, β¦, a_n on Bob's homework paper, where n is always an even number. Bob is asked to divide those numbers into groups, where each group must contain at least 2 numbers. Suppose the numbers are divided into m groups, and the sum of the numbers in the j-th group is s_j. Bob's aim is to minimize the sum of the square of s_j, that is $$$β_{j = 1}^{m} s_j^2.$$$
Bob is puzzled by this hard problem. Could you please help him solve it?
Input
The first line contains an even integer n (2 β€ n β€ 3 β
10^5), denoting that there are n integers on Bob's homework paper.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^4), describing the numbers you need to deal with.
Output
A single line containing one integer, denoting the minimum of the sum of the square of s_j, which is $$$β_{i = j}^{m} s_j^2, where m$$$ is the number of groups.
Examples
Input
4
8 5 2 3
Output
164
Input
6
1 1 1 2 2 2
Output
27
Note
In the first sample, one of the optimal solutions is to divide those 4 numbers into 2 groups \{2, 8\}, \{5, 3\}. Thus the answer is (2 + 8)^2 + (5 + 3)^2 = 164.
In the second sample, one of the optimal solutions is to divide those 6 numbers into 3 groups \{1, 2\}, \{1, 2\}, \{1, 2\}. Thus the answer is (1 + 2)^2 + (1 + 2)^2 + (1 + 2)^2 = 27.
Submitted Solution:
```
n = int(input())
l = [*map(int, input().split())]
l.sort()
i, j = 0, n - 1
res = 0
while i <= j:
res += (l[i] + l[j]) ** 2
i += 1
j -= 1
print(res)
```
Yes
| 13,352 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lunar New Year is approaching, and Bob is struggling with his homework β a number division problem.
There are n positive integers a_1, a_2, β¦, a_n on Bob's homework paper, where n is always an even number. Bob is asked to divide those numbers into groups, where each group must contain at least 2 numbers. Suppose the numbers are divided into m groups, and the sum of the numbers in the j-th group is s_j. Bob's aim is to minimize the sum of the square of s_j, that is $$$β_{j = 1}^{m} s_j^2.$$$
Bob is puzzled by this hard problem. Could you please help him solve it?
Input
The first line contains an even integer n (2 β€ n β€ 3 β
10^5), denoting that there are n integers on Bob's homework paper.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^4), describing the numbers you need to deal with.
Output
A single line containing one integer, denoting the minimum of the sum of the square of s_j, which is $$$β_{i = j}^{m} s_j^2, where m$$$ is the number of groups.
Examples
Input
4
8 5 2 3
Output
164
Input
6
1 1 1 2 2 2
Output
27
Note
In the first sample, one of the optimal solutions is to divide those 4 numbers into 2 groups \{2, 8\}, \{5, 3\}. Thus the answer is (2 + 8)^2 + (5 + 3)^2 = 164.
In the second sample, one of the optimal solutions is to divide those 6 numbers into 3 groups \{1, 2\}, \{1, 2\}, \{1, 2\}. Thus the answer is (1 + 2)^2 + (1 + 2)^2 + (1 + 2)^2 = 27.
Submitted Solution:
```
# import sys
# sys.stdin = open("test.in","r")
# sys.stdout = open("test.out","w")
n=int(input())
a=sorted(map(int,input().split()))
c=0
for i in range(n//2):
c+=(a[i]+a[n-1-i])**2
print(c)
```
Yes
| 13,353 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lunar New Year is approaching, and Bob is struggling with his homework β a number division problem.
There are n positive integers a_1, a_2, β¦, a_n on Bob's homework paper, where n is always an even number. Bob is asked to divide those numbers into groups, where each group must contain at least 2 numbers. Suppose the numbers are divided into m groups, and the sum of the numbers in the j-th group is s_j. Bob's aim is to minimize the sum of the square of s_j, that is $$$β_{j = 1}^{m} s_j^2.$$$
Bob is puzzled by this hard problem. Could you please help him solve it?
Input
The first line contains an even integer n (2 β€ n β€ 3 β
10^5), denoting that there are n integers on Bob's homework paper.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^4), describing the numbers you need to deal with.
Output
A single line containing one integer, denoting the minimum of the sum of the square of s_j, which is $$$β_{i = j}^{m} s_j^2, where m$$$ is the number of groups.
Examples
Input
4
8 5 2 3
Output
164
Input
6
1 1 1 2 2 2
Output
27
Note
In the first sample, one of the optimal solutions is to divide those 4 numbers into 2 groups \{2, 8\}, \{5, 3\}. Thus the answer is (2 + 8)^2 + (5 + 3)^2 = 164.
In the second sample, one of the optimal solutions is to divide those 6 numbers into 3 groups \{1, 2\}, \{1, 2\}, \{1, 2\}. Thus the answer is (1 + 2)^2 + (1 + 2)^2 + (1 + 2)^2 = 27.
Submitted Solution:
```
n=int(input())
a=[*map(int,input().split())]
a.sort()
i=0
ans=0
while i<n:
ans+=(a[i]+a[n-i-1])*(a[i]+a[n-i-1])
i+=1
ans=int(ans/2)
print(ans)
```
Yes
| 13,354 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lunar New Year is approaching, and Bob is struggling with his homework β a number division problem.
There are n positive integers a_1, a_2, β¦, a_n on Bob's homework paper, where n is always an even number. Bob is asked to divide those numbers into groups, where each group must contain at least 2 numbers. Suppose the numbers are divided into m groups, and the sum of the numbers in the j-th group is s_j. Bob's aim is to minimize the sum of the square of s_j, that is $$$β_{j = 1}^{m} s_j^2.$$$
Bob is puzzled by this hard problem. Could you please help him solve it?
Input
The first line contains an even integer n (2 β€ n β€ 3 β
10^5), denoting that there are n integers on Bob's homework paper.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^4), describing the numbers you need to deal with.
Output
A single line containing one integer, denoting the minimum of the sum of the square of s_j, which is $$$β_{i = j}^{m} s_j^2, where m$$$ is the number of groups.
Examples
Input
4
8 5 2 3
Output
164
Input
6
1 1 1 2 2 2
Output
27
Note
In the first sample, one of the optimal solutions is to divide those 4 numbers into 2 groups \{2, 8\}, \{5, 3\}. Thus the answer is (2 + 8)^2 + (5 + 3)^2 = 164.
In the second sample, one of the optimal solutions is to divide those 6 numbers into 3 groups \{1, 2\}, \{1, 2\}, \{1, 2\}. Thus the answer is (1 + 2)^2 + (1 + 2)^2 + (1 + 2)^2 = 27.
Submitted Solution:
```
# Collaborated with Rudransh
inp = input()
n = int(inp)
inp = input().split(" ")
a = []
sum = 0
for i in inp:
a.append(int(i))
a.sort()
for i in range(n):
sum += pow(a[i] + a[n - i - 1], 2)
print(sum/2)
```
No
| 13,355 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lunar New Year is approaching, and Bob is struggling with his homework β a number division problem.
There are n positive integers a_1, a_2, β¦, a_n on Bob's homework paper, where n is always an even number. Bob is asked to divide those numbers into groups, where each group must contain at least 2 numbers. Suppose the numbers are divided into m groups, and the sum of the numbers in the j-th group is s_j. Bob's aim is to minimize the sum of the square of s_j, that is $$$β_{j = 1}^{m} s_j^2.$$$
Bob is puzzled by this hard problem. Could you please help him solve it?
Input
The first line contains an even integer n (2 β€ n β€ 3 β
10^5), denoting that there are n integers on Bob's homework paper.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^4), describing the numbers you need to deal with.
Output
A single line containing one integer, denoting the minimum of the sum of the square of s_j, which is $$$β_{i = j}^{m} s_j^2, where m$$$ is the number of groups.
Examples
Input
4
8 5 2 3
Output
164
Input
6
1 1 1 2 2 2
Output
27
Note
In the first sample, one of the optimal solutions is to divide those 4 numbers into 2 groups \{2, 8\}, \{5, 3\}. Thus the answer is (2 + 8)^2 + (5 + 3)^2 = 164.
In the second sample, one of the optimal solutions is to divide those 6 numbers into 3 groups \{1, 2\}, \{1, 2\}, \{1, 2\}. Thus the answer is (1 + 2)^2 + (1 + 2)^2 + (1 + 2)^2 = 27.
Submitted Solution:
```
n=int(input())
s=input().split()
s.sort()
summ=0
for i in range(n//2):
summ=summ+(int(s[i])+int(s[n-1-i]))**2
print(summ)
```
No
| 13,356 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lunar New Year is approaching, and Bob is struggling with his homework β a number division problem.
There are n positive integers a_1, a_2, β¦, a_n on Bob's homework paper, where n is always an even number. Bob is asked to divide those numbers into groups, where each group must contain at least 2 numbers. Suppose the numbers are divided into m groups, and the sum of the numbers in the j-th group is s_j. Bob's aim is to minimize the sum of the square of s_j, that is $$$β_{j = 1}^{m} s_j^2.$$$
Bob is puzzled by this hard problem. Could you please help him solve it?
Input
The first line contains an even integer n (2 β€ n β€ 3 β
10^5), denoting that there are n integers on Bob's homework paper.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^4), describing the numbers you need to deal with.
Output
A single line containing one integer, denoting the minimum of the sum of the square of s_j, which is $$$β_{i = j}^{m} s_j^2, where m$$$ is the number of groups.
Examples
Input
4
8 5 2 3
Output
164
Input
6
1 1 1 2 2 2
Output
27
Note
In the first sample, one of the optimal solutions is to divide those 4 numbers into 2 groups \{2, 8\}, \{5, 3\}. Thus the answer is (2 + 8)^2 + (5 + 3)^2 = 164.
In the second sample, one of the optimal solutions is to divide those 6 numbers into 3 groups \{1, 2\}, \{1, 2\}, \{1, 2\}. Thus the answer is (1 + 2)^2 + (1 + 2)^2 + (1 + 2)^2 = 27.
Submitted Solution:
```
import heapq
n = int(input())
g = n//2
nums = list(map(int,input().split()))
nums = sorted(nums,reverse = True)
groups = [float("inf")]+[0 for _ in range(g)]+[float("inf")]
i = 1
h = [(0,_+1) for _ in range(g)]
heapq.heapify(h)
for num in nums:
total,i = heapq.heappop(h)
groups[i] += num
heapq.heappush(h,(groups[i],i))
print (sum(x**2 for x in groups[1:-1]))
```
No
| 13,357 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Lunar New Year is approaching, and Bob is struggling with his homework β a number division problem.
There are n positive integers a_1, a_2, β¦, a_n on Bob's homework paper, where n is always an even number. Bob is asked to divide those numbers into groups, where each group must contain at least 2 numbers. Suppose the numbers are divided into m groups, and the sum of the numbers in the j-th group is s_j. Bob's aim is to minimize the sum of the square of s_j, that is $$$β_{j = 1}^{m} s_j^2.$$$
Bob is puzzled by this hard problem. Could you please help him solve it?
Input
The first line contains an even integer n (2 β€ n β€ 3 β
10^5), denoting that there are n integers on Bob's homework paper.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 10^4), describing the numbers you need to deal with.
Output
A single line containing one integer, denoting the minimum of the sum of the square of s_j, which is $$$β_{i = j}^{m} s_j^2, where m$$$ is the number of groups.
Examples
Input
4
8 5 2 3
Output
164
Input
6
1 1 1 2 2 2
Output
27
Note
In the first sample, one of the optimal solutions is to divide those 4 numbers into 2 groups \{2, 8\}, \{5, 3\}. Thus the answer is (2 + 8)^2 + (5 + 3)^2 = 164.
In the second sample, one of the optimal solutions is to divide those 6 numbers into 3 groups \{1, 2\}, \{1, 2\}, \{1, 2\}. Thus the answer is (1 + 2)^2 + (1 + 2)^2 + (1 + 2)^2 = 27.
Submitted Solution:
```
n=int(input())
s=input().split()
s=list(s)
for i in range(n):
s[i]=int(s[i])
s.sort()
sum=0
print(*s)
for i in range(n//2):
sum+=((s[i]) + s[n-1-i])**2
print(sum)
```
No
| 13,358 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two arrays a and b, each contains n integers.
You want to create a new array c as follows: choose some real (i.e. not necessarily integer) number d, and then for every i β [1, n] let c_i := d β
a_i + b_i.
Your goal is to maximize the number of zeroes in array c. What is the largest possible answer, if you choose d optimally?
Input
The first line contains one integer n (1 β€ n β€ 2 β
10^5) β the number of elements in both arrays.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 β€ a_i β€ 10^9).
The third line contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9).
Output
Print one integer β the maximum number of zeroes in array c, if you choose d optimally.
Examples
Input
5
1 2 3 4 5
2 4 7 11 3
Output
2
Input
3
13 37 39
1 2 3
Output
2
Input
4
0 0 0 0
1 2 3 4
Output
0
Input
3
1 2 -1
-6 -12 6
Output
3
Note
In the first example, we may choose d = -2.
In the second example, we may choose d = -1/13.
In the third example, we cannot obtain any zero in array c, no matter which d we choose.
In the fourth example, we may choose d = 6.
Tags: hashing, math, number theory
Correct Solution:
```
from fractions import Fraction
from collections import Counter
inf = 10**9+7
n = int(input())
a = [int(s) for s in input().split()]
b = [int(s) for s in input().split()]
d = []
zeros = 0
for i in range(n):
if a[i] != 0:
d.append(Fraction(-b[i], a[i]))
else:
if b[i] == 0:
zeros += 1
de = Counter(d)
if de == {}:
print(0+zeros)
else:
ans = max(list(de.values()))+zeros
print(ans)
```
| 13,359 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two arrays a and b, each contains n integers.
You want to create a new array c as follows: choose some real (i.e. not necessarily integer) number d, and then for every i β [1, n] let c_i := d β
a_i + b_i.
Your goal is to maximize the number of zeroes in array c. What is the largest possible answer, if you choose d optimally?
Input
The first line contains one integer n (1 β€ n β€ 2 β
10^5) β the number of elements in both arrays.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 β€ a_i β€ 10^9).
The third line contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9).
Output
Print one integer β the maximum number of zeroes in array c, if you choose d optimally.
Examples
Input
5
1 2 3 4 5
2 4 7 11 3
Output
2
Input
3
13 37 39
1 2 3
Output
2
Input
4
0 0 0 0
1 2 3 4
Output
0
Input
3
1 2 -1
-6 -12 6
Output
3
Note
In the first example, we may choose d = -2.
In the second example, we may choose d = -1/13.
In the third example, we cannot obtain any zero in array c, no matter which d we choose.
In the fourth example, we may choose d = 6.
Tags: hashing, math, number theory
Correct Solution:
```
from decimal import getcontext, Decimal
getcontext().prec = 50
n = int(input())
arr1 = list(map(int, input().split()))
arr2 = list(map(int, input().split()))
mp = {}
appendix = 0
for i in range(n):
if arr1[i] == arr2[i] == 0:
appendix += 1
elif arr1[i] != 0:
tmp = Decimal(arr2[i]) / Decimal(arr1[i])
mp.setdefault(tmp, 0)
mp[tmp] += 1
if len(mp):
print(max(mp.values()) + appendix)
else:
print(appendix)
```
| 13,360 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two arrays a and b, each contains n integers.
You want to create a new array c as follows: choose some real (i.e. not necessarily integer) number d, and then for every i β [1, n] let c_i := d β
a_i + b_i.
Your goal is to maximize the number of zeroes in array c. What is the largest possible answer, if you choose d optimally?
Input
The first line contains one integer n (1 β€ n β€ 2 β
10^5) β the number of elements in both arrays.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 β€ a_i β€ 10^9).
The third line contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9).
Output
Print one integer β the maximum number of zeroes in array c, if you choose d optimally.
Examples
Input
5
1 2 3 4 5
2 4 7 11 3
Output
2
Input
3
13 37 39
1 2 3
Output
2
Input
4
0 0 0 0
1 2 3 4
Output
0
Input
3
1 2 -1
-6 -12 6
Output
3
Note
In the first example, we may choose d = -2.
In the second example, we may choose d = -1/13.
In the third example, we cannot obtain any zero in array c, no matter which d we choose.
In the fourth example, we may choose d = 6.
Tags: hashing, math, number theory
Correct Solution:
```
from fractions import gcd
n = int(input())
ai = list(map(int,input().split()))
bi = list(map(int,input().split()))
s = {}
num = 0
mod = 2 * 10**9 + 1
for i in range(n):
if ai[i] != 0:
temp = gcd(ai[i],bi[i])
try :
s[-ai[i] // temp + mod * (bi[i] // temp)] += 1
except :
s[-ai[i] // temp + mod * (bi[i] // temp)] = 1
elif bi[i] == ai[i]:
num += 1
ans = 0
for i in s:
ans = max(ans,s[i])
print(ans + num)
```
| 13,361 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two arrays a and b, each contains n integers.
You want to create a new array c as follows: choose some real (i.e. not necessarily integer) number d, and then for every i β [1, n] let c_i := d β
a_i + b_i.
Your goal is to maximize the number of zeroes in array c. What is the largest possible answer, if you choose d optimally?
Input
The first line contains one integer n (1 β€ n β€ 2 β
10^5) β the number of elements in both arrays.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 β€ a_i β€ 10^9).
The third line contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9).
Output
Print one integer β the maximum number of zeroes in array c, if you choose d optimally.
Examples
Input
5
1 2 3 4 5
2 4 7 11 3
Output
2
Input
3
13 37 39
1 2 3
Output
2
Input
4
0 0 0 0
1 2 3 4
Output
0
Input
3
1 2 -1
-6 -12 6
Output
3
Note
In the first example, we may choose d = -2.
In the second example, we may choose d = -1/13.
In the third example, we cannot obtain any zero in array c, no matter which d we choose.
In the fourth example, we may choose d = 6.
Tags: hashing, math, number theory
Correct Solution:
```
from decimal import Decimal
from collections import *
n = int(input())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
c = []
d = defaultdict(int)
ans = 0
value = 0
for i in range(n):
if (a[i]==0 and b[i]!=0):
continue
if a[i]==0 and b[i]==0:
ans+=1
continue
val = Decimal(b[i])/Decimal(a[i])
d[val]+=1
value = max(value,d[val])
print(value+ans)
```
| 13,362 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two arrays a and b, each contains n integers.
You want to create a new array c as follows: choose some real (i.e. not necessarily integer) number d, and then for every i β [1, n] let c_i := d β
a_i + b_i.
Your goal is to maximize the number of zeroes in array c. What is the largest possible answer, if you choose d optimally?
Input
The first line contains one integer n (1 β€ n β€ 2 β
10^5) β the number of elements in both arrays.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 β€ a_i β€ 10^9).
The third line contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9).
Output
Print one integer β the maximum number of zeroes in array c, if you choose d optimally.
Examples
Input
5
1 2 3 4 5
2 4 7 11 3
Output
2
Input
3
13 37 39
1 2 3
Output
2
Input
4
0 0 0 0
1 2 3 4
Output
0
Input
3
1 2 -1
-6 -12 6
Output
3
Note
In the first example, we may choose d = -2.
In the second example, we may choose d = -1/13.
In the third example, we cannot obtain any zero in array c, no matter which d we choose.
In the fourth example, we may choose d = 6.
Tags: hashing, math, number theory
Correct Solution:
```
from collections import Counter
from fractions import Fraction
N = int(input())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
D = []
indef_num = 0
for (a, b) in zip(A, B):
if a == 0:
if b == 0:
indef_num += 1
continue
D.append(Fraction(-b, a))
if not D:
print(indef_num)
else:
d = Counter(D).most_common()
print(D.count(d[0][0]) + indef_num)
```
| 13,363 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two arrays a and b, each contains n integers.
You want to create a new array c as follows: choose some real (i.e. not necessarily integer) number d, and then for every i β [1, n] let c_i := d β
a_i + b_i.
Your goal is to maximize the number of zeroes in array c. What is the largest possible answer, if you choose d optimally?
Input
The first line contains one integer n (1 β€ n β€ 2 β
10^5) β the number of elements in both arrays.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 β€ a_i β€ 10^9).
The third line contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9).
Output
Print one integer β the maximum number of zeroes in array c, if you choose d optimally.
Examples
Input
5
1 2 3 4 5
2 4 7 11 3
Output
2
Input
3
13 37 39
1 2 3
Output
2
Input
4
0 0 0 0
1 2 3 4
Output
0
Input
3
1 2 -1
-6 -12 6
Output
3
Note
In the first example, we may choose d = -2.
In the second example, we may choose d = -1/13.
In the third example, we cannot obtain any zero in array c, no matter which d we choose.
In the fourth example, we may choose d = 6.
Tags: hashing, math, number theory
Correct Solution:
```
from math import gcd
def get_sign(_u, _b):
if (_u > 0 and _b > 0) or (_u < 0 and _b < 0):
return ""
else:
return "-"
if __name__ == '__main__':
input()
u = list(map(int, input().split(" ")))
b = list(map(int, input().split(" ")))
acc = 0
reduced = []
extra = 0
for i in range(len(u)):
_u = abs(u[i])
_b = abs(b[i])
sign = get_sign(u[i], b[i])
if _u == 0 or _b == 0:
if _u == 0 and _b == 0:
acc += 1
continue
elif _u != 0 and _b == 0:
extra += 1
continue
else:
continue
while True:
_gcd = gcd(_u, _b)
if _gcd == 1:
break
_u = _u // _gcd
_b = _b // _gcd
reduced.append(sign + str(_u) + ":" + str(_b))
m = {}
for i in range(len(reduced)):
if reduced[i] in m:
m[reduced[i]] += 1
else:
m[reduced[i]] = 1
_max = 0
for key in m:
if m[key] > _max:
_max = m[key]
print(max(acc + _max, acc + extra))
```
| 13,364 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two arrays a and b, each contains n integers.
You want to create a new array c as follows: choose some real (i.e. not necessarily integer) number d, and then for every i β [1, n] let c_i := d β
a_i + b_i.
Your goal is to maximize the number of zeroes in array c. What is the largest possible answer, if you choose d optimally?
Input
The first line contains one integer n (1 β€ n β€ 2 β
10^5) β the number of elements in both arrays.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 β€ a_i β€ 10^9).
The third line contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9).
Output
Print one integer β the maximum number of zeroes in array c, if you choose d optimally.
Examples
Input
5
1 2 3 4 5
2 4 7 11 3
Output
2
Input
3
13 37 39
1 2 3
Output
2
Input
4
0 0 0 0
1 2 3 4
Output
0
Input
3
1 2 -1
-6 -12 6
Output
3
Note
In the first example, we may choose d = -2.
In the second example, we may choose d = -1/13.
In the third example, we cannot obtain any zero in array c, no matter which d we choose.
In the fourth example, we may choose d = 6.
Tags: hashing, math, number theory
Correct Solution:
```
n=int(input())
from decimal import *
getcontext().prec = 100
a=[int(i) for i in input().split()]
b=[int(i) for i in input().split()]
sm=0
d1=[]
d2=[]
cnt=0
for i in range(n):
if a[i]==0 and b[i]==0:
cnt+=1
continue
if a[i]==0 :
continue
d1.append((-Decimal(b[i]))/Decimal(a[i]))
from collections import Counter
c=Counter(d1)
maxi=0
for i in c:
maxi=max(maxi,c[i])
print(maxi+cnt)
```
| 13,365 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two arrays a and b, each contains n integers.
You want to create a new array c as follows: choose some real (i.e. not necessarily integer) number d, and then for every i β [1, n] let c_i := d β
a_i + b_i.
Your goal is to maximize the number of zeroes in array c. What is the largest possible answer, if you choose d optimally?
Input
The first line contains one integer n (1 β€ n β€ 2 β
10^5) β the number of elements in both arrays.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 β€ a_i β€ 10^9).
The third line contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9).
Output
Print one integer β the maximum number of zeroes in array c, if you choose d optimally.
Examples
Input
5
1 2 3 4 5
2 4 7 11 3
Output
2
Input
3
13 37 39
1 2 3
Output
2
Input
4
0 0 0 0
1 2 3 4
Output
0
Input
3
1 2 -1
-6 -12 6
Output
3
Note
In the first example, we may choose d = -2.
In the second example, we may choose d = -1/13.
In the third example, we cannot obtain any zero in array c, no matter which d we choose.
In the fourth example, we may choose d = 6.
Tags: hashing, math, number theory
Correct Solution:
```
'''
i see two problems now:
is 1000000/17 = 2000000/34 something along that line, involving multiple decimals
solution is do sedsuonyangtam.
a = 0, b = 0
'''
def gcd(a, b):
if b == 0:
return a
else:
return gcd(b, a % b)
n = int(input())
a = [int(e) for e in input().split()]
b = [int(e) for e in input().split()]
d_nonneg, d_neg = {}, {}
m, v = 0, 0
for i in range(n):
if a[i] == 0 and b[i] == 0:
v = v + 1
if a[i] != 0:
aa, bb = a[i], b[i]
if bb == 0:
if (0, 1) not in d_nonneg:
d_nonneg[(0, 1)] = 1
else:
d_nonneg[(0, 1)] += 1
p = d_nonneg[(0, 1)]
if aa < 0 and bb < 0:
aa, bb = aa * (-1), bb * (-1)
if aa > 0 and bb > 0:
x = gcd(aa, bb)
aa, bb = aa // x, bb // x
if (aa, bb) not in d_nonneg:
d_nonneg[(aa, bb)] = 1
else:
d_nonneg[(aa, bb)] += 1
p = d_nonneg[(aa, bb)]
else:
if aa < 0:
aa = aa * (-1)
else:
bb = bb * (-1)
x = gcd(aa, bb)
aa, bb = aa // x, bb // x
if (aa, bb) not in d_neg:
d_neg[(aa, bb)] = 1
else:
d_neg[(aa, bb)] += 1
p = d_neg[(aa, bb)]
if p > m:
m = p
print(m + v)
```
| 13,366 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two arrays a and b, each contains n integers.
You want to create a new array c as follows: choose some real (i.e. not necessarily integer) number d, and then for every i β [1, n] let c_i := d β
a_i + b_i.
Your goal is to maximize the number of zeroes in array c. What is the largest possible answer, if you choose d optimally?
Input
The first line contains one integer n (1 β€ n β€ 2 β
10^5) β the number of elements in both arrays.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 β€ a_i β€ 10^9).
The third line contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9).
Output
Print one integer β the maximum number of zeroes in array c, if you choose d optimally.
Examples
Input
5
1 2 3 4 5
2 4 7 11 3
Output
2
Input
3
13 37 39
1 2 3
Output
2
Input
4
0 0 0 0
1 2 3 4
Output
0
Input
3
1 2 -1
-6 -12 6
Output
3
Note
In the first example, we may choose d = -2.
In the second example, we may choose d = -1/13.
In the third example, we cannot obtain any zero in array c, no matter which d we choose.
In the fourth example, we may choose d = 6.
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
dct={}
import math
def simple_fraction(a,b):
x=math.gcd(a,b)
a//=x
b//=x
if a==0:
return (0,1)
if (a<0 and b>0) or (a>0 and b<0):
return (-abs(a),abs(b))
return (abs(a),abs(b))
allcase=0
ans=0
for i in range(n):
if a[i]==0:
if b[i]==0:
allcase+=1
continue
d=simple_fraction(-b[i],a[i])
if d in dct:
dct[d]+=1
ans=max(ans,dct[d])
else:
dct[d]=1
ans=max(ans,1)
print(allcase+ans)
```
Yes
| 13,367 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two arrays a and b, each contains n integers.
You want to create a new array c as follows: choose some real (i.e. not necessarily integer) number d, and then for every i β [1, n] let c_i := d β
a_i + b_i.
Your goal is to maximize the number of zeroes in array c. What is the largest possible answer, if you choose d optimally?
Input
The first line contains one integer n (1 β€ n β€ 2 β
10^5) β the number of elements in both arrays.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 β€ a_i β€ 10^9).
The third line contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9).
Output
Print one integer β the maximum number of zeroes in array c, if you choose d optimally.
Examples
Input
5
1 2 3 4 5
2 4 7 11 3
Output
2
Input
3
13 37 39
1 2 3
Output
2
Input
4
0 0 0 0
1 2 3 4
Output
0
Input
3
1 2 -1
-6 -12 6
Output
3
Note
In the first example, we may choose d = -2.
In the second example, we may choose d = -1/13.
In the third example, we cannot obtain any zero in array c, no matter which d we choose.
In the fourth example, we may choose d = 6.
Submitted Solution:
```
# https://codeforces.com/problemset/problem/1133/D
from sys import stdin, exit
from typing import List, Tuple
import math
stdin.readline() # ignore
a_arr = [int(v) for v in stdin.readline().rstrip().split()]
b_arr = [int(v) for v in stdin.readline().rstrip().split()]
def fractionize(numerator: int, denominator: int):
if denominator < 0:
numerator = -numerator
denominator = -denominator
elif denominator == 0:
raise ZeroDivisionError()
gcd = math.gcd(abs(numerator), denominator)
return numerator // gcd, denominator // gcd
class FractionN:
numerator: int
denominator: int
def __init__(self, numerator: int, denominator=1):
self.numerator, self.denominator = fractionize(numerator, denominator)
def __eq__(self, other):
return self.numerator == other.numerator and self.denominator == other.denominator
def __hash__(self):
return hash((self.numerator, self.denominator))
def __repr__(self):
return f"<Fraction({self.numerator} / {self.denominator})>"
frac_occurs = dict()
wildcards = 0
not_divisible = 0
for a, b in zip(a_arr, b_arr):
if a == 0 and b != 0:
not_divisible += 1
continue
elif a == 0 and b == 0:
wildcards += 1
continue
else:
# key = fractionize(-b, a)
key = FractionN(-b, a)
if key in frac_occurs:
frac_occurs[key] += 1
else:
frac_occurs[key] = 1
if wildcards == 0 and not_divisible == len(a_arr):
print(0)
else:
print((max(frac_occurs.values()) if len(frac_occurs) > 0 else 0) + wildcards)
```
Yes
| 13,368 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two arrays a and b, each contains n integers.
You want to create a new array c as follows: choose some real (i.e. not necessarily integer) number d, and then for every i β [1, n] let c_i := d β
a_i + b_i.
Your goal is to maximize the number of zeroes in array c. What is the largest possible answer, if you choose d optimally?
Input
The first line contains one integer n (1 β€ n β€ 2 β
10^5) β the number of elements in both arrays.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 β€ a_i β€ 10^9).
The third line contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9).
Output
Print one integer β the maximum number of zeroes in array c, if you choose d optimally.
Examples
Input
5
1 2 3 4 5
2 4 7 11 3
Output
2
Input
3
13 37 39
1 2 3
Output
2
Input
4
0 0 0 0
1 2 3 4
Output
0
Input
3
1 2 -1
-6 -12 6
Output
3
Note
In the first example, we may choose d = -2.
In the second example, we may choose d = -1/13.
In the third example, we cannot obtain any zero in array c, no matter which d we choose.
In the fourth example, we may choose d = 6.
Submitted Solution:
```
from sys import stdin,stdout
from itertools import combinations
from collections import defaultdict
from decimal import *
getcontext().prec = 50
def listIn():
return list((map(int,stdin.readline().strip().split())))
def stringListIn():
return([x for x in stdin.readline().split()])
def intIn():
return (int(stdin.readline()))
def stringIn():
return (stdin.readline().strip())
def solve(a,b,n):
d=[0]*n
li={}
cnt=0
for i in range(n):
if a[i]!=0:
d[i]=-(Decimal(b[i])/Decimal(a[i]))
else:
if b[i]==0:
cnt+=1
continue
ele=d[i]
if ele not in li:
li[ele]=1
else:
li[ele]+=1
#print(li)
if len(li)>0:
return (max(li.values())+cnt)
else:
return cnt
if __name__=="__main__":
n=intIn()
a=listIn()
b=listIn()
ans=solve(a,b,n)
print(ans)
```
Yes
| 13,369 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two arrays a and b, each contains n integers.
You want to create a new array c as follows: choose some real (i.e. not necessarily integer) number d, and then for every i β [1, n] let c_i := d β
a_i + b_i.
Your goal is to maximize the number of zeroes in array c. What is the largest possible answer, if you choose d optimally?
Input
The first line contains one integer n (1 β€ n β€ 2 β
10^5) β the number of elements in both arrays.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 β€ a_i β€ 10^9).
The third line contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9).
Output
Print one integer β the maximum number of zeroes in array c, if you choose d optimally.
Examples
Input
5
1 2 3 4 5
2 4 7 11 3
Output
2
Input
3
13 37 39
1 2 3
Output
2
Input
4
0 0 0 0
1 2 3 4
Output
0
Input
3
1 2 -1
-6 -12 6
Output
3
Note
In the first example, we may choose d = -2.
In the second example, we may choose d = -1/13.
In the third example, we cannot obtain any zero in array c, no matter which d we choose.
In the fourth example, we may choose d = 6.
Submitted Solution:
```
#!/usr/bin/env python
# coding: utf-8
# In[17]:
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
from collections import defaultdict
def lcd(a, b):
while b:
a, b = b, a % b
return a
c = [0] * n
nums = defaultdict(int)
res_max = 0
res_zero = 0
for i in range(n):
if a[i] == 0:
if not b[i]:
res_zero += 1
continue
r = -b[i] * 1e10 / a[i]
nums[r] += 1
res_max = max(res_max, nums[r])
print(res_max + res_zero)
```
Yes
| 13,370 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two arrays a and b, each contains n integers.
You want to create a new array c as follows: choose some real (i.e. not necessarily integer) number d, and then for every i β [1, n] let c_i := d β
a_i + b_i.
Your goal is to maximize the number of zeroes in array c. What is the largest possible answer, if you choose d optimally?
Input
The first line contains one integer n (1 β€ n β€ 2 β
10^5) β the number of elements in both arrays.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 β€ a_i β€ 10^9).
The third line contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9).
Output
Print one integer β the maximum number of zeroes in array c, if you choose d optimally.
Examples
Input
5
1 2 3 4 5
2 4 7 11 3
Output
2
Input
3
13 37 39
1 2 3
Output
2
Input
4
0 0 0 0
1 2 3 4
Output
0
Input
3
1 2 -1
-6 -12 6
Output
3
Note
In the first example, we may choose d = -2.
In the second example, we may choose d = -1/13.
In the third example, we cannot obtain any zero in array c, no matter which d we choose.
In the fourth example, we may choose d = 6.
Submitted Solution:
```
n = int(input())
b = list(map(int, input(). split()))
c = list(map(int, input(). split()))
k = {}
mx = 0
q = 0
br = 0
for i in range(n):
if b[i] == 0 and c[i] == 0:
q += 1
continue
if b[i] == 0:
br = 1
break
s = -c[i] / b[i]
if s in k:
k[s] += 1
if k[s] > mx:
mx = k[s]
else:
k[s] = 1
if k[s] > mx:
mx = k[s]
if br == 1:
print(0)
else:
print(mx + q)
```
No
| 13,371 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two arrays a and b, each contains n integers.
You want to create a new array c as follows: choose some real (i.e. not necessarily integer) number d, and then for every i β [1, n] let c_i := d β
a_i + b_i.
Your goal is to maximize the number of zeroes in array c. What is the largest possible answer, if you choose d optimally?
Input
The first line contains one integer n (1 β€ n β€ 2 β
10^5) β the number of elements in both arrays.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 β€ a_i β€ 10^9).
The third line contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9).
Output
Print one integer β the maximum number of zeroes in array c, if you choose d optimally.
Examples
Input
5
1 2 3 4 5
2 4 7 11 3
Output
2
Input
3
13 37 39
1 2 3
Output
2
Input
4
0 0 0 0
1 2 3 4
Output
0
Input
3
1 2 -1
-6 -12 6
Output
3
Note
In the first example, we may choose d = -2.
In the second example, we may choose d = -1/13.
In the third example, we cannot obtain any zero in array c, no matter which d we choose.
In the fourth example, we may choose d = 6.
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
c=[-99999999999999999999999999]*n
for i in range (n):
if(a[i]!=0):
c[i]= (-b[i]/a[i])
#print(c)
c.sort()
co=0
for i in range (n):
if(a[i]==0 and b[i]!=0):
co+=1
if(co==n):
print(0)
exit()
maxc=0
co=1
for i in range (1,n):
if(c[i]==c[i-1] and c[i]!=-99999999999999999999999999 ):
co+=1
if(co>maxc):
maxc=co
else:
co=1
for i in range (n):
if(a[i]==0 and b[i]==0):
maxc+=1
print(maxc)
```
No
| 13,372 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two arrays a and b, each contains n integers.
You want to create a new array c as follows: choose some real (i.e. not necessarily integer) number d, and then for every i β [1, n] let c_i := d β
a_i + b_i.
Your goal is to maximize the number of zeroes in array c. What is the largest possible answer, if you choose d optimally?
Input
The first line contains one integer n (1 β€ n β€ 2 β
10^5) β the number of elements in both arrays.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 β€ a_i β€ 10^9).
The third line contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9).
Output
Print one integer β the maximum number of zeroes in array c, if you choose d optimally.
Examples
Input
5
1 2 3 4 5
2 4 7 11 3
Output
2
Input
3
13 37 39
1 2 3
Output
2
Input
4
0 0 0 0
1 2 3 4
Output
0
Input
3
1 2 -1
-6 -12 6
Output
3
Note
In the first example, we may choose d = -2.
In the second example, we may choose d = -1/13.
In the third example, we cannot obtain any zero in array c, no matter which d we choose.
In the fourth example, we may choose d = 6.
Submitted Solution:
```
N=int(input())
A=list(map(int,input().split()))
B=list(map(int,input().split()))
D={'':0}
count = 0
for a,b in zip(A,B):
if a !=0:
x=b/a
if x in D:
D[x] += 1
else:
D[x] = 1
else:
count += 1
print(max(list(D.values()))+count)
```
No
| 13,373 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two arrays a and b, each contains n integers.
You want to create a new array c as follows: choose some real (i.e. not necessarily integer) number d, and then for every i β [1, n] let c_i := d β
a_i + b_i.
Your goal is to maximize the number of zeroes in array c. What is the largest possible answer, if you choose d optimally?
Input
The first line contains one integer n (1 β€ n β€ 2 β
10^5) β the number of elements in both arrays.
The second line contains n integers a_1, a_2, ..., a_n (-10^9 β€ a_i β€ 10^9).
The third line contains n integers b_1, b_2, ..., b_n (-10^9 β€ b_i β€ 10^9).
Output
Print one integer β the maximum number of zeroes in array c, if you choose d optimally.
Examples
Input
5
1 2 3 4 5
2 4 7 11 3
Output
2
Input
3
13 37 39
1 2 3
Output
2
Input
4
0 0 0 0
1 2 3 4
Output
0
Input
3
1 2 -1
-6 -12 6
Output
3
Note
In the first example, we may choose d = -2.
In the second example, we may choose d = -1/13.
In the third example, we cannot obtain any zero in array c, no matter which d we choose.
In the fourth example, we may choose d = 6.
Submitted Solution:
```
import sys
from fractions import gcd
input = sys.stdin.readline
n = int(input())
arr1 = [int(i) for i in input().split()]
arr2 = [int(i) for i in input().split()]
arr3 = dict()
for i in range(len(arr1)):
if arr1[i] is not 0:
if arr2[i] is 0:
arr3["0"] = arr3.get("0",0)+1
else:
g = gcd(arr2[i],arr1[i])
s = "%d/%d"%(arr2[i]//g,arr1[i]//g)
arr3[s] = arr3.get(s,0) + 1
if arr1[i]==0 and arr2[i]==0:
arr3[0] = arr3.get(0,0) + 1
res = list(arr3.values())
if len(res)==0:
print(0)
else:
print(max(res))
```
No
| 13,374 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has guessed three positive integers a, b and c. He keeps these numbers in secret, but he writes down four numbers on a board in arbitrary order β their pairwise sums (three numbers) and sum of all three numbers (one number). So, there are four numbers on a board in random order: a+b, a+c, b+c and a+b+c.
You have to guess three numbers a, b and c using given numbers. Print three guessed integers in any order.
Pay attention that some given numbers a, b and c can be equal (it is also possible that a=b=c).
Input
The only line of the input contains four positive integers x_1, x_2, x_3, x_4 (2 β€ x_i β€ 10^9) β numbers written on a board in random order. It is guaranteed that the answer exists for the given number x_1, x_2, x_3, x_4.
Output
Print such positive integers a, b and c that four numbers written on a board are values a+b, a+c, b+c and a+b+c written in some order. Print a, b and c in any order. If there are several answers, you can print any. It is guaranteed that the answer exists.
Examples
Input
3 6 5 4
Output
2 1 3
Input
40 40 40 60
Output
20 20 20
Input
201 101 101 200
Output
1 100 100
Tags: math
Correct Solution:
```
a,b,c,d=map(int,input().split())
l=[a,b,c,d]
l.sort()
print(l[-1]-l[0],l[-1]-l[1],l[-1]-l[2])
```
| 13,375 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has guessed three positive integers a, b and c. He keeps these numbers in secret, but he writes down four numbers on a board in arbitrary order β their pairwise sums (three numbers) and sum of all three numbers (one number). So, there are four numbers on a board in random order: a+b, a+c, b+c and a+b+c.
You have to guess three numbers a, b and c using given numbers. Print three guessed integers in any order.
Pay attention that some given numbers a, b and c can be equal (it is also possible that a=b=c).
Input
The only line of the input contains four positive integers x_1, x_2, x_3, x_4 (2 β€ x_i β€ 10^9) β numbers written on a board in random order. It is guaranteed that the answer exists for the given number x_1, x_2, x_3, x_4.
Output
Print such positive integers a, b and c that four numbers written on a board are values a+b, a+c, b+c and a+b+c written in some order. Print a, b and c in any order. If there are several answers, you can print any. It is guaranteed that the answer exists.
Examples
Input
3 6 5 4
Output
2 1 3
Input
40 40 40 60
Output
20 20 20
Input
201 101 101 200
Output
1 100 100
Tags: math
Correct Solution:
```
arr = list(map(int,input().split()))
suma = max(arr)
for i in arr:
if(i == suma):
continue
print(suma - i,end = ' ')
```
| 13,376 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has guessed three positive integers a, b and c. He keeps these numbers in secret, but he writes down four numbers on a board in arbitrary order β their pairwise sums (three numbers) and sum of all three numbers (one number). So, there are four numbers on a board in random order: a+b, a+c, b+c and a+b+c.
You have to guess three numbers a, b and c using given numbers. Print three guessed integers in any order.
Pay attention that some given numbers a, b and c can be equal (it is also possible that a=b=c).
Input
The only line of the input contains four positive integers x_1, x_2, x_3, x_4 (2 β€ x_i β€ 10^9) β numbers written on a board in random order. It is guaranteed that the answer exists for the given number x_1, x_2, x_3, x_4.
Output
Print such positive integers a, b and c that four numbers written on a board are values a+b, a+c, b+c and a+b+c written in some order. Print a, b and c in any order. If there are several answers, you can print any. It is guaranteed that the answer exists.
Examples
Input
3 6 5 4
Output
2 1 3
Input
40 40 40 60
Output
20 20 20
Input
201 101 101 200
Output
1 100 100
Tags: math
Correct Solution:
```
'''Rajat pal'''
from collections import defaultdict
from sys import stdin
from bisect import bisect_left as bl, bisect_right as br,insort
import heapq
from collections import deque
import math
####### Stay away from negative people. They have problem for every solution.########
a=list(map(int,input().rstrip().split()))
maxi=max(a)
index=a.index(maxi)
for i in range(4):
if i==index:
pass
else:
print(maxi-a[i],end=' ')
```
| 13,377 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has guessed three positive integers a, b and c. He keeps these numbers in secret, but he writes down four numbers on a board in arbitrary order β their pairwise sums (three numbers) and sum of all three numbers (one number). So, there are four numbers on a board in random order: a+b, a+c, b+c and a+b+c.
You have to guess three numbers a, b and c using given numbers. Print three guessed integers in any order.
Pay attention that some given numbers a, b and c can be equal (it is also possible that a=b=c).
Input
The only line of the input contains four positive integers x_1, x_2, x_3, x_4 (2 β€ x_i β€ 10^9) β numbers written on a board in random order. It is guaranteed that the answer exists for the given number x_1, x_2, x_3, x_4.
Output
Print such positive integers a, b and c that four numbers written on a board are values a+b, a+c, b+c and a+b+c written in some order. Print a, b and c in any order. If there are several answers, you can print any. It is guaranteed that the answer exists.
Examples
Input
3 6 5 4
Output
2 1 3
Input
40 40 40 60
Output
20 20 20
Input
201 101 101 200
Output
1 100 100
Tags: math
Correct Solution:
```
abcd=list(map(int,input().split()))
abcd=sorted(abcd)
ans=[]
m=max(abcd)
for i in abcd:
if i!=m:
ans.append(m-i)
print(*ans[::-1])
```
| 13,378 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has guessed three positive integers a, b and c. He keeps these numbers in secret, but he writes down four numbers on a board in arbitrary order β their pairwise sums (three numbers) and sum of all three numbers (one number). So, there are four numbers on a board in random order: a+b, a+c, b+c and a+b+c.
You have to guess three numbers a, b and c using given numbers. Print three guessed integers in any order.
Pay attention that some given numbers a, b and c can be equal (it is also possible that a=b=c).
Input
The only line of the input contains four positive integers x_1, x_2, x_3, x_4 (2 β€ x_i β€ 10^9) β numbers written on a board in random order. It is guaranteed that the answer exists for the given number x_1, x_2, x_3, x_4.
Output
Print such positive integers a, b and c that four numbers written on a board are values a+b, a+c, b+c and a+b+c written in some order. Print a, b and c in any order. If there are several answers, you can print any. It is guaranteed that the answer exists.
Examples
Input
3 6 5 4
Output
2 1 3
Input
40 40 40 60
Output
20 20 20
Input
201 101 101 200
Output
1 100 100
Tags: math
Correct Solution:
```
arr = sorted(list(map(int, input().split())))
print(' '.join([str(arr[3] - arr[i]) for i in range(3)]))
```
| 13,379 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has guessed three positive integers a, b and c. He keeps these numbers in secret, but he writes down four numbers on a board in arbitrary order β their pairwise sums (three numbers) and sum of all three numbers (one number). So, there are four numbers on a board in random order: a+b, a+c, b+c and a+b+c.
You have to guess three numbers a, b and c using given numbers. Print three guessed integers in any order.
Pay attention that some given numbers a, b and c can be equal (it is also possible that a=b=c).
Input
The only line of the input contains four positive integers x_1, x_2, x_3, x_4 (2 β€ x_i β€ 10^9) β numbers written on a board in random order. It is guaranteed that the answer exists for the given number x_1, x_2, x_3, x_4.
Output
Print such positive integers a, b and c that four numbers written on a board are values a+b, a+c, b+c and a+b+c written in some order. Print a, b and c in any order. If there are several answers, you can print any. It is guaranteed that the answer exists.
Examples
Input
3 6 5 4
Output
2 1 3
Input
40 40 40 60
Output
20 20 20
Input
201 101 101 200
Output
1 100 100
Tags: math
Correct Solution:
```
import math
l=[int(i) for i in input().split()]
l.sort(reverse=True)
p=l[1]
q=l[2]
r=l[3]
a=l[0]-l[1]
b=l[0]-l[2]
c=l[0]-l[3]
print(a,b,c)
```
| 13,380 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has guessed three positive integers a, b and c. He keeps these numbers in secret, but he writes down four numbers on a board in arbitrary order β their pairwise sums (three numbers) and sum of all three numbers (one number). So, there are four numbers on a board in random order: a+b, a+c, b+c and a+b+c.
You have to guess three numbers a, b and c using given numbers. Print three guessed integers in any order.
Pay attention that some given numbers a, b and c can be equal (it is also possible that a=b=c).
Input
The only line of the input contains four positive integers x_1, x_2, x_3, x_4 (2 β€ x_i β€ 10^9) β numbers written on a board in random order. It is guaranteed that the answer exists for the given number x_1, x_2, x_3, x_4.
Output
Print such positive integers a, b and c that four numbers written on a board are values a+b, a+c, b+c and a+b+c written in some order. Print a, b and c in any order. If there are several answers, you can print any. It is guaranteed that the answer exists.
Examples
Input
3 6 5 4
Output
2 1 3
Input
40 40 40 60
Output
20 20 20
Input
201 101 101 200
Output
1 100 100
Tags: math
Correct Solution:
```
a, b, c, d = map(int, input().split(' '))
x = int(max(a, b, c, d))
s = ''
if x != a:
s += str(x-a)+' '
if x != b:
s += str(x-b)+' '
if x != c:
s += str(x-c)+' '
if x != d:
s += str(x-d)+' '
print(s.strip())
```
| 13,381 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Polycarp has guessed three positive integers a, b and c. He keeps these numbers in secret, but he writes down four numbers on a board in arbitrary order β their pairwise sums (three numbers) and sum of all three numbers (one number). So, there are four numbers on a board in random order: a+b, a+c, b+c and a+b+c.
You have to guess three numbers a, b and c using given numbers. Print three guessed integers in any order.
Pay attention that some given numbers a, b and c can be equal (it is also possible that a=b=c).
Input
The only line of the input contains four positive integers x_1, x_2, x_3, x_4 (2 β€ x_i β€ 10^9) β numbers written on a board in random order. It is guaranteed that the answer exists for the given number x_1, x_2, x_3, x_4.
Output
Print such positive integers a, b and c that four numbers written on a board are values a+b, a+c, b+c and a+b+c written in some order. Print a, b and c in any order. If there are several answers, you can print any. It is guaranteed that the answer exists.
Examples
Input
3 6 5 4
Output
2 1 3
Input
40 40 40 60
Output
20 20 20
Input
201 101 101 200
Output
1 100 100
Tags: math
Correct Solution:
```
x1,x2,x3,x4 = map(int,input().split())
m = max(x1,x2,x3,x4)
if m == x1:
print(x1 - x2, x1 - x3, x1 - x4)
elif m == x2:
print(x2 - x1, x2 - x3, x2 - x4)
elif m == x3:
print(x3 - x1, x3 - x2, x3 - x4)
elif m == x4:
print(x4 - x1, x4 - x2, x4 - x3)
```
| 13,382 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has guessed three positive integers a, b and c. He keeps these numbers in secret, but he writes down four numbers on a board in arbitrary order β their pairwise sums (three numbers) and sum of all three numbers (one number). So, there are four numbers on a board in random order: a+b, a+c, b+c and a+b+c.
You have to guess three numbers a, b and c using given numbers. Print three guessed integers in any order.
Pay attention that some given numbers a, b and c can be equal (it is also possible that a=b=c).
Input
The only line of the input contains four positive integers x_1, x_2, x_3, x_4 (2 β€ x_i β€ 10^9) β numbers written on a board in random order. It is guaranteed that the answer exists for the given number x_1, x_2, x_3, x_4.
Output
Print such positive integers a, b and c that four numbers written on a board are values a+b, a+c, b+c and a+b+c written in some order. Print a, b and c in any order. If there are several answers, you can print any. It is guaranteed that the answer exists.
Examples
Input
3 6 5 4
Output
2 1 3
Input
40 40 40 60
Output
20 20 20
Input
201 101 101 200
Output
1 100 100
Submitted Solution:
```
x1,x2,x3,x4=sorted(map(int,input().split(' ')))
print(x4-x3,x4-x2,x4-x1)
```
Yes
| 13,383 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has guessed three positive integers a, b and c. He keeps these numbers in secret, but he writes down four numbers on a board in arbitrary order β their pairwise sums (three numbers) and sum of all three numbers (one number). So, there are four numbers on a board in random order: a+b, a+c, b+c and a+b+c.
You have to guess three numbers a, b and c using given numbers. Print three guessed integers in any order.
Pay attention that some given numbers a, b and c can be equal (it is also possible that a=b=c).
Input
The only line of the input contains four positive integers x_1, x_2, x_3, x_4 (2 β€ x_i β€ 10^9) β numbers written on a board in random order. It is guaranteed that the answer exists for the given number x_1, x_2, x_3, x_4.
Output
Print such positive integers a, b and c that four numbers written on a board are values a+b, a+c, b+c and a+b+c written in some order. Print a, b and c in any order. If there are several answers, you can print any. It is guaranteed that the answer exists.
Examples
Input
3 6 5 4
Output
2 1 3
Input
40 40 40 60
Output
20 20 20
Input
201 101 101 200
Output
1 100 100
Submitted Solution:
```
# https://codeforces.com/problemset/problem/1154/A
def main():
nums = sorted(list(map(int, input().split())))
print(nums[3] - nums[0], nums[3] - nums[1], nums[3] - nums[2])
if __name__ == '__main__':
main()
```
Yes
| 13,384 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has guessed three positive integers a, b and c. He keeps these numbers in secret, but he writes down four numbers on a board in arbitrary order β their pairwise sums (three numbers) and sum of all three numbers (one number). So, there are four numbers on a board in random order: a+b, a+c, b+c and a+b+c.
You have to guess three numbers a, b and c using given numbers. Print three guessed integers in any order.
Pay attention that some given numbers a, b and c can be equal (it is also possible that a=b=c).
Input
The only line of the input contains four positive integers x_1, x_2, x_3, x_4 (2 β€ x_i β€ 10^9) β numbers written on a board in random order. It is guaranteed that the answer exists for the given number x_1, x_2, x_3, x_4.
Output
Print such positive integers a, b and c that four numbers written on a board are values a+b, a+c, b+c and a+b+c written in some order. Print a, b and c in any order. If there are several answers, you can print any. It is guaranteed that the answer exists.
Examples
Input
3 6 5 4
Output
2 1 3
Input
40 40 40 60
Output
20 20 20
Input
201 101 101 200
Output
1 100 100
Submitted Solution:
```
sums = list(input().split(' '))
sums = list(map(int, sums))
sums.sort()
abc = max(sums)
print(abc-sums[0],abc-sums[1],abc-sums[2])
```
Yes
| 13,385 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has guessed three positive integers a, b and c. He keeps these numbers in secret, but he writes down four numbers on a board in arbitrary order β their pairwise sums (three numbers) and sum of all three numbers (one number). So, there are four numbers on a board in random order: a+b, a+c, b+c and a+b+c.
You have to guess three numbers a, b and c using given numbers. Print three guessed integers in any order.
Pay attention that some given numbers a, b and c can be equal (it is also possible that a=b=c).
Input
The only line of the input contains four positive integers x_1, x_2, x_3, x_4 (2 β€ x_i β€ 10^9) β numbers written on a board in random order. It is guaranteed that the answer exists for the given number x_1, x_2, x_3, x_4.
Output
Print such positive integers a, b and c that four numbers written on a board are values a+b, a+c, b+c and a+b+c written in some order. Print a, b and c in any order. If there are several answers, you can print any. It is guaranteed that the answer exists.
Examples
Input
3 6 5 4
Output
2 1 3
Input
40 40 40 60
Output
20 20 20
Input
201 101 101 200
Output
1 100 100
Submitted Solution:
```
l=list(map(int,input().split()))
d=sorted(l)
m=max(d)
a=m-d[0]
b=m-d[1]
c=m-d[2]
print(a,b,c)
```
Yes
| 13,386 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has guessed three positive integers a, b and c. He keeps these numbers in secret, but he writes down four numbers on a board in arbitrary order β their pairwise sums (three numbers) and sum of all three numbers (one number). So, there are four numbers on a board in random order: a+b, a+c, b+c and a+b+c.
You have to guess three numbers a, b and c using given numbers. Print three guessed integers in any order.
Pay attention that some given numbers a, b and c can be equal (it is also possible that a=b=c).
Input
The only line of the input contains four positive integers x_1, x_2, x_3, x_4 (2 β€ x_i β€ 10^9) β numbers written on a board in random order. It is guaranteed that the answer exists for the given number x_1, x_2, x_3, x_4.
Output
Print such positive integers a, b and c that four numbers written on a board are values a+b, a+c, b+c and a+b+c written in some order. Print a, b and c in any order. If there are several answers, you can print any. It is guaranteed that the answer exists.
Examples
Input
3 6 5 4
Output
2 1 3
Input
40 40 40 60
Output
20 20 20
Input
201 101 101 200
Output
1 100 100
Submitted Solution:
```
arr = sorted(list(map(int, input().split())))
li = []
for index in range(len(arr) - 1):
li.append(arr[-1] - arr[index])
print(sorted(li))
```
No
| 13,387 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has guessed three positive integers a, b and c. He keeps these numbers in secret, but he writes down four numbers on a board in arbitrary order β their pairwise sums (three numbers) and sum of all three numbers (one number). So, there are four numbers on a board in random order: a+b, a+c, b+c and a+b+c.
You have to guess three numbers a, b and c using given numbers. Print three guessed integers in any order.
Pay attention that some given numbers a, b and c can be equal (it is also possible that a=b=c).
Input
The only line of the input contains four positive integers x_1, x_2, x_3, x_4 (2 β€ x_i β€ 10^9) β numbers written on a board in random order. It is guaranteed that the answer exists for the given number x_1, x_2, x_3, x_4.
Output
Print such positive integers a, b and c that four numbers written on a board are values a+b, a+c, b+c and a+b+c written in some order. Print a, b and c in any order. If there are several answers, you can print any. It is guaranteed that the answer exists.
Examples
Input
3 6 5 4
Output
2 1 3
Input
40 40 40 60
Output
20 20 20
Input
201 101 101 200
Output
1 100 100
Submitted Solution:
```
#!/usr/bin/python3
# https://codeforces.com/problemset/problem/1154/A
input_str = input().split(" ")
num_list = [int(x) for x in input_str]
abc = max(num_list)
pop_flag = False
res = list()
for x in num_list:
if x == abc and not pop_flag:
pop_flag = True
pass
else:
print(x, end=" ")
```
No
| 13,388 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has guessed three positive integers a, b and c. He keeps these numbers in secret, but he writes down four numbers on a board in arbitrary order β their pairwise sums (three numbers) and sum of all three numbers (one number). So, there are four numbers on a board in random order: a+b, a+c, b+c and a+b+c.
You have to guess three numbers a, b and c using given numbers. Print three guessed integers in any order.
Pay attention that some given numbers a, b and c can be equal (it is also possible that a=b=c).
Input
The only line of the input contains four positive integers x_1, x_2, x_3, x_4 (2 β€ x_i β€ 10^9) β numbers written on a board in random order. It is guaranteed that the answer exists for the given number x_1, x_2, x_3, x_4.
Output
Print such positive integers a, b and c that four numbers written on a board are values a+b, a+c, b+c and a+b+c written in some order. Print a, b and c in any order. If there are several answers, you can print any. It is guaranteed that the answer exists.
Examples
Input
3 6 5 4
Output
2 1 3
Input
40 40 40 60
Output
20 20 20
Input
201 101 101 200
Output
1 100 100
Submitted Solution:
```
nums = list(sorted(map(int, input().split())))
print(f'{nums[0] - nums[3]} {nums[1] - nums[3]} {nums[2] - nums[3]}')
```
No
| 13,389 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Polycarp has guessed three positive integers a, b and c. He keeps these numbers in secret, but he writes down four numbers on a board in arbitrary order β their pairwise sums (three numbers) and sum of all three numbers (one number). So, there are four numbers on a board in random order: a+b, a+c, b+c and a+b+c.
You have to guess three numbers a, b and c using given numbers. Print three guessed integers in any order.
Pay attention that some given numbers a, b and c can be equal (it is also possible that a=b=c).
Input
The only line of the input contains four positive integers x_1, x_2, x_3, x_4 (2 β€ x_i β€ 10^9) β numbers written on a board in random order. It is guaranteed that the answer exists for the given number x_1, x_2, x_3, x_4.
Output
Print such positive integers a, b and c that four numbers written on a board are values a+b, a+c, b+c and a+b+c written in some order. Print a, b and c in any order. If there are several answers, you can print any. It is guaranteed that the answer exists.
Examples
Input
3 6 5 4
Output
2 1 3
Input
40 40 40 60
Output
20 20 20
Input
201 101 101 200
Output
1 100 100
Submitted Solution:
```
a,b,c,d=map(int,input().split())
e=[a,b,c,d]
e.sort(reverse = True)
tam=0
f=[]
for i in range(0,len(e)-1):
tam = e[i] - e[3]
f.append(tam)
for i in range(0,len(f)):
print(f[i],end=' ')
```
No
| 13,390 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have an array a_1, a_2, ..., a_n.
Let's call some subarray a_l, a_{l + 1}, ... , a_r of this array a subpermutation if it contains all integers from 1 to r-l+1 exactly once. For example, array a = [2, 2, 1, 3, 2, 3, 1] contains 6 subarrays which are subpermutations: [a_2 ... a_3], [a_2 ... a_4], [a_3 ... a_3], [a_3 ... a_5], [a_5 ... a_7], [a_7 ... a_7].
You are asked to calculate the number of subpermutations.
Input
The first line contains one integer n (1 β€ n β€ 3 β
10^5).
The second line contains n integers a_1, a_2, ... , a_n (1 β€ a_i β€ n).
This array can contain the same integers.
Output
Print the number of subpermutations of the array a.
Examples
Input
8
2 4 1 3 4 2 1 2
Output
7
Input
5
1 1 2 1 2
Output
6
Note
There are 7 subpermutations in the first test case. Their segments of indices are [1, 4], [3, 3], [3, 6], [4, 7], [6, 7], [7, 7] and [7, 8].
In the second test case 6 subpermutations exist: [1, 1], [2, 2], [2, 3], [3, 4], [4, 4] and [4, 5].
Tags: brute force, data structures, divide and conquer, hashing, math
Correct Solution:
```
import sys
import math
input=sys.stdin.readline
#sys.setrecursionlimit(1000000)
mod=int(1000000007)
i=lambda :map(int,input().split())
n=int(input())
a=[int(x) for x in input().split()]
t=[[0]*21 for i in range(300005)]
for i in range(n):
t[i][0]=a[i]
def build(n):
for j in range(1,20):
for i in range(n):
if i+(1<<j)-1>n-1:
break;
t[i][j]=max(t[i][j-1],t[i+(1<<(j-1))][j-1])
def query(p,q):
p,q=int(p),int(q)
log=int(math.log2(q-p+1))
m=t[p][log]
n=t[q-(1<<log)+1][log]
return max(m,n)
b=[-1]*(n+2)
build(n)
max1=-1
ans=0
for i in range(n):
max1=max(max1,b[a[i]])
b[a[i]]=i
x=b[1]
while x>max1:
if x<=max1:
break
p=query(x,i)
if p==i-x+1:
ans+=1
x=b[p+1]
else:
x=i-p+1
print(ans)
```
| 13,391 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have an array a_1, a_2, ..., a_n.
Let's call some subarray a_l, a_{l + 1}, ... , a_r of this array a subpermutation if it contains all integers from 1 to r-l+1 exactly once. For example, array a = [2, 2, 1, 3, 2, 3, 1] contains 6 subarrays which are subpermutations: [a_2 ... a_3], [a_2 ... a_4], [a_3 ... a_3], [a_3 ... a_5], [a_5 ... a_7], [a_7 ... a_7].
You are asked to calculate the number of subpermutations.
Input
The first line contains one integer n (1 β€ n β€ 3 β
10^5).
The second line contains n integers a_1, a_2, ... , a_n (1 β€ a_i β€ n).
This array can contain the same integers.
Output
Print the number of subpermutations of the array a.
Examples
Input
8
2 4 1 3 4 2 1 2
Output
7
Input
5
1 1 2 1 2
Output
6
Note
There are 7 subpermutations in the first test case. Their segments of indices are [1, 4], [3, 3], [3, 6], [4, 7], [6, 7], [7, 7] and [7, 8].
In the second test case 6 subpermutations exist: [1, 1], [2, 2], [2, 3], [3, 4], [4, 4] and [4, 5].
Tags: brute force, data structures, divide and conquer, hashing, math
Correct Solution:
```
#import sys
import math
#input=sys.stdin.readline
#sys.setrecursionlimit(1000000)
mod=int(1000000007)
i=lambda :map(int,input().split())
n=int(input())
a=[int(x) for x in input().split()]
t=[[0]*21 for i in range(300005)]
for i in range(n):
t[i][0]=a[i]
def build(n):
for j in range(1,20):
for i in range(n):
if i+(1<<j)-1>n-1:
break;
t[i][j]=max(t[i][j-1],t[i+(1<<(j-1))][j-1])
def query(p,q):
p,q=int(p),int(q)
log=int(math.log2(q-p+1))
m=t[p][log]
n=t[q-(1<<log)+1][log]
return max(m,n)
b=[-1]*(n+2)
build(n)
max1=-1
ans=0
for i in range(n):
max1=max(max1,b[a[i]])
b[a[i]]=i
x=b[1]
while x>max1:
if x<=max1:
break
p=query(x,i)
if p==i-x+1:
ans+=1
x=b[p+1]
else:
x=i-p+1
print(ans)
```
| 13,392 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have an array a_1, a_2, ..., a_n.
Let's call some subarray a_l, a_{l + 1}, ... , a_r of this array a subpermutation if it contains all integers from 1 to r-l+1 exactly once. For example, array a = [2, 2, 1, 3, 2, 3, 1] contains 6 subarrays which are subpermutations: [a_2 ... a_3], [a_2 ... a_4], [a_3 ... a_3], [a_3 ... a_5], [a_5 ... a_7], [a_7 ... a_7].
You are asked to calculate the number of subpermutations.
Input
The first line contains one integer n (1 β€ n β€ 3 β
10^5).
The second line contains n integers a_1, a_2, ... , a_n (1 β€ a_i β€ n).
This array can contain the same integers.
Output
Print the number of subpermutations of the array a.
Examples
Input
8
2 4 1 3 4 2 1 2
Output
7
Input
5
1 1 2 1 2
Output
6
Note
There are 7 subpermutations in the first test case. Their segments of indices are [1, 4], [3, 3], [3, 6], [4, 7], [6, 7], [7, 7] and [7, 8].
In the second test case 6 subpermutations exist: [1, 1], [2, 2], [2, 3], [3, 4], [4, 4] and [4, 5].
Submitted Solution:
```
from sys import stdin
def inp():
return stdin.buffer.readline().rstrip().decode('utf8')
def itg():
return int(stdin.buffer.readline())
def mpint():
return map(int, stdin.buffer.readline().split())
# ############################## import
# ############################## main
# for __ in range(itg()):
n = itg()
arr = tuple(mpint())
ans = 0
def dfs(mx, start, end, st):
global ans
ans += 1
s = st.copy()
m = mx
for i in range(start - 1, -1, -1):
a = arr[i]
if a in s:
break
s.add(a)
m = max(m, a)
if a == mx + 1:
dfs(m, i, end, s)
s = st.copy()
m = mx
for i in range(end + 1, n):
a = arr[i]
if a in s:
break
s.add(a)
m = max(m, a)
if a == mx + 1:
dfs(m, start, i, s)
for index, num in enumerate(arr):
if num == 1:
dfs(1, index, index, {1})
print(ans)
# Please check!
```
No
| 13,393 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n segments drawn on a plane; the i-th segment connects two points (x_{i, 1}, y_{i, 1}) and (x_{i, 2}, y_{i, 2}). Each segment is non-degenerate, and is either horizontal or vertical β formally, for every i β [1, n] either x_{i, 1} = x_{i, 2} or y_{i, 1} = y_{i, 2} (but only one of these conditions holds). Only segments of different types may intersect: no pair of horizontal segments shares any common points, and no pair of vertical segments shares any common points.
We say that four segments having indices h_1, h_2, v_1 and v_2 such that h_1 < h_2 and v_1 < v_2 form a rectangle if the following conditions hold:
* segments h_1 and h_2 are horizontal;
* segments v_1 and v_2 are vertical;
* segment h_1 intersects with segment v_1;
* segment h_2 intersects with segment v_1;
* segment h_1 intersects with segment v_2;
* segment h_2 intersects with segment v_2.
Please calculate the number of ways to choose four segments so they form a rectangle. Note that the conditions h_1 < h_2 and v_1 < v_2 should hold.
Input
The first line contains one integer n (1 β€ n β€ 5000) β the number of segments.
Then n lines follow. The i-th line contains four integers x_{i, 1}, y_{i, 1}, x_{i, 2} and y_{i, 2} denoting the endpoints of the i-th segment. All coordinates of the endpoints are in the range [-5000, 5000].
It is guaranteed that each segment is non-degenerate and is either horizontal or vertical. Furthermore, if two segments share a common point, one of these segments is horizontal, and another one is vertical.
Output
Print one integer β the number of ways to choose four segments so they form a rectangle.
Examples
Input
7
-1 4 -1 -2
6 -1 -2 -1
-2 3 6 3
2 -2 2 4
4 -1 4 3
5 3 5 1
5 2 1 2
Output
7
Input
5
1 5 1 0
0 1 5 1
5 4 0 4
4 2 4 0
4 3 4 5
Output
0
Note
The following pictures represent sample cases:
<image> <image>
Submitted Solution:
```
def bisect(arr,val):
l=0
r=len(arr)
while r-l!=1:
m=(l+r)//2
if arr[m][0]>val:
r=m
else:
l=m
return l
n=int(input())
v=[]
h=[]
for _ in range(n):
a,b,c,d=map(int,input().split())
a,b,c,d=a+5000,b+5000,c+5000,d+5000
if a==c:
v.append((a,min(b,d),abs(d-b)))
else:
h.append((b,min(a,c),abs(c-a)))
v=sorted(v,key=lambda x:x[0])
h=sorted(h,key=lambda x:x[0])
lv=len(v)
lh=len(h)
ans=0
for i in range(lv):
for j in range(i+1,lv):
if v[i][1]>(v[j][1]+v[j][2]):
continue
elif (v[i][1]+v[i][2])<v[j][1]:
continue
else:
bottom=max(v[i][1],v[j][1])
top=min((v[i][1]+v[i][2]),(v[j][1]+v[j][2]))
left=v[i][0]
right=v[j][0]
l=bisect(h,bottom)
r=bisect(h,top)
if h[l][0]<bottom:
l+=1
cnt=0
for k in range(l,r+1):
tleft=h[k][1]
tright=h[k][1]+h[k][2]
print(tleft,tright,left,right)
if tleft<=left and right<=tright:
cnt+=1
ans+=(cnt*(cnt-1))//2
print(ans)
```
No
| 13,394 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 100) β the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 β€ n β€ 100) β the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it β the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1
Tags: math
Correct Solution:
```
import math
test=int(input())
for i in range(test):
n=int(input())
l=list(map(int,input().split()))
suml=sum(l)
print(math.ceil(suml/n))
```
| 13,395 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 100) β the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 β€ n β€ 100) β the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it β the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1
Tags: math
Correct Solution:
```
t = int(input())
import math
for _ in range(t):
n = int(input())
A = list(map(int,input().split()))
s =sum(A)
print(math.ceil(s/n))
```
| 13,396 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 100) β the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 β€ n β€ 100) β the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it β the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1
Tags: math
Correct Solution:
```
import math
q=int(input())
for i in range(q):
n=int(input())
l=list(map(int,input().split()))
p=sum(l)
p=math.ceil(p/n)
print(p)
```
| 13,397 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 100) β the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 β€ n β€ 100) β the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it β the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1
Tags: math
Correct Solution:
```
from collections import Counter
if __name__ == '__main__':
q = int(input())
for t in range(q):
n = int(input())
l = [int(i) for i in input().split(" ")]
s = sum(l)
if s%n==0:
print(s//n)
else:
print(s//n+1)
```
| 13,398 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are both a shop keeper and a shop assistant at a small nearby shop. You have n goods, the i-th good costs a_i coins.
You got tired of remembering the price of each product when customers ask for it, thus you decided to simplify your life. More precisely you decided to set the same price for all n goods you have.
However, you don't want to lose any money so you want to choose the price in such a way that the sum of new prices is not less than the sum of the initial prices. It means that if you sell all n goods for the new price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
On the other hand, you don't want to lose customers because of big prices so among all prices you can choose you need to choose the minimum one.
So you need to find the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 β€ q β€ 100) β the number of queries. Then q queries follow.
The first line of the query contains one integer n (1 β€ n β€ 100) β the number of goods. The second line of the query contains n integers a_1, a_2, ..., a_n (1 β€ a_i β€ 10^7), where a_i is the price of the i-th good.
Output
For each query, print the answer for it β the minimum possible equal price of all n goods so if you sell them for this price, you will receive at least the same (or greater) amount of money as if you sell them for their initial prices.
Example
Input
3
5
1 2 3 4 5
3
1 2 2
4
1 1 1 1
Output
3
2
1
Tags: math
Correct Solution:
```
q=int(input())
for i in range(q):
n=int(input())
total=sum([int(x) for x in input().split()])
if total%n>0:
print(int(total/n)+1)
else:
print(int(total/n))
```
| 13,399 |
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