message stringlengths 2 22.7k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 145 109k | cluster float64 9 9 | __index_level_0__ int64 290 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easier version of the problem. In this version, 1 ≤ n ≤ 10^5 and 0 ≤ a_i ≤ 1. You can hack this problem only if you solve and lock both problems.
Christmas is coming, and our protagonist, Bob, is preparing a spectacular present for his long-time best friend Alice. This year, he decides to prepare n boxes of chocolate, numbered from 1 to n. Initially, the i-th box contains a_i chocolate pieces.
Since Bob is a typical nice guy, he will not send Alice n empty boxes. In other words, at least one of a_1, a_2, …, a_n is positive. Since Alice dislikes coprime sets, she will be happy only if there exists some integer k > 1 such that the number of pieces in each box is divisible by k. Note that Alice won't mind if there exists some empty boxes.
Charlie, Alice's boyfriend, also is Bob's second best friend, so he decides to help Bob by rearranging the chocolate pieces. In one second, Charlie can pick up a piece in box i and put it into either box i-1 or box i+1 (if such boxes exist). Of course, he wants to help his friend as quickly as possible. Therefore, he asks you to calculate the minimum number of seconds he would need to make Alice happy.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of chocolate boxes.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1) — the number of chocolate pieces in the i-th box.
It is guaranteed that at least one of a_1, a_2, …, a_n is positive.
Output
If there is no way for Charlie to make Alice happy, print -1.
Otherwise, print a single integer x — the minimum number of seconds for Charlie to help Bob make Alice happy.
Examples
Input
3
1 0 1
Output
2
Input
1
1
Output
-1 | instruction | 0 | 30,180 | 9 | 60,360 |
Tags: constructive algorithms, greedy, math, number theory, ternary search, two pointers
Correct Solution:
```
import math
n = int(input())
l = list(map(int, input().split(" ")))
s = sum(l)
if s == 1:
print(-1)
else:
sff = {s}
if s % 2 == 0:
sff.add(2)
for i in range(2, int(math.sqrt(s)) + 1, 1):
if s % i == 0:
sff.add(i)
sff.add(s // i)
x = []
for i in range(n):
if l[i] == 1:
x.append(i)
anss = 10**18
for sf in sff:
ans = 0
for i in range(0, len(x), sf):
mid = x[i+sf//2]
for j in range(i, i+sf):
ans += abs(mid- x[j])
anss = min(ans, anss)
print(anss)
``` | output | 1 | 30,180 | 9 | 60,361 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easier version of the problem. In this version, 1 ≤ n ≤ 10^5 and 0 ≤ a_i ≤ 1. You can hack this problem only if you solve and lock both problems.
Christmas is coming, and our protagonist, Bob, is preparing a spectacular present for his long-time best friend Alice. This year, he decides to prepare n boxes of chocolate, numbered from 1 to n. Initially, the i-th box contains a_i chocolate pieces.
Since Bob is a typical nice guy, he will not send Alice n empty boxes. In other words, at least one of a_1, a_2, …, a_n is positive. Since Alice dislikes coprime sets, she will be happy only if there exists some integer k > 1 such that the number of pieces in each box is divisible by k. Note that Alice won't mind if there exists some empty boxes.
Charlie, Alice's boyfriend, also is Bob's second best friend, so he decides to help Bob by rearranging the chocolate pieces. In one second, Charlie can pick up a piece in box i and put it into either box i-1 or box i+1 (if such boxes exist). Of course, he wants to help his friend as quickly as possible. Therefore, he asks you to calculate the minimum number of seconds he would need to make Alice happy.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of chocolate boxes.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1) — the number of chocolate pieces in the i-th box.
It is guaranteed that at least one of a_1, a_2, …, a_n is positive.
Output
If there is no way for Charlie to make Alice happy, print -1.
Otherwise, print a single integer x — the minimum number of seconds for Charlie to help Bob make Alice happy.
Examples
Input
3
1 0 1
Output
2
Input
1
1
Output
-1 | instruction | 0 | 30,181 | 9 | 60,362 |
Tags: constructive algorithms, greedy, math, number theory, ternary search, two pointers
Correct Solution:
```
def simple_div(x):
if not x & 1:
yield 2
while not x & 1:
x >>= 1
i = 3
while i * i <= x:
if x % i == 0:
yield i
while x % i == 0:
x //= i
i += 2
if x != 1:
yield x
def __main__():
n = int(input())
a = list(map(int, input().split()))
sa = sum(a)
a.pop()
if sa == 1:
print(-1)
return
res = 2**64
for d in simple_div(sa):
tmp = 0
m = 0
for x in a:
m = (x + m) % d
tmp += m if m * 2 <= d else d - m
if tmp < res:
res = tmp
print(res)
__main__()
``` | output | 1 | 30,181 | 9 | 60,363 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easier version of the problem. In this version, 1 ≤ n ≤ 10^5 and 0 ≤ a_i ≤ 1. You can hack this problem only if you solve and lock both problems.
Christmas is coming, and our protagonist, Bob, is preparing a spectacular present for his long-time best friend Alice. This year, he decides to prepare n boxes of chocolate, numbered from 1 to n. Initially, the i-th box contains a_i chocolate pieces.
Since Bob is a typical nice guy, he will not send Alice n empty boxes. In other words, at least one of a_1, a_2, …, a_n is positive. Since Alice dislikes coprime sets, she will be happy only if there exists some integer k > 1 such that the number of pieces in each box is divisible by k. Note that Alice won't mind if there exists some empty boxes.
Charlie, Alice's boyfriend, also is Bob's second best friend, so he decides to help Bob by rearranging the chocolate pieces. In one second, Charlie can pick up a piece in box i and put it into either box i-1 or box i+1 (if such boxes exist). Of course, he wants to help his friend as quickly as possible. Therefore, he asks you to calculate the minimum number of seconds he would need to make Alice happy.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of chocolate boxes.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1) — the number of chocolate pieces in the i-th box.
It is guaranteed that at least one of a_1, a_2, …, a_n is positive.
Output
If there is no way for Charlie to make Alice happy, print -1.
Otherwise, print a single integer x — the minimum number of seconds for Charlie to help Bob make Alice happy.
Examples
Input
3
1 0 1
Output
2
Input
1
1
Output
-1 | instruction | 0 | 30,182 | 9 | 60,364 |
Tags: constructive algorithms, greedy, math, number theory, ternary search, two pointers
Correct Solution:
```
from math import ceil
n=int(input())
x=list(map(int,input().split()))
count,mod,mi=0,0,1234567890645725645376267536754173
v,y=[],[0]*(n+1)
ans=123456712220335762365463526564561564534565546326
for i in range(n): count+=x[i]
t=count
for i in range(2,t+1):
if t%i==0:
v.append(i)
while(t%i==0):t//=i
for i2 in v:
c=0
for i in range(n):
y[i]=x[i]
for i in range(n):
if y[i]<=0:
y[i+1]-=abs(y[i])
c+=abs(y[i])
elif y[i]%i2<i2-y[i]%i2:
y[i+1]+=y[i]%i2
c+=y[i]%i2
else:
y[i+1]-=i2-y[i]%i2
c+=i2-y[i]%i2
ans=min(ans,c)
if count==1: print(-1)
else: print(ans)
``` | output | 1 | 30,182 | 9 | 60,365 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easier version of the problem. In this version, 1 ≤ n ≤ 10^5 and 0 ≤ a_i ≤ 1. You can hack this problem only if you solve and lock both problems.
Christmas is coming, and our protagonist, Bob, is preparing a spectacular present for his long-time best friend Alice. This year, he decides to prepare n boxes of chocolate, numbered from 1 to n. Initially, the i-th box contains a_i chocolate pieces.
Since Bob is a typical nice guy, he will not send Alice n empty boxes. In other words, at least one of a_1, a_2, …, a_n is positive. Since Alice dislikes coprime sets, she will be happy only if there exists some integer k > 1 such that the number of pieces in each box is divisible by k. Note that Alice won't mind if there exists some empty boxes.
Charlie, Alice's boyfriend, also is Bob's second best friend, so he decides to help Bob by rearranging the chocolate pieces. In one second, Charlie can pick up a piece in box i and put it into either box i-1 or box i+1 (if such boxes exist). Of course, he wants to help his friend as quickly as possible. Therefore, he asks you to calculate the minimum number of seconds he would need to make Alice happy.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of chocolate boxes.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1) — the number of chocolate pieces in the i-th box.
It is guaranteed that at least one of a_1, a_2, …, a_n is positive.
Output
If there is no way for Charlie to make Alice happy, print -1.
Otherwise, print a single integer x — the minimum number of seconds for Charlie to help Bob make Alice happy.
Examples
Input
3
1 0 1
Output
2
Input
1
1
Output
-1 | instruction | 0 | 30,183 | 9 | 60,366 |
Tags: constructive algorithms, greedy, math, number theory, ternary search, two pointers
Correct Solution:
```
mod = 1000000007
eps = 10**-9
def main():
import sys
input = sys.stdin.readline
N = int(input())
A = list(map(int, input().split()))
S = sum(A)
if S == 1:
print(-1)
exit()
div_list = [S]
for d in range(2, int(S ** 0.5) + 1):
if S % d == 0:
div_list.append(d)
div_list.append(S // d)
if len(div_list) > 2:
if div_list[-1] == div_list[-2]:
div_list.pop()
ans_best = 10**10
for D in div_list:
ans = 0
cnt = 0
i_list = []
for i, a in enumerate(A):
if a == 1:
i_list.append(i)
cnt += 1
if cnt == D:
cnt = 0
j = i_list[D // 2]
for ii in i_list:
ans += abs(ii - j)
i_list = []
ans_best = min(ans_best, ans)
print(ans_best)
if __name__ == '__main__':
main()
``` | output | 1 | 30,183 | 9 | 60,367 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easier version of the problem. In this version, 1 ≤ n ≤ 10^5 and 0 ≤ a_i ≤ 1. You can hack this problem only if you solve and lock both problems.
Christmas is coming, and our protagonist, Bob, is preparing a spectacular present for his long-time best friend Alice. This year, he decides to prepare n boxes of chocolate, numbered from 1 to n. Initially, the i-th box contains a_i chocolate pieces.
Since Bob is a typical nice guy, he will not send Alice n empty boxes. In other words, at least one of a_1, a_2, …, a_n is positive. Since Alice dislikes coprime sets, she will be happy only if there exists some integer k > 1 such that the number of pieces in each box is divisible by k. Note that Alice won't mind if there exists some empty boxes.
Charlie, Alice's boyfriend, also is Bob's second best friend, so he decides to help Bob by rearranging the chocolate pieces. In one second, Charlie can pick up a piece in box i and put it into either box i-1 or box i+1 (if such boxes exist). Of course, he wants to help his friend as quickly as possible. Therefore, he asks you to calculate the minimum number of seconds he would need to make Alice happy.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of chocolate boxes.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1) — the number of chocolate pieces in the i-th box.
It is guaranteed that at least one of a_1, a_2, …, a_n is positive.
Output
If there is no way for Charlie to make Alice happy, print -1.
Otherwise, print a single integer x — the minimum number of seconds for Charlie to help Bob make Alice happy.
Examples
Input
3
1 0 1
Output
2
Input
1
1
Output
-1 | instruction | 0 | 30,184 | 9 | 60,368 |
Tags: constructive algorithms, greedy, math, number theory, ternary search, two pointers
Correct Solution:
```
from collections import deque
import sys
def input():
return sys.stdin.readline().rstrip()
def find_minimum_divisors(M):
res = set([])
for i in range(1, M + 1):
if i * i > M:
break
if M % i == 0:
res.add(i)
res.add(M//i)
res.remove(1)
return list(res)
def slv():
n = int(input())
a = list(map(int, input().split()))
b = deque()
for i in range(n):
if a[i] == 1:
b.append(i)
if len(b) == 1:
print(-1)
return
final_ans = 1 << 128
for div in find_minimum_divisors(len(b)):
B = b.copy()
ans = 0
for _ in range(len(b)//div):
array = []
for _ in range(div):
array.append(B.popleft())
cost = sum(array) - array[0] * div
tmpmincost = cost
for i in range(div - 1):
d = array[i + 1] - array[i]
cost += (2 * (i + 1) - div) * d
tmpmincost = min(tmpmincost, cost)
assert tmpmincost >= 0
ans += tmpmincost
final_ans = min(final_ans, ans)
print(final_ans)
return
def main():
t = 1
for i in range(t):
slv()
return
if __name__ == "__main__":
main()
``` | output | 1 | 30,184 | 9 | 60,369 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easier version of the problem. In this version, 1 ≤ n ≤ 10^5 and 0 ≤ a_i ≤ 1. You can hack this problem only if you solve and lock both problems.
Christmas is coming, and our protagonist, Bob, is preparing a spectacular present for his long-time best friend Alice. This year, he decides to prepare n boxes of chocolate, numbered from 1 to n. Initially, the i-th box contains a_i chocolate pieces.
Since Bob is a typical nice guy, he will not send Alice n empty boxes. In other words, at least one of a_1, a_2, …, a_n is positive. Since Alice dislikes coprime sets, she will be happy only if there exists some integer k > 1 such that the number of pieces in each box is divisible by k. Note that Alice won't mind if there exists some empty boxes.
Charlie, Alice's boyfriend, also is Bob's second best friend, so he decides to help Bob by rearranging the chocolate pieces. In one second, Charlie can pick up a piece in box i and put it into either box i-1 or box i+1 (if such boxes exist). Of course, he wants to help his friend as quickly as possible. Therefore, he asks you to calculate the minimum number of seconds he would need to make Alice happy.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of chocolate boxes.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1) — the number of chocolate pieces in the i-th box.
It is guaranteed that at least one of a_1, a_2, …, a_n is positive.
Output
If there is no way for Charlie to make Alice happy, print -1.
Otherwise, print a single integer x — the minimum number of seconds for Charlie to help Bob make Alice happy.
Examples
Input
3
1 0 1
Output
2
Input
1
1
Output
-1
Submitted Solution:
```
def get_prime_set(ub):
from itertools import chain
from math import sqrt
if ub < 4:
return ({}, {}, {2}, {2, 3})[ub]
ub, ub_sqrt = ub+1, int(sqrt(ub))+1
primes = {2, 3} | set(chain(range(5, ub, 6), range(7, ub, 6)))
du = primes.difference_update
for n in chain(range(5, ub_sqrt, 6), range(7, ub_sqrt, 6)):
if n in primes:
du(range(n*3, ub, n*2))
return primes
n = int(input())
index_a = []
for i, x in enumerate(map(int, input().split())):
if x:
index_a.append(i)
count = len(index_a)
if count == 1:
print(-1)
exit()
primes = get_prime_set(count)
ans = 10**18
for bucket_size in primes:
if count % bucket_size:
continue
st = 0
for i in range(count//bucket_size):
bucket = index_a[i*bucket_size:(i+1)*bucket_size]
mid = bucket[bucket_size//2]
st += sum(abs(mid-x) for x in bucket)
ans = min(ans, st)
print(ans)
``` | instruction | 0 | 30,185 | 9 | 60,370 |
Yes | output | 1 | 30,185 | 9 | 60,371 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easier version of the problem. In this version, 1 ≤ n ≤ 10^5 and 0 ≤ a_i ≤ 1. You can hack this problem only if you solve and lock both problems.
Christmas is coming, and our protagonist, Bob, is preparing a spectacular present for his long-time best friend Alice. This year, he decides to prepare n boxes of chocolate, numbered from 1 to n. Initially, the i-th box contains a_i chocolate pieces.
Since Bob is a typical nice guy, he will not send Alice n empty boxes. In other words, at least one of a_1, a_2, …, a_n is positive. Since Alice dislikes coprime sets, she will be happy only if there exists some integer k > 1 such that the number of pieces in each box is divisible by k. Note that Alice won't mind if there exists some empty boxes.
Charlie, Alice's boyfriend, also is Bob's second best friend, so he decides to help Bob by rearranging the chocolate pieces. In one second, Charlie can pick up a piece in box i and put it into either box i-1 or box i+1 (if such boxes exist). Of course, he wants to help his friend as quickly as possible. Therefore, he asks you to calculate the minimum number of seconds he would need to make Alice happy.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of chocolate boxes.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1) — the number of chocolate pieces in the i-th box.
It is guaranteed that at least one of a_1, a_2, …, a_n is positive.
Output
If there is no way for Charlie to make Alice happy, print -1.
Otherwise, print a single integer x — the minimum number of seconds for Charlie to help Bob make Alice happy.
Examples
Input
3
1 0 1
Output
2
Input
1
1
Output
-1
Submitted Solution:
```
def count(x):
ans = 0
for i in range(0, m, x):
st = (2 * i + x - 1) // 2
for j in range(i, x + i):
ans += abs(a[j] - a[st])
return ans
n = int(input())
data = list(map(int, input().split()))
a = []
m = 0
for i in range(n):
if data[i] == 1:
a.append(i)
m += 1
k = []
for i in range(2, m + 1):
if m % i == 0:
k.append(i)
l, r = 0, len(k) - 1
while r - l > 7:
m1 = l + (r - l) // 3
m2 = r - (r - l) // 3
if count(k[m1]) > count(k[m2]):
l = m1
else:
r = m2
t = 10 ** 18
for i in range(l, r + 1):
t = min(t, count(k[i]))
if t == 10 ** 18:
print(-1)
else:
print(t)
``` | instruction | 0 | 30,186 | 9 | 60,372 |
Yes | output | 1 | 30,186 | 9 | 60,373 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easier version of the problem. In this version, 1 ≤ n ≤ 10^5 and 0 ≤ a_i ≤ 1. You can hack this problem only if you solve and lock both problems.
Christmas is coming, and our protagonist, Bob, is preparing a spectacular present for his long-time best friend Alice. This year, he decides to prepare n boxes of chocolate, numbered from 1 to n. Initially, the i-th box contains a_i chocolate pieces.
Since Bob is a typical nice guy, he will not send Alice n empty boxes. In other words, at least one of a_1, a_2, …, a_n is positive. Since Alice dislikes coprime sets, she will be happy only if there exists some integer k > 1 such that the number of pieces in each box is divisible by k. Note that Alice won't mind if there exists some empty boxes.
Charlie, Alice's boyfriend, also is Bob's second best friend, so he decides to help Bob by rearranging the chocolate pieces. In one second, Charlie can pick up a piece in box i and put it into either box i-1 or box i+1 (if such boxes exist). Of course, he wants to help his friend as quickly as possible. Therefore, he asks you to calculate the minimum number of seconds he would need to make Alice happy.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of chocolate boxes.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1) — the number of chocolate pieces in the i-th box.
It is guaranteed that at least one of a_1, a_2, …, a_n is positive.
Output
If there is no way for Charlie to make Alice happy, print -1.
Otherwise, print a single integer x — the minimum number of seconds for Charlie to help Bob make Alice happy.
Examples
Input
3
1 0 1
Output
2
Input
1
1
Output
-1
Submitted Solution:
```
def get(x):
cnt=x.count(1)
cnt//=2;cnt+=1
mid=0
tmp=0
for i in range(len(x)):
tmp+=x[i]
if(tmp==cnt):
mid=i
break
ret=0
for i in range(len(x)):
if(x[i]):
ret+=abs(i-mid)
return ret
n=int(input())
a=list(map(int,input().split()))
b=a
sm=sum(a)
if(sm==1):
print(-1)
exit()
if(sm==0):
print(0)
exit()
l=[]
i=1
while(i*i<=sm):
if(sm%i==0):
l.append(i)
l.append(sm//i)
i+=1
l.remove(1)
ans=10**20
for k in l:
c=0
a=b[:]
i=0
while(i<n):
j=i
sa=0
while(sa<k and j<n):
sa+=a[j]
j+=1
now=a[i:j]
i=j
c+=get(now)
ans=min(ans,c)
print(ans)
``` | instruction | 0 | 30,187 | 9 | 60,374 |
Yes | output | 1 | 30,187 | 9 | 60,375 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easier version of the problem. In this version, 1 ≤ n ≤ 10^5 and 0 ≤ a_i ≤ 1. You can hack this problem only if you solve and lock both problems.
Christmas is coming, and our protagonist, Bob, is preparing a spectacular present for his long-time best friend Alice. This year, he decides to prepare n boxes of chocolate, numbered from 1 to n. Initially, the i-th box contains a_i chocolate pieces.
Since Bob is a typical nice guy, he will not send Alice n empty boxes. In other words, at least one of a_1, a_2, …, a_n is positive. Since Alice dislikes coprime sets, she will be happy only if there exists some integer k > 1 such that the number of pieces in each box is divisible by k. Note that Alice won't mind if there exists some empty boxes.
Charlie, Alice's boyfriend, also is Bob's second best friend, so he decides to help Bob by rearranging the chocolate pieces. In one second, Charlie can pick up a piece in box i and put it into either box i-1 or box i+1 (if such boxes exist). Of course, he wants to help his friend as quickly as possible. Therefore, he asks you to calculate the minimum number of seconds he would need to make Alice happy.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of chocolate boxes.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1) — the number of chocolate pieces in the i-th box.
It is guaranteed that at least one of a_1, a_2, …, a_n is positive.
Output
If there is no way for Charlie to make Alice happy, print -1.
Otherwise, print a single integer x — the minimum number of seconds for Charlie to help Bob make Alice happy.
Examples
Input
3
1 0 1
Output
2
Input
1
1
Output
-1
Submitted Solution:
```
import math
arr = [0] * 1000100
n = 0
# Calculate amount of move for prime number x
def gogo(x):
global n
global arr
cv = 0
ans = 0
for i in range(n):
cv = (cv + arr[i]) % x
ans += min(cv, x - cv)
# print(x, i, arr[i], cv, ans)
return ans
def main():
global n
global arr
n = int(input())
csum = 0
arr = list(map(int, input().split()))
csum = sum(arr)
# v stores the prime
v = []
for i in range(2, math.ceil(math.sqrt(csum))):
if csum % i == 0:
v.append(i)
while csum % i == 0:
csum //= i
if csum > 1:
v.append(csum)
# print(v)
ans = 3e18
for f in v:
# Find minimum amount of moves
ans = min(ans, gogo(f))
# print(f, ans)
if ans > 2e18:
print(-1)
else:
print(ans)
main()
``` | instruction | 0 | 30,188 | 9 | 60,376 |
Yes | output | 1 | 30,188 | 9 | 60,377 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easier version of the problem. In this version, 1 ≤ n ≤ 10^5 and 0 ≤ a_i ≤ 1. You can hack this problem only if you solve and lock both problems.
Christmas is coming, and our protagonist, Bob, is preparing a spectacular present for his long-time best friend Alice. This year, he decides to prepare n boxes of chocolate, numbered from 1 to n. Initially, the i-th box contains a_i chocolate pieces.
Since Bob is a typical nice guy, he will not send Alice n empty boxes. In other words, at least one of a_1, a_2, …, a_n is positive. Since Alice dislikes coprime sets, she will be happy only if there exists some integer k > 1 such that the number of pieces in each box is divisible by k. Note that Alice won't mind if there exists some empty boxes.
Charlie, Alice's boyfriend, also is Bob's second best friend, so he decides to help Bob by rearranging the chocolate pieces. In one second, Charlie can pick up a piece in box i and put it into either box i-1 or box i+1 (if such boxes exist). Of course, he wants to help his friend as quickly as possible. Therefore, he asks you to calculate the minimum number of seconds he would need to make Alice happy.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of chocolate boxes.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1) — the number of chocolate pieces in the i-th box.
It is guaranteed that at least one of a_1, a_2, …, a_n is positive.
Output
If there is no way for Charlie to make Alice happy, print -1.
Otherwise, print a single integer x — the minimum number of seconds for Charlie to help Bob make Alice happy.
Examples
Input
3
1 0 1
Output
2
Input
1
1
Output
-1
Submitted Solution:
```
#---------------------------iye ha aam zindegi---------------------------------------------
import math
import heapq,bisect
import sys
from collections import deque,defaultdict
from fractions import Fraction
mod=10**9+7
mod1=998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class SegmentTree1:
def __init__(self, data, default=9999999, func=lambda a, b: min(a , b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: a + b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
#-------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
#--------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
#--------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
#--------------------------------------------------product----------------------------------------
def product(l):
por=1
for i in range(len(l)):
por*=l[i]
return por
#--------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] <= key):
# At least (mid + 1) elements are there
# whose values are less than
# or equal to key
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
#--------------------------------------------------binary----------------------------------------
def countdig(n):
c=0
while(n>0):
n//=10
c+=1
return c
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
# If mid element is greater than
# k update leftGreater and r
if (arr[m] > k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
#--------------------------------------------------binary------------------------------------
n=int(input())
l=list(map(int,input().split()))
s=sum(l)
if s==1:
print(-1)
sys.exit()
if s==0:
print(0)
sys.exit()
tot=s
for i in range(2,int(s**0.5)+1):
if s%i==0:
tot=i
break
cur=0
e=0
f=0
ans=0
#print(tot,s)
for i in range(n):
cur+=l[i]
#print(cur)
if f==1:
e+=1
if cur==tot:
cur=0
ans+=e
#print(ans)
f=0
e=0
continue
if f==0 and l[i]==1:
f=1
if ans==99999:
print(s)
print(ans)
``` | instruction | 0 | 30,189 | 9 | 60,378 |
No | output | 1 | 30,189 | 9 | 60,379 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easier version of the problem. In this version, 1 ≤ n ≤ 10^5 and 0 ≤ a_i ≤ 1. You can hack this problem only if you solve and lock both problems.
Christmas is coming, and our protagonist, Bob, is preparing a spectacular present for his long-time best friend Alice. This year, he decides to prepare n boxes of chocolate, numbered from 1 to n. Initially, the i-th box contains a_i chocolate pieces.
Since Bob is a typical nice guy, he will not send Alice n empty boxes. In other words, at least one of a_1, a_2, …, a_n is positive. Since Alice dislikes coprime sets, she will be happy only if there exists some integer k > 1 such that the number of pieces in each box is divisible by k. Note that Alice won't mind if there exists some empty boxes.
Charlie, Alice's boyfriend, also is Bob's second best friend, so he decides to help Bob by rearranging the chocolate pieces. In one second, Charlie can pick up a piece in box i and put it into either box i-1 or box i+1 (if such boxes exist). Of course, he wants to help his friend as quickly as possible. Therefore, he asks you to calculate the minimum number of seconds he would need to make Alice happy.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of chocolate boxes.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1) — the number of chocolate pieces in the i-th box.
It is guaranteed that at least one of a_1, a_2, …, a_n is positive.
Output
If there is no way for Charlie to make Alice happy, print -1.
Otherwise, print a single integer x — the minimum number of seconds for Charlie to help Bob make Alice happy.
Examples
Input
3
1 0 1
Output
2
Input
1
1
Output
-1
Submitted Solution:
```
import sys
a = int(input())
b = list(map(int, input().split()))
s = sum(b)
if s == 1:
print(-1)
sys.exit(0)
k = 1
t = 1
while t <= s and k == 1:
if s % t == 0:
k = t
t += 1
if k == 1:
k = s
l = 0
m = 0
mas = [[] for i in range(s // k)]
for i in range(a):
if b[i] == 1 and l != k:
l += 1
mas[m].append(i)
elif b[i] == 1 and l == k:
l = 1
m += 1
mas[m].append(i)
else:
pass
h = 0
for i in range(s // k):
for g in range(k):
if g != k // 2:
h += abs(mas[i][g] - mas[i][k // 2])
print(h)
``` | instruction | 0 | 30,190 | 9 | 60,380 |
No | output | 1 | 30,190 | 9 | 60,381 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easier version of the problem. In this version, 1 ≤ n ≤ 10^5 and 0 ≤ a_i ≤ 1. You can hack this problem only if you solve and lock both problems.
Christmas is coming, and our protagonist, Bob, is preparing a spectacular present for his long-time best friend Alice. This year, he decides to prepare n boxes of chocolate, numbered from 1 to n. Initially, the i-th box contains a_i chocolate pieces.
Since Bob is a typical nice guy, he will not send Alice n empty boxes. In other words, at least one of a_1, a_2, …, a_n is positive. Since Alice dislikes coprime sets, she will be happy only if there exists some integer k > 1 such that the number of pieces in each box is divisible by k. Note that Alice won't mind if there exists some empty boxes.
Charlie, Alice's boyfriend, also is Bob's second best friend, so he decides to help Bob by rearranging the chocolate pieces. In one second, Charlie can pick up a piece in box i and put it into either box i-1 or box i+1 (if such boxes exist). Of course, he wants to help his friend as quickly as possible. Therefore, he asks you to calculate the minimum number of seconds he would need to make Alice happy.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of chocolate boxes.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1) — the number of chocolate pieces in the i-th box.
It is guaranteed that at least one of a_1, a_2, …, a_n is positive.
Output
If there is no way for Charlie to make Alice happy, print -1.
Otherwise, print a single integer x — the minimum number of seconds for Charlie to help Bob make Alice happy.
Examples
Input
3
1 0 1
Output
2
Input
1
1
Output
-1
Submitted Solution:
```
#---------------------------iye ha aam zindegi---------------------------------------------
import math
import heapq,bisect
import sys
from collections import deque,defaultdict
from fractions import Fraction
mod=10**9+7
mod1=998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class SegmentTree1:
def __init__(self, data, default=9999999, func=lambda a, b: min(a , b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: a + b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
#-------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
#--------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
#--------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
#--------------------------------------------------product----------------------------------------
def product(l):
por=1
for i in range(len(l)):
por*=l[i]
return por
#--------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] <= key):
# At least (mid + 1) elements are there
# whose values are less than
# or equal to key
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
#--------------------------------------------------binary----------------------------------------
def countdig(n):
c=0
while(n>0):
n//=10
c+=1
return c
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
# If mid element is greater than
# k update leftGreater and r
if (arr[m] > k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
#--------------------------------------------------binary------------------------------------
n=int(input())
l=list(map(int,input().split()))
s=sum(l)
if s==1:
print(-1)
sys.exit()
if s==0:
print(0)
sys.exit()
tot=s
for i in range(2,int(s**0.5)+1):
if s%i==0:
tot=i
cur=0
e=0
f=0
ans=0
if n==100000:
print(tot)
for i in range(n):
cur+=l[i]
#print(cur)
if f==1:
e+=1
if cur==tot:
cur=0
ans+=e
#print(ans)
f=0
e=0
continue
if f==0 and l[i]==1:
f=1
print(ans)
``` | instruction | 0 | 30,191 | 9 | 60,382 |
No | output | 1 | 30,191 | 9 | 60,383 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easier version of the problem. In this version, 1 ≤ n ≤ 10^5 and 0 ≤ a_i ≤ 1. You can hack this problem only if you solve and lock both problems.
Christmas is coming, and our protagonist, Bob, is preparing a spectacular present for his long-time best friend Alice. This year, he decides to prepare n boxes of chocolate, numbered from 1 to n. Initially, the i-th box contains a_i chocolate pieces.
Since Bob is a typical nice guy, he will not send Alice n empty boxes. In other words, at least one of a_1, a_2, …, a_n is positive. Since Alice dislikes coprime sets, she will be happy only if there exists some integer k > 1 such that the number of pieces in each box is divisible by k. Note that Alice won't mind if there exists some empty boxes.
Charlie, Alice's boyfriend, also is Bob's second best friend, so he decides to help Bob by rearranging the chocolate pieces. In one second, Charlie can pick up a piece in box i and put it into either box i-1 or box i+1 (if such boxes exist). Of course, he wants to help his friend as quickly as possible. Therefore, he asks you to calculate the minimum number of seconds he would need to make Alice happy.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of chocolate boxes.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1) — the number of chocolate pieces in the i-th box.
It is guaranteed that at least one of a_1, a_2, …, a_n is positive.
Output
If there is no way for Charlie to make Alice happy, print -1.
Otherwise, print a single integer x — the minimum number of seconds for Charlie to help Bob make Alice happy.
Examples
Input
3
1 0 1
Output
2
Input
1
1
Output
-1
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
if(a.count(1)&1):
print(-1)
exit()
c=0
for i in range(n):
if(a[i]&1):
a[i+1]+=1
c+=1
print(c)
``` | instruction | 0 | 30,192 | 9 | 60,384 |
No | output | 1 | 30,192 | 9 | 60,385 |
Provide a correct Python 3 solution for this coding contest problem.
I’m planning to have a party on my birthday. Many of my friends will come to the party. Some of them will come with one or more pieces of cakes, but it is not certain if the number of the cakes is a multiple of the number of people coming.
I wish to enjoy the cakes equally among the partiers. So, I decided to apply the following rules. First, all the party attendants are given the same number of cakes. If some remainder occurs, a piece goes on a priority basis to the party host (that’s me!). How many pieces of cake can I enjoy?
Given the number of my friends and cake information, make a program to calculate how many pieces of cake I can enjoy. Note that I am not counted in the number of my friends.
Input
The input is given in the following format.
$N$ $C$
$p_1$ $p_2$ ... $p_C$
The first line provides the number of my friends $N$ ($1 \leq N \leq 100$) and the number of those among them who brought one or more pieces of cake with them $C$ ($1 \leq C \leq N$). The second line provides an array of integers $p_i$ ($1 \leq p_i \leq100$), each of which shows the number of cakes of the $i$-th friend of mine who was willing to come up with one or more pieces of cake.
Output
Output the number of cakes I can enjoy.
Examples
Input
5 4
5 5 6 5
Output
4
Input
7 5
8 8 8 8 8
Output
5
Input
100 3
3 3 3
Output
1 | instruction | 0 | 30,875 | 9 | 61,750 |
"Correct Solution:
```
# coding: utf-8
import math
N, C = list(map(int, input().split(' ')))
cakes = list(map(int, input().split(' ')))
num_cakes = sum(cakes)
mycake = math.ceil(num_cakes/(N+1))
print(mycake)
``` | output | 1 | 30,875 | 9 | 61,751 |
Provide a correct Python 3 solution for this coding contest problem.
I’m planning to have a party on my birthday. Many of my friends will come to the party. Some of them will come with one or more pieces of cakes, but it is not certain if the number of the cakes is a multiple of the number of people coming.
I wish to enjoy the cakes equally among the partiers. So, I decided to apply the following rules. First, all the party attendants are given the same number of cakes. If some remainder occurs, a piece goes on a priority basis to the party host (that’s me!). How many pieces of cake can I enjoy?
Given the number of my friends and cake information, make a program to calculate how many pieces of cake I can enjoy. Note that I am not counted in the number of my friends.
Input
The input is given in the following format.
$N$ $C$
$p_1$ $p_2$ ... $p_C$
The first line provides the number of my friends $N$ ($1 \leq N \leq 100$) and the number of those among them who brought one or more pieces of cake with them $C$ ($1 \leq C \leq N$). The second line provides an array of integers $p_i$ ($1 \leq p_i \leq100$), each of which shows the number of cakes of the $i$-th friend of mine who was willing to come up with one or more pieces of cake.
Output
Output the number of cakes I can enjoy.
Examples
Input
5 4
5 5 6 5
Output
4
Input
7 5
8 8 8 8 8
Output
5
Input
100 3
3 3 3
Output
1 | instruction | 0 | 30,876 | 9 | 61,752 |
"Correct Solution:
```
n,c = map(int, input().split())
pi = list(map(int, input().split()))
n += 1
total_p = 0
for p in pi: total_p += p
p = int(total_p/n)
if total_p%n != 0 : p += 1
print(p)
``` | output | 1 | 30,876 | 9 | 61,753 |
Provide a correct Python 3 solution for this coding contest problem.
I’m planning to have a party on my birthday. Many of my friends will come to the party. Some of them will come with one or more pieces of cakes, but it is not certain if the number of the cakes is a multiple of the number of people coming.
I wish to enjoy the cakes equally among the partiers. So, I decided to apply the following rules. First, all the party attendants are given the same number of cakes. If some remainder occurs, a piece goes on a priority basis to the party host (that’s me!). How many pieces of cake can I enjoy?
Given the number of my friends and cake information, make a program to calculate how many pieces of cake I can enjoy. Note that I am not counted in the number of my friends.
Input
The input is given in the following format.
$N$ $C$
$p_1$ $p_2$ ... $p_C$
The first line provides the number of my friends $N$ ($1 \leq N \leq 100$) and the number of those among them who brought one or more pieces of cake with them $C$ ($1 \leq C \leq N$). The second line provides an array of integers $p_i$ ($1 \leq p_i \leq100$), each of which shows the number of cakes of the $i$-th friend of mine who was willing to come up with one or more pieces of cake.
Output
Output the number of cakes I can enjoy.
Examples
Input
5 4
5 5 6 5
Output
4
Input
7 5
8 8 8 8 8
Output
5
Input
100 3
3 3 3
Output
1 | instruction | 0 | 30,877 | 9 | 61,754 |
"Correct Solution:
```
n,c = map(int, input().split())
p = list(map(int, input().split()))
v = 0
for i in p:
v += i
ans = v//(n+1)
if v % (n+1) > 0: ans += 1
print(ans)
``` | output | 1 | 30,877 | 9 | 61,755 |
Provide a correct Python 3 solution for this coding contest problem.
I’m planning to have a party on my birthday. Many of my friends will come to the party. Some of them will come with one or more pieces of cakes, but it is not certain if the number of the cakes is a multiple of the number of people coming.
I wish to enjoy the cakes equally among the partiers. So, I decided to apply the following rules. First, all the party attendants are given the same number of cakes. If some remainder occurs, a piece goes on a priority basis to the party host (that’s me!). How many pieces of cake can I enjoy?
Given the number of my friends and cake information, make a program to calculate how many pieces of cake I can enjoy. Note that I am not counted in the number of my friends.
Input
The input is given in the following format.
$N$ $C$
$p_1$ $p_2$ ... $p_C$
The first line provides the number of my friends $N$ ($1 \leq N \leq 100$) and the number of those among them who brought one or more pieces of cake with them $C$ ($1 \leq C \leq N$). The second line provides an array of integers $p_i$ ($1 \leq p_i \leq100$), each of which shows the number of cakes of the $i$-th friend of mine who was willing to come up with one or more pieces of cake.
Output
Output the number of cakes I can enjoy.
Examples
Input
5 4
5 5 6 5
Output
4
Input
7 5
8 8 8 8 8
Output
5
Input
100 3
3 3 3
Output
1 | instruction | 0 | 30,878 | 9 | 61,756 |
"Correct Solution:
```
n,c=map(int,input().split())
n+=1
s=sum(map(int,input().split()))
print(s//n+bool(s%n))
``` | output | 1 | 30,878 | 9 | 61,757 |
Provide a correct Python 3 solution for this coding contest problem.
I’m planning to have a party on my birthday. Many of my friends will come to the party. Some of them will come with one or more pieces of cakes, but it is not certain if the number of the cakes is a multiple of the number of people coming.
I wish to enjoy the cakes equally among the partiers. So, I decided to apply the following rules. First, all the party attendants are given the same number of cakes. If some remainder occurs, a piece goes on a priority basis to the party host (that’s me!). How many pieces of cake can I enjoy?
Given the number of my friends and cake information, make a program to calculate how many pieces of cake I can enjoy. Note that I am not counted in the number of my friends.
Input
The input is given in the following format.
$N$ $C$
$p_1$ $p_2$ ... $p_C$
The first line provides the number of my friends $N$ ($1 \leq N \leq 100$) and the number of those among them who brought one or more pieces of cake with them $C$ ($1 \leq C \leq N$). The second line provides an array of integers $p_i$ ($1 \leq p_i \leq100$), each of which shows the number of cakes of the $i$-th friend of mine who was willing to come up with one or more pieces of cake.
Output
Output the number of cakes I can enjoy.
Examples
Input
5 4
5 5 6 5
Output
4
Input
7 5
8 8 8 8 8
Output
5
Input
100 3
3 3 3
Output
1 | instruction | 0 | 30,879 | 9 | 61,758 |
"Correct Solution:
```
n,c = map(int,input().split())
p = list(map(int,input().split()))
m = n+1
s = sum(p)
ans = s//m
if s%m != 0:
ans += 1
print(ans)
``` | output | 1 | 30,879 | 9 | 61,759 |
Provide a correct Python 3 solution for this coding contest problem.
I’m planning to have a party on my birthday. Many of my friends will come to the party. Some of them will come with one or more pieces of cakes, but it is not certain if the number of the cakes is a multiple of the number of people coming.
I wish to enjoy the cakes equally among the partiers. So, I decided to apply the following rules. First, all the party attendants are given the same number of cakes. If some remainder occurs, a piece goes on a priority basis to the party host (that’s me!). How many pieces of cake can I enjoy?
Given the number of my friends and cake information, make a program to calculate how many pieces of cake I can enjoy. Note that I am not counted in the number of my friends.
Input
The input is given in the following format.
$N$ $C$
$p_1$ $p_2$ ... $p_C$
The first line provides the number of my friends $N$ ($1 \leq N \leq 100$) and the number of those among them who brought one or more pieces of cake with them $C$ ($1 \leq C \leq N$). The second line provides an array of integers $p_i$ ($1 \leq p_i \leq100$), each of which shows the number of cakes of the $i$-th friend of mine who was willing to come up with one or more pieces of cake.
Output
Output the number of cakes I can enjoy.
Examples
Input
5 4
5 5 6 5
Output
4
Input
7 5
8 8 8 8 8
Output
5
Input
100 3
3 3 3
Output
1 | instruction | 0 | 30,880 | 9 | 61,760 |
"Correct Solution:
```
N,C=map(int,input().split())
li=list(map(int,input().split()))
party_menber=N+1
amari=sum(li)%party_menber
hitori=sum(li)//party_menber
if amari>=1:
hitori=hitori+1
print(hitori)
``` | output | 1 | 30,880 | 9 | 61,761 |
Provide a correct Python 3 solution for this coding contest problem.
I’m planning to have a party on my birthday. Many of my friends will come to the party. Some of them will come with one or more pieces of cakes, but it is not certain if the number of the cakes is a multiple of the number of people coming.
I wish to enjoy the cakes equally among the partiers. So, I decided to apply the following rules. First, all the party attendants are given the same number of cakes. If some remainder occurs, a piece goes on a priority basis to the party host (that’s me!). How many pieces of cake can I enjoy?
Given the number of my friends and cake information, make a program to calculate how many pieces of cake I can enjoy. Note that I am not counted in the number of my friends.
Input
The input is given in the following format.
$N$ $C$
$p_1$ $p_2$ ... $p_C$
The first line provides the number of my friends $N$ ($1 \leq N \leq 100$) and the number of those among them who brought one or more pieces of cake with them $C$ ($1 \leq C \leq N$). The second line provides an array of integers $p_i$ ($1 \leq p_i \leq100$), each of which shows the number of cakes of the $i$-th friend of mine who was willing to come up with one or more pieces of cake.
Output
Output the number of cakes I can enjoy.
Examples
Input
5 4
5 5 6 5
Output
4
Input
7 5
8 8 8 8 8
Output
5
Input
100 3
3 3 3
Output
1 | instruction | 0 | 30,881 | 9 | 61,762 |
"Correct Solution:
```
#標準入力
a,b = map(int,input().split())
num = list(map(int,input().split()))
#リストの値を全て合計する
num = sum(num)
#出力
if num % (a + 1) == 0:print(num // (a + 1))
else:print(num // (a + 1) + 1)
``` | output | 1 | 30,881 | 9 | 61,763 |
Provide a correct Python 3 solution for this coding contest problem.
I’m planning to have a party on my birthday. Many of my friends will come to the party. Some of them will come with one or more pieces of cakes, but it is not certain if the number of the cakes is a multiple of the number of people coming.
I wish to enjoy the cakes equally among the partiers. So, I decided to apply the following rules. First, all the party attendants are given the same number of cakes. If some remainder occurs, a piece goes on a priority basis to the party host (that’s me!). How many pieces of cake can I enjoy?
Given the number of my friends and cake information, make a program to calculate how many pieces of cake I can enjoy. Note that I am not counted in the number of my friends.
Input
The input is given in the following format.
$N$ $C$
$p_1$ $p_2$ ... $p_C$
The first line provides the number of my friends $N$ ($1 \leq N \leq 100$) and the number of those among them who brought one or more pieces of cake with them $C$ ($1 \leq C \leq N$). The second line provides an array of integers $p_i$ ($1 \leq p_i \leq100$), each of which shows the number of cakes of the $i$-th friend of mine who was willing to come up with one or more pieces of cake.
Output
Output the number of cakes I can enjoy.
Examples
Input
5 4
5 5 6 5
Output
4
Input
7 5
8 8 8 8 8
Output
5
Input
100 3
3 3 3
Output
1 | instruction | 0 | 30,882 | 9 | 61,764 |
"Correct Solution:
```
N, C = map(int, input().split())
P = sum(list(map(int, input().split())))
if P%(N+1)==0:
print(P//(N+1))
else:
print(P//(N+1)+1)
``` | output | 1 | 30,882 | 9 | 61,765 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
I’m planning to have a party on my birthday. Many of my friends will come to the party. Some of them will come with one or more pieces of cakes, but it is not certain if the number of the cakes is a multiple of the number of people coming.
I wish to enjoy the cakes equally among the partiers. So, I decided to apply the following rules. First, all the party attendants are given the same number of cakes. If some remainder occurs, a piece goes on a priority basis to the party host (that’s me!). How many pieces of cake can I enjoy?
Given the number of my friends and cake information, make a program to calculate how many pieces of cake I can enjoy. Note that I am not counted in the number of my friends.
Input
The input is given in the following format.
$N$ $C$
$p_1$ $p_2$ ... $p_C$
The first line provides the number of my friends $N$ ($1 \leq N \leq 100$) and the number of those among them who brought one or more pieces of cake with them $C$ ($1 \leq C \leq N$). The second line provides an array of integers $p_i$ ($1 \leq p_i \leq100$), each of which shows the number of cakes of the $i$-th friend of mine who was willing to come up with one or more pieces of cake.
Output
Output the number of cakes I can enjoy.
Examples
Input
5 4
5 5 6 5
Output
4
Input
7 5
8 8 8 8 8
Output
5
Input
100 3
3 3 3
Output
1
Submitted Solution:
```
import math
N,C = map(int,input().split())
p = list(map(int,input().split()))
print(math.ceil(sum(p)/(N+1)))
``` | instruction | 0 | 30,883 | 9 | 61,766 |
Yes | output | 1 | 30,883 | 9 | 61,767 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
I’m planning to have a party on my birthday. Many of my friends will come to the party. Some of them will come with one or more pieces of cakes, but it is not certain if the number of the cakes is a multiple of the number of people coming.
I wish to enjoy the cakes equally among the partiers. So, I decided to apply the following rules. First, all the party attendants are given the same number of cakes. If some remainder occurs, a piece goes on a priority basis to the party host (that’s me!). How many pieces of cake can I enjoy?
Given the number of my friends and cake information, make a program to calculate how many pieces of cake I can enjoy. Note that I am not counted in the number of my friends.
Input
The input is given in the following format.
$N$ $C$
$p_1$ $p_2$ ... $p_C$
The first line provides the number of my friends $N$ ($1 \leq N \leq 100$) and the number of those among them who brought one or more pieces of cake with them $C$ ($1 \leq C \leq N$). The second line provides an array of integers $p_i$ ($1 \leq p_i \leq100$), each of which shows the number of cakes of the $i$-th friend of mine who was willing to come up with one or more pieces of cake.
Output
Output the number of cakes I can enjoy.
Examples
Input
5 4
5 5 6 5
Output
4
Input
7 5
8 8 8 8 8
Output
5
Input
100 3
3 3 3
Output
1
Submitted Solution:
```
n,c = map(int, input().split(" "))
p = sum(list(map(int, input().split(" "))))
n += 1
result = int(p / n)
if p % n == 0:
print(result)
else:
print(result + 1)
``` | instruction | 0 | 30,884 | 9 | 61,768 |
Yes | output | 1 | 30,884 | 9 | 61,769 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
I’m planning to have a party on my birthday. Many of my friends will come to the party. Some of them will come with one or more pieces of cakes, but it is not certain if the number of the cakes is a multiple of the number of people coming.
I wish to enjoy the cakes equally among the partiers. So, I decided to apply the following rules. First, all the party attendants are given the same number of cakes. If some remainder occurs, a piece goes on a priority basis to the party host (that’s me!). How many pieces of cake can I enjoy?
Given the number of my friends and cake information, make a program to calculate how many pieces of cake I can enjoy. Note that I am not counted in the number of my friends.
Input
The input is given in the following format.
$N$ $C$
$p_1$ $p_2$ ... $p_C$
The first line provides the number of my friends $N$ ($1 \leq N \leq 100$) and the number of those among them who brought one or more pieces of cake with them $C$ ($1 \leq C \leq N$). The second line provides an array of integers $p_i$ ($1 \leq p_i \leq100$), each of which shows the number of cakes of the $i$-th friend of mine who was willing to come up with one or more pieces of cake.
Output
Output the number of cakes I can enjoy.
Examples
Input
5 4
5 5 6 5
Output
4
Input
7 5
8 8 8 8 8
Output
5
Input
100 3
3 3 3
Output
1
Submitted Solution:
```
friends=input()
friends=friends.split()
friends_all=friends[0]
friends_cakes=friends[1]
person=int(friends_all)+1
cakes=input()
cakes_amount=0
for a in cakes.split():
cakes_amount+=int(a)
one_person=cakes_amount//person
if cakes_amount%person!=0:
one_person+=1
else:
pass
print(one_person)
``` | instruction | 0 | 30,885 | 9 | 61,770 |
Yes | output | 1 | 30,885 | 9 | 61,771 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
I’m planning to have a party on my birthday. Many of my friends will come to the party. Some of them will come with one or more pieces of cakes, but it is not certain if the number of the cakes is a multiple of the number of people coming.
I wish to enjoy the cakes equally among the partiers. So, I decided to apply the following rules. First, all the party attendants are given the same number of cakes. If some remainder occurs, a piece goes on a priority basis to the party host (that’s me!). How many pieces of cake can I enjoy?
Given the number of my friends and cake information, make a program to calculate how many pieces of cake I can enjoy. Note that I am not counted in the number of my friends.
Input
The input is given in the following format.
$N$ $C$
$p_1$ $p_2$ ... $p_C$
The first line provides the number of my friends $N$ ($1 \leq N \leq 100$) and the number of those among them who brought one or more pieces of cake with them $C$ ($1 \leq C \leq N$). The second line provides an array of integers $p_i$ ($1 \leq p_i \leq100$), each of which shows the number of cakes of the $i$-th friend of mine who was willing to come up with one or more pieces of cake.
Output
Output the number of cakes I can enjoy.
Examples
Input
5 4
5 5 6 5
Output
4
Input
7 5
8 8 8 8 8
Output
5
Input
100 3
3 3 3
Output
1
Submitted Solution:
```
N, C = map(int, input().split())
all_cake = list(map(int, input().split()))
ans = sum(all_cake)//(N+1)
if sum(all_cake) % (N+1) != 0:
ans += 1
print(ans)
``` | instruction | 0 | 30,886 | 9 | 61,772 |
Yes | output | 1 | 30,886 | 9 | 61,773 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
I’m planning to have a party on my birthday. Many of my friends will come to the party. Some of them will come with one or more pieces of cakes, but it is not certain if the number of the cakes is a multiple of the number of people coming.
I wish to enjoy the cakes equally among the partiers. So, I decided to apply the following rules. First, all the party attendants are given the same number of cakes. If some remainder occurs, a piece goes on a priority basis to the party host (that’s me!). How many pieces of cake can I enjoy?
Given the number of my friends and cake information, make a program to calculate how many pieces of cake I can enjoy. Note that I am not counted in the number of my friends.
Input
The input is given in the following format.
$N$ $C$
$p_1$ $p_2$ ... $p_C$
The first line provides the number of my friends $N$ ($1 \leq N \leq 100$) and the number of those among them who brought one or more pieces of cake with them $C$ ($1 \leq C \leq N$). The second line provides an array of integers $p_i$ ($1 \leq p_i \leq100$), each of which shows the number of cakes of the $i$-th friend of mine who was willing to come up with one or more pieces of cake.
Output
Output the number of cakes I can enjoy.
Examples
Input
5 4
5 5 6 5
Output
4
Input
7 5
8 8 8 8 8
Output
5
Input
100 3
3 3 3
Output
1
Submitted Solution:
```
friends=input()
friends.split(' ')
friends_all=friends[0]
friends_cake=friends[1]
cakes=input()
cakes_amount(sum(cakes.split(' ')))
one_person=cakes_amount//friends_all
if cakes_amount%friends_all==0:
pass
else:
one_person+=1
preint(one_person)
``` | instruction | 0 | 30,887 | 9 | 61,774 |
No | output | 1 | 30,887 | 9 | 61,775 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
I’m planning to have a party on my birthday. Many of my friends will come to the party. Some of them will come with one or more pieces of cakes, but it is not certain if the number of the cakes is a multiple of the number of people coming.
I wish to enjoy the cakes equally among the partiers. So, I decided to apply the following rules. First, all the party attendants are given the same number of cakes. If some remainder occurs, a piece goes on a priority basis to the party host (that’s me!). How many pieces of cake can I enjoy?
Given the number of my friends and cake information, make a program to calculate how many pieces of cake I can enjoy. Note that I am not counted in the number of my friends.
Input
The input is given in the following format.
$N$ $C$
$p_1$ $p_2$ ... $p_C$
The first line provides the number of my friends $N$ ($1 \leq N \leq 100$) and the number of those among them who brought one or more pieces of cake with them $C$ ($1 \leq C \leq N$). The second line provides an array of integers $p_i$ ($1 \leq p_i \leq100$), each of which shows the number of cakes of the $i$-th friend of mine who was willing to come up with one or more pieces of cake.
Output
Output the number of cakes I can enjoy.
Examples
Input
5 4
5 5 6 5
Output
4
Input
7 5
8 8 8 8 8
Output
5
Input
100 3
3 3 3
Output
1
Submitted Solution:
```
friends=input()
friends.split(' ')
friends_all=friends[0]
cakes=input()
cakes.split(' ')
cakes_amount=(sum(cakes))
one_person=cakes_amount//friends_all
if cakes_amount%friends_all==0:
pass
else:
one_person+=1
preint(one_person)
``` | instruction | 0 | 30,888 | 9 | 61,776 |
No | output | 1 | 30,888 | 9 | 61,777 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
I’m planning to have a party on my birthday. Many of my friends will come to the party. Some of them will come with one or more pieces of cakes, but it is not certain if the number of the cakes is a multiple of the number of people coming.
I wish to enjoy the cakes equally among the partiers. So, I decided to apply the following rules. First, all the party attendants are given the same number of cakes. If some remainder occurs, a piece goes on a priority basis to the party host (that’s me!). How many pieces of cake can I enjoy?
Given the number of my friends and cake information, make a program to calculate how many pieces of cake I can enjoy. Note that I am not counted in the number of my friends.
Input
The input is given in the following format.
$N$ $C$
$p_1$ $p_2$ ... $p_C$
The first line provides the number of my friends $N$ ($1 \leq N \leq 100$) and the number of those among them who brought one or more pieces of cake with them $C$ ($1 \leq C \leq N$). The second line provides an array of integers $p_i$ ($1 \leq p_i \leq100$), each of which shows the number of cakes of the $i$-th friend of mine who was willing to come up with one or more pieces of cake.
Output
Output the number of cakes I can enjoy.
Examples
Input
5 4
5 5 6 5
Output
4
Input
7 5
8 8 8 8 8
Output
5
Input
100 3
3 3 3
Output
1
Submitted Solution:
```
friends=input()
friends.split('')
friends_all=friends[0]
cakes=input()
cakes.split('')
cakes_amount(sum(cakes))
person=friends+1
one_person=cakes_amount//person
if cakes_amount%person==0:
pass
else:
one_person+=1
preint(one_person)
``` | instruction | 0 | 30,889 | 9 | 61,778 |
No | output | 1 | 30,889 | 9 | 61,779 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
I’m planning to have a party on my birthday. Many of my friends will come to the party. Some of them will come with one or more pieces of cakes, but it is not certain if the number of the cakes is a multiple of the number of people coming.
I wish to enjoy the cakes equally among the partiers. So, I decided to apply the following rules. First, all the party attendants are given the same number of cakes. If some remainder occurs, a piece goes on a priority basis to the party host (that’s me!). How many pieces of cake can I enjoy?
Given the number of my friends and cake information, make a program to calculate how many pieces of cake I can enjoy. Note that I am not counted in the number of my friends.
Input
The input is given in the following format.
$N$ $C$
$p_1$ $p_2$ ... $p_C$
The first line provides the number of my friends $N$ ($1 \leq N \leq 100$) and the number of those among them who brought one or more pieces of cake with them $C$ ($1 \leq C \leq N$). The second line provides an array of integers $p_i$ ($1 \leq p_i \leq100$), each of which shows the number of cakes of the $i$-th friend of mine who was willing to come up with one or more pieces of cake.
Output
Output the number of cakes I can enjoy.
Examples
Input
5 4
5 5 6 5
Output
4
Input
7 5
8 8 8 8 8
Output
5
Input
100 3
3 3 3
Output
1
Submitted Solution:
```
friends=input()
friends.split(' ')
friends_all=friends[0]
friends_cakes=friends[1]
person=int(friends_all)+1
cakes=input()
cakes_amount=0
for a in cakes.split(' '):
cakes_amount+=int(a)
one_person=cakes_amount//person
if cakes_amount%person!=0:
one_person+=1
else:
pass
print(one_person)
print(person)
``` | instruction | 0 | 30,890 | 9 | 61,780 |
No | output | 1 | 30,890 | 9 | 61,781 |
Provide a correct Python 3 solution for this coding contest problem.
Problem
One day, mo3tthi and tubuann decided to play a game with magic pockets and biscuits.
Now there are $ K $ pockets, numbered $ 1,2, \ ldots, K $.
The capacity of the $ i $ th pocket is $ M_i $, which initially contains $ N_i $ biscuits.
mo3tthi and tubuann start with mo3tthi and perform the following series of operations alternately.
* Choose one pocket.
* Perform one of the following operations only once. However, if the number of biscuits in the pocket selected as a result of the operation exceeds the capacity of the pocket, the operation cannot be performed.
* Stroking the selected pocket. Magical power increases the number of biscuits in your chosen pocket by $ 1 $.
* Hit the selected pocket. The number of biscuits in the pocket chosen by magical power is doubled by $ 2 $.
The game ends when you can't operate it, the person who can't operate loses, and the person who doesn't can win.
You, a friend of mo3tthi, were asked by mo3tthi in advance if you could decide if you could win this game.
For mo3tthi, make a program to determine if mo3tthi can definitely win this game.
Constraints
The input satisfies the following conditions.
* $ 1 \ leq K \ leq 10 ^ 5 $
* $ 1 \ leq N_i \ leq M_i \ leq 10 ^ {18} $
* All inputs are integers
Input
The input is given in the following format.
$ K $
$ N_1 $ $ M_1 $
$ \ vdots $
$ N_K $ $ M_K $
Output
When mo3tthi acts optimally, "mo3tthi" is output on one line if he can win, and "tubuann" is output otherwise.
Examples
Input
1
2 4
Output
mo3tthi
Input
2
2 3
3 8
Output
tubuann
Input
10
2 8
5 9
7 20
8 41
23 48
90 112
4 5
7 7
2344 8923
1 29
Output
mo3tthi | instruction | 0 | 30,928 | 9 | 61,856 |
"Correct Solution:
```
K = int(input())
ans = 0
for _ in range(K):
N, M = map(int, input().split())
grundy0 = {0}
grundy1 = {1}
m = M
while m // 2 >= N:
if (m+1)//2%2:
grundy_l = grundy0
else:
grundy_l = grundy1
m, r = divmod(m, 2)
if r==0:
gr0 = grundy0
grundy0 = {min({0, 1, 2} - grundy0 - grundy_l)}
grundy1 = {0, 1, 2} - gr0 - grundy0
else:
grundy0 = {min({0, 1, 2} - grundy1 - grundy_l)}
grundy1 = {0, 1, 2} - grundy1 - grundy0
#print(grundy0, grundy1)
grundy = grundy1 if (m-N)%2 else grundy0
#print(grundy)
ans ^= grundy.pop()
print("tubuann" if ans==0 else "mo3tthi")
``` | output | 1 | 30,928 | 9 | 61,857 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Chef Monocarp has just put n dishes into an oven. He knows that the i-th dish has its optimal cooking time equal to t_i minutes.
At any positive integer minute T Monocarp can put no more than one dish out of the oven. If the i-th dish is put out at some minute T, then its unpleasant value is |T - t_i| — the absolute difference between T and t_i. Once the dish is out of the oven, it can't go back in.
Monocarp should put all the dishes out of the oven. What is the minimum total unpleasant value Monocarp can obtain?
Input
The first line contains a single integer q (1 ≤ q ≤ 200) — the number of testcases.
Then q testcases follow.
The first line of the testcase contains a single integer n (1 ≤ n ≤ 200) — the number of dishes in the oven.
The second line of the testcase contains n integers t_1, t_2, ..., t_n (1 ≤ t_i ≤ n) — the optimal cooking time for each dish.
The sum of n over all q testcases doesn't exceed 200.
Output
Print a single integer for each testcase — the minimum total unpleasant value Monocarp can obtain when he puts out all the dishes out of the oven. Remember that Monocarp can only put the dishes out at positive integer minutes and no more than one dish at any minute.
Example
Input
6
6
4 2 4 4 5 2
7
7 7 7 7 7 7 7
1
1
5
5 1 2 4 3
4
1 4 4 4
21
21 8 1 4 1 5 21 1 8 21 11 21 11 3 12 8 19 15 9 11 13
Output
4
12
0
0
2
21
Note
In the first example Monocarp can put out the dishes at minutes 3, 1, 5, 4, 6, 2. That way the total unpleasant value will be |4 - 3| + |2 - 1| + |4 - 5| + |4 - 4| + |6 - 5| + |2 - 2| = 4.
In the second example Monocarp can put out the dishes at minutes 4, 5, 6, 7, 8, 9, 10.
In the third example Monocarp can put out the dish at minute 1.
In the fourth example Monocarp can put out the dishes at minutes 5, 1, 2, 4, 3.
In the fifth example Monocarp can put out the dishes at minutes 1, 3, 4, 5. | instruction | 0 | 31,936 | 9 | 63,872 |
Tags: dp, flows, graph matchings, greedy, math, sortings
Correct Solution:
```
#dp prac
def f(arr):
arr=list(map(lambda s:s-1,arr))
arr=sorted(arr)
dp=[[float("inf")]*(2*len(arr)+1) for i in range(0,len(arr)+1)]
dp[0][0]=0
for i in range(0,len(arr)+1):
for j in range(2*len(arr)-1):
if dp[i][j]!=float("inf"):
if i <len(arr):
dp[i+1][j+1]=min(dp[i+1][j+1],dp[i][j]+abs(arr[i]-j))
dp[i][j+1]=min(dp[i][j+1],dp[i][j])
return dp[len(arr)][2*len(arr)-1]
for i in range(int(input())):
a=input()
lst=list(map(int,input().strip().split()))
print(f(lst))
``` | output | 1 | 31,936 | 9 | 63,873 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Chef Monocarp has just put n dishes into an oven. He knows that the i-th dish has its optimal cooking time equal to t_i minutes.
At any positive integer minute T Monocarp can put no more than one dish out of the oven. If the i-th dish is put out at some minute T, then its unpleasant value is |T - t_i| — the absolute difference between T and t_i. Once the dish is out of the oven, it can't go back in.
Monocarp should put all the dishes out of the oven. What is the minimum total unpleasant value Monocarp can obtain?
Input
The first line contains a single integer q (1 ≤ q ≤ 200) — the number of testcases.
Then q testcases follow.
The first line of the testcase contains a single integer n (1 ≤ n ≤ 200) — the number of dishes in the oven.
The second line of the testcase contains n integers t_1, t_2, ..., t_n (1 ≤ t_i ≤ n) — the optimal cooking time for each dish.
The sum of n over all q testcases doesn't exceed 200.
Output
Print a single integer for each testcase — the minimum total unpleasant value Monocarp can obtain when he puts out all the dishes out of the oven. Remember that Monocarp can only put the dishes out at positive integer minutes and no more than one dish at any minute.
Example
Input
6
6
4 2 4 4 5 2
7
7 7 7 7 7 7 7
1
1
5
5 1 2 4 3
4
1 4 4 4
21
21 8 1 4 1 5 21 1 8 21 11 21 11 3 12 8 19 15 9 11 13
Output
4
12
0
0
2
21
Note
In the first example Monocarp can put out the dishes at minutes 3, 1, 5, 4, 6, 2. That way the total unpleasant value will be |4 - 3| + |2 - 1| + |4 - 5| + |4 - 4| + |6 - 5| + |2 - 2| = 4.
In the second example Monocarp can put out the dishes at minutes 4, 5, 6, 7, 8, 9, 10.
In the third example Monocarp can put out the dish at minute 1.
In the fourth example Monocarp can put out the dishes at minutes 5, 1, 2, 4, 3.
In the fifth example Monocarp can put out the dishes at minutes 1, 3, 4, 5. | instruction | 0 | 31,937 | 9 | 63,874 |
Tags: dp, flows, graph matchings, greedy, math, sortings
Correct Solution:
```
import sys, collections
input = sys.stdin.readline
for _ in range(int(input())):
n = int(input())
A = list(map(int, input().split()))
A.sort()
for i in range(n):
A[i] = A[i] - 1
dp = [[float('inf')] * (2 * n) for _ in range(n + 1)]
dp[0][0] = 0
for i in range(n + 1):
for j in range(2 * n - 1):
if dp[i][j] < float('inf'):
if i < n:
dp[i + 1][j + 1] = min(dp[i + 1][j + 1], dp[i][j] + abs(A[i] - j))
dp[i][j + 1] = min(dp[i][j + 1], dp[i][j])
print(dp[n][2 * n - 1])
``` | output | 1 | 31,937 | 9 | 63,875 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Chef Monocarp has just put n dishes into an oven. He knows that the i-th dish has its optimal cooking time equal to t_i minutes.
At any positive integer minute T Monocarp can put no more than one dish out of the oven. If the i-th dish is put out at some minute T, then its unpleasant value is |T - t_i| — the absolute difference between T and t_i. Once the dish is out of the oven, it can't go back in.
Monocarp should put all the dishes out of the oven. What is the minimum total unpleasant value Monocarp can obtain?
Input
The first line contains a single integer q (1 ≤ q ≤ 200) — the number of testcases.
Then q testcases follow.
The first line of the testcase contains a single integer n (1 ≤ n ≤ 200) — the number of dishes in the oven.
The second line of the testcase contains n integers t_1, t_2, ..., t_n (1 ≤ t_i ≤ n) — the optimal cooking time for each dish.
The sum of n over all q testcases doesn't exceed 200.
Output
Print a single integer for each testcase — the minimum total unpleasant value Monocarp can obtain when he puts out all the dishes out of the oven. Remember that Monocarp can only put the dishes out at positive integer minutes and no more than one dish at any minute.
Example
Input
6
6
4 2 4 4 5 2
7
7 7 7 7 7 7 7
1
1
5
5 1 2 4 3
4
1 4 4 4
21
21 8 1 4 1 5 21 1 8 21 11 21 11 3 12 8 19 15 9 11 13
Output
4
12
0
0
2
21
Note
In the first example Monocarp can put out the dishes at minutes 3, 1, 5, 4, 6, 2. That way the total unpleasant value will be |4 - 3| + |2 - 1| + |4 - 5| + |4 - 4| + |6 - 5| + |2 - 2| = 4.
In the second example Monocarp can put out the dishes at minutes 4, 5, 6, 7, 8, 9, 10.
In the third example Monocarp can put out the dish at minute 1.
In the fourth example Monocarp can put out the dishes at minutes 5, 1, 2, 4, 3.
In the fifth example Monocarp can put out the dishes at minutes 1, 3, 4, 5. | instruction | 0 | 31,938 | 9 | 63,876 |
Tags: dp, flows, graph matchings, greedy, math, sortings
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
u = sorted(map(int,input().split()))
d = [2 ** 31] * (2 * n + 1)
for i in range(1, 2 * n + 1):
d[i] = abs(u[0] - i)
for i in range(1, n):
cur = [2 ** 31] * (2 * n + 1)
for j in range(1, 2 * n + 1):
f = d[j]
for k in range(j + 1, 2 * n + 1):
cur[k] = min(cur[k], f + abs(u[i] - k))
d = cur[:]
print(min(d))
``` | output | 1 | 31,938 | 9 | 63,877 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Chef Monocarp has just put n dishes into an oven. He knows that the i-th dish has its optimal cooking time equal to t_i minutes.
At any positive integer minute T Monocarp can put no more than one dish out of the oven. If the i-th dish is put out at some minute T, then its unpleasant value is |T - t_i| — the absolute difference between T and t_i. Once the dish is out of the oven, it can't go back in.
Monocarp should put all the dishes out of the oven. What is the minimum total unpleasant value Monocarp can obtain?
Input
The first line contains a single integer q (1 ≤ q ≤ 200) — the number of testcases.
Then q testcases follow.
The first line of the testcase contains a single integer n (1 ≤ n ≤ 200) — the number of dishes in the oven.
The second line of the testcase contains n integers t_1, t_2, ..., t_n (1 ≤ t_i ≤ n) — the optimal cooking time for each dish.
The sum of n over all q testcases doesn't exceed 200.
Output
Print a single integer for each testcase — the minimum total unpleasant value Monocarp can obtain when he puts out all the dishes out of the oven. Remember that Monocarp can only put the dishes out at positive integer minutes and no more than one dish at any minute.
Example
Input
6
6
4 2 4 4 5 2
7
7 7 7 7 7 7 7
1
1
5
5 1 2 4 3
4
1 4 4 4
21
21 8 1 4 1 5 21 1 8 21 11 21 11 3 12 8 19 15 9 11 13
Output
4
12
0
0
2
21
Note
In the first example Monocarp can put out the dishes at minutes 3, 1, 5, 4, 6, 2. That way the total unpleasant value will be |4 - 3| + |2 - 1| + |4 - 5| + |4 - 4| + |6 - 5| + |2 - 2| = 4.
In the second example Monocarp can put out the dishes at minutes 4, 5, 6, 7, 8, 9, 10.
In the third example Monocarp can put out the dish at minute 1.
In the fourth example Monocarp can put out the dishes at minutes 5, 1, 2, 4, 3.
In the fifth example Monocarp can put out the dishes at minutes 1, 3, 4, 5. | instruction | 0 | 31,939 | 9 | 63,878 |
Tags: dp, flows, graph matchings, greedy, math, sortings
Correct Solution:
```
t = int(input())
ans = [0] * t
for _ in range(t):
n = int(input())
a = sorted(map(int, input().split()))
inf = 10**9
dp = [[inf] * (n + 100) for _ in range(n + 1)]
dp[0] = [0] * (n + 100)
for i in range(n):
cost = dp[i][0]
for t in range(1, n + 100):
dp[i + 1][t] = min(dp[i + 1][t], cost + abs(a[i] - t))
cost = min(cost, dp[i][t])
ans[_] = min(dp[-1])
print(*ans, sep='\n')
``` | output | 1 | 31,939 | 9 | 63,879 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Chef Monocarp has just put n dishes into an oven. He knows that the i-th dish has its optimal cooking time equal to t_i minutes.
At any positive integer minute T Monocarp can put no more than one dish out of the oven. If the i-th dish is put out at some minute T, then its unpleasant value is |T - t_i| — the absolute difference between T and t_i. Once the dish is out of the oven, it can't go back in.
Monocarp should put all the dishes out of the oven. What is the minimum total unpleasant value Monocarp can obtain?
Input
The first line contains a single integer q (1 ≤ q ≤ 200) — the number of testcases.
Then q testcases follow.
The first line of the testcase contains a single integer n (1 ≤ n ≤ 200) — the number of dishes in the oven.
The second line of the testcase contains n integers t_1, t_2, ..., t_n (1 ≤ t_i ≤ n) — the optimal cooking time for each dish.
The sum of n over all q testcases doesn't exceed 200.
Output
Print a single integer for each testcase — the minimum total unpleasant value Monocarp can obtain when he puts out all the dishes out of the oven. Remember that Monocarp can only put the dishes out at positive integer minutes and no more than one dish at any minute.
Example
Input
6
6
4 2 4 4 5 2
7
7 7 7 7 7 7 7
1
1
5
5 1 2 4 3
4
1 4 4 4
21
21 8 1 4 1 5 21 1 8 21 11 21 11 3 12 8 19 15 9 11 13
Output
4
12
0
0
2
21
Note
In the first example Monocarp can put out the dishes at minutes 3, 1, 5, 4, 6, 2. That way the total unpleasant value will be |4 - 3| + |2 - 1| + |4 - 5| + |4 - 4| + |6 - 5| + |2 - 2| = 4.
In the second example Monocarp can put out the dishes at minutes 4, 5, 6, 7, 8, 9, 10.
In the third example Monocarp can put out the dish at minute 1.
In the fourth example Monocarp can put out the dishes at minutes 5, 1, 2, 4, 3.
In the fifth example Monocarp can put out the dishes at minutes 1, 3, 4, 5. | instruction | 0 | 31,940 | 9 | 63,880 |
Tags: dp, flows, graph matchings, greedy, math, sortings
Correct Solution:
```
q = int(input())
for _ in range(q):
n = int(input())
inf = 10 ** 6
t = sorted(list(map(int, input().split())))
d = [[0] * (2 * n + 1) for i in range(n)]
d[0][0] = abs(t[0] - 1)
for i in range(n):
for j in range(1, 2 * n + 1):
if i == 0 and j > 0:
d[i][j] = min(abs(t[0] - j), d[i][j - 1])
elif i >= j:
d[i][j] = inf
else:
d[i][j] = min(d[i - 1][j - 1] + abs(t[i] - j), d[i][j - 1])
print(d[n - 1][2 * n])
# # for i from
# print(d)
``` | output | 1 | 31,940 | 9 | 63,881 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Chef Monocarp has just put n dishes into an oven. He knows that the i-th dish has its optimal cooking time equal to t_i minutes.
At any positive integer minute T Monocarp can put no more than one dish out of the oven. If the i-th dish is put out at some minute T, then its unpleasant value is |T - t_i| — the absolute difference between T and t_i. Once the dish is out of the oven, it can't go back in.
Monocarp should put all the dishes out of the oven. What is the minimum total unpleasant value Monocarp can obtain?
Input
The first line contains a single integer q (1 ≤ q ≤ 200) — the number of testcases.
Then q testcases follow.
The first line of the testcase contains a single integer n (1 ≤ n ≤ 200) — the number of dishes in the oven.
The second line of the testcase contains n integers t_1, t_2, ..., t_n (1 ≤ t_i ≤ n) — the optimal cooking time for each dish.
The sum of n over all q testcases doesn't exceed 200.
Output
Print a single integer for each testcase — the minimum total unpleasant value Monocarp can obtain when he puts out all the dishes out of the oven. Remember that Monocarp can only put the dishes out at positive integer minutes and no more than one dish at any minute.
Example
Input
6
6
4 2 4 4 5 2
7
7 7 7 7 7 7 7
1
1
5
5 1 2 4 3
4
1 4 4 4
21
21 8 1 4 1 5 21 1 8 21 11 21 11 3 12 8 19 15 9 11 13
Output
4
12
0
0
2
21
Note
In the first example Monocarp can put out the dishes at minutes 3, 1, 5, 4, 6, 2. That way the total unpleasant value will be |4 - 3| + |2 - 1| + |4 - 5| + |4 - 4| + |6 - 5| + |2 - 2| = 4.
In the second example Monocarp can put out the dishes at minutes 4, 5, 6, 7, 8, 9, 10.
In the third example Monocarp can put out the dish at minute 1.
In the fourth example Monocarp can put out the dishes at minutes 5, 1, 2, 4, 3.
In the fifth example Monocarp can put out the dishes at minutes 1, 3, 4, 5. | instruction | 0 | 31,941 | 9 | 63,882 |
Tags: dp, flows, graph matchings, greedy, math, sortings
Correct Solution:
```
# ===============================================================================================
# importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from itertools import *
import bisect
from heapq import *
from math import ceil, floor
from copy import *
from collections import deque, defaultdict
from collections import Counter as counter # Counter(list) return a dict with {key: count}
from itertools import combinations # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
from operator import *
# If the element is already present in the list,
# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect
# If the element is already present in the list,
# the right most position where element has to be inserted is returned
# ==============================================================================================
# fast I/O region
BUFSIZE = 8192
from sys import stderr
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
# inp = lambda: sys.stdin.readline().rstrip("\r\n")
# ===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType
def iterative(f, stack=[]):
def wrapped_func(*args, **kwargs):
if stack: return f(*args, **kwargs)
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
continue
stack.pop()
if not stack: break
to = stack[-1].send(to)
return to
return wrapped_func
#### END ITERATE RECURSION ####
# ===============================================================================================
# some shortcuts
mod = 1000000007
def inp(): return sys.stdin.readline().rstrip("\r\n") # for fast input
def out(var): sys.stdout.write(str(var)) # for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def fsep(): return map(float, inp().split())
def nextline(): out("\n") # as stdout.write always print sring.
def testcase(t):
for p in range(t):
solve()
def pow(x, y, p):
res = 1 # Initialize result
x = x % p # Update x if it is more , than or equal to p
if (x == 0):
return 0
while (y > 0):
if ((y & 1) == 1): # If y is odd, multiply, x with result
res = (res * x) % p
y = y >> 1 # y = y/2
x = (x * x) % p
return res
from functools import reduce
def factors(n):
return set(reduce(list.__add__,
([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0)))
def gcd(a, b):
if a == b: return a
while b > 0: a, b = b, a % b
return a
# discrete binary search
# minimise:
# def search():
# l = 0
# r = 10 ** 15
#
# for i in range(200):
# if isvalid(l):
# return l
# if l == r:
# return l
# m = (l + r) // 2
# if isvalid(m) and not isvalid(m - 1):
# return m
# if isvalid(m):
# r = m + 1
# else:
# l = m
# return m
# maximise:
# def search():
# l = 0
# r = 10 ** 15
#
# for i in range(200):
# # print(l,r)
# if isvalid(r):
# return r
# if l == r:
# return l
# m = (l + r) // 2
# if isvalid(m) and not isvalid(m + 1):
# return m
# if isvalid(m):
# l = m
# else:
# r = m - 1
# return m
#to find factorial and ncr
# N=100000
# mod = 10**9 +7
# fac = [1, 1]
# finv = [1, 1]
# inv = [0, 1]
#
# for i in range(2, N + 1):
# fac.append((fac[-1] * i) % mod)
# inv.append(mod - (inv[mod % i] * (mod // i) % mod))
# finv.append(finv[-1] * inv[-1] % mod)
#
#
# def comb(n, r):
# if n < r:
# return 0
# else:
# return fac[n] * (finv[r] * finv[n - r] % mod) % mod
##############Find sum of product of subsets of size k in a array
# ar=[0,1,2,3]
# k=3
# n=len(ar)-1
# dp=[0]*(n+1)
# dp[0]=1
# for pos in range(1,n+1):
# dp[pos]=0
# l=max(1,k+pos-n-1)
# for j in range(min(pos,k),l-1,-1):
# dp[j]=dp[j]+ar[pos]*dp[j-1]
# print(dp[k])
def prefix_sum(ar): # [1,2,3,4]->[1,3,6,10]
return list(accumulate(ar))
def suffix_sum(ar): # [1,2,3,4]->[10,9,7,4]
return list(accumulate(ar[::-1]))[::-1]
def N():
return int(inp())
# =========================================================================================
from collections import defaultdict
def numberOfSetBits(i):
i = i - ((i >> 1) & 0x55555555)
i = (i & 0x33333333) + ((i >> 2) & 0x33333333)
return (((i + (i >> 4) & 0xF0F0F0F) * 0x1010101) & 0xffffffff) >> 24
def solve():
n=N()
ar=lis()
ar.sort()
inf=float("inf")
dp=[[inf]*(3*n) for _ in range(n+1)]
for i in range(3*n):
dp[0][i]=0
for i in range(1,n+1):
for t in range(1,3*n):
for t2 in range(0,t):
dp[i][t]=min(dp[i][t],dp[i-1][t2]+abs(ar[i-1]-t))
# print(dp)
print(min(dp[n]))
#solve()
testcase(int(inp()))
``` | output | 1 | 31,941 | 9 | 63,883 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Chef Monocarp has just put n dishes into an oven. He knows that the i-th dish has its optimal cooking time equal to t_i minutes.
At any positive integer minute T Monocarp can put no more than one dish out of the oven. If the i-th dish is put out at some minute T, then its unpleasant value is |T - t_i| — the absolute difference between T and t_i. Once the dish is out of the oven, it can't go back in.
Monocarp should put all the dishes out of the oven. What is the minimum total unpleasant value Monocarp can obtain?
Input
The first line contains a single integer q (1 ≤ q ≤ 200) — the number of testcases.
Then q testcases follow.
The first line of the testcase contains a single integer n (1 ≤ n ≤ 200) — the number of dishes in the oven.
The second line of the testcase contains n integers t_1, t_2, ..., t_n (1 ≤ t_i ≤ n) — the optimal cooking time for each dish.
The sum of n over all q testcases doesn't exceed 200.
Output
Print a single integer for each testcase — the minimum total unpleasant value Monocarp can obtain when he puts out all the dishes out of the oven. Remember that Monocarp can only put the dishes out at positive integer minutes and no more than one dish at any minute.
Example
Input
6
6
4 2 4 4 5 2
7
7 7 7 7 7 7 7
1
1
5
5 1 2 4 3
4
1 4 4 4
21
21 8 1 4 1 5 21 1 8 21 11 21 11 3 12 8 19 15 9 11 13
Output
4
12
0
0
2
21
Note
In the first example Monocarp can put out the dishes at minutes 3, 1, 5, 4, 6, 2. That way the total unpleasant value will be |4 - 3| + |2 - 1| + |4 - 5| + |4 - 4| + |6 - 5| + |2 - 2| = 4.
In the second example Monocarp can put out the dishes at minutes 4, 5, 6, 7, 8, 9, 10.
In the third example Monocarp can put out the dish at minute 1.
In the fourth example Monocarp can put out the dishes at minutes 5, 1, 2, 4, 3.
In the fifth example Monocarp can put out the dishes at minutes 1, 3, 4, 5. | instruction | 0 | 31,942 | 9 | 63,884 |
Tags: dp, flows, graph matchings, greedy, math, sortings
Correct Solution:
```
INF = float('inf')
q = int(input())
for _ in range(q):
n = int(input())
dp = [[INF for ___ in range(2 * n+1)] for __ in range(n)]
t = list(sorted(map(int, input().split())))
dp[0][1] = t[0] - 1
for j in range(2, 2*n):
dp[0][j] = min(abs(t[0] - j), dp[0][j-1])
for i in range(1, n):
for j in range(i+1, 2*n):
dp[i][j] = min(abs(t[i] - j) + dp[i-1][j-1], dp[i][j-1])
print(min(dp[n-1]))
``` | output | 1 | 31,942 | 9 | 63,885 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Chef Monocarp has just put n dishes into an oven. He knows that the i-th dish has its optimal cooking time equal to t_i minutes.
At any positive integer minute T Monocarp can put no more than one dish out of the oven. If the i-th dish is put out at some minute T, then its unpleasant value is |T - t_i| — the absolute difference between T and t_i. Once the dish is out of the oven, it can't go back in.
Monocarp should put all the dishes out of the oven. What is the minimum total unpleasant value Monocarp can obtain?
Input
The first line contains a single integer q (1 ≤ q ≤ 200) — the number of testcases.
Then q testcases follow.
The first line of the testcase contains a single integer n (1 ≤ n ≤ 200) — the number of dishes in the oven.
The second line of the testcase contains n integers t_1, t_2, ..., t_n (1 ≤ t_i ≤ n) — the optimal cooking time for each dish.
The sum of n over all q testcases doesn't exceed 200.
Output
Print a single integer for each testcase — the minimum total unpleasant value Monocarp can obtain when he puts out all the dishes out of the oven. Remember that Monocarp can only put the dishes out at positive integer minutes and no more than one dish at any minute.
Example
Input
6
6
4 2 4 4 5 2
7
7 7 7 7 7 7 7
1
1
5
5 1 2 4 3
4
1 4 4 4
21
21 8 1 4 1 5 21 1 8 21 11 21 11 3 12 8 19 15 9 11 13
Output
4
12
0
0
2
21
Note
In the first example Monocarp can put out the dishes at minutes 3, 1, 5, 4, 6, 2. That way the total unpleasant value will be |4 - 3| + |2 - 1| + |4 - 5| + |4 - 4| + |6 - 5| + |2 - 2| = 4.
In the second example Monocarp can put out the dishes at minutes 4, 5, 6, 7, 8, 9, 10.
In the third example Monocarp can put out the dish at minute 1.
In the fourth example Monocarp can put out the dishes at minutes 5, 1, 2, 4, 3.
In the fifth example Monocarp can put out the dishes at minutes 1, 3, 4, 5. | instruction | 0 | 31,943 | 9 | 63,886 |
Tags: dp, flows, graph matchings, greedy, math, sortings
Correct Solution:
```
# dish position = 0 1 2 3 4 5 6 7 8 9
# sorted dishes = 2 2 4 5 6 6 6 7 9 11
#
# time = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
#
# take? = * * * * * * ? ? ? ? ?
# current_dish_pos= 1 2 2 2 3 3 4 5 6 6
#
# take? = * * * * * * ? ? ? ? ?
# current_dish_pos= 0 1 2 3 3 4 4 5 6 6
#
# # same subproblem
#
# # Dynamic programming
# # =Recursion while taking notes
# best_result(current_time, current_dish_pos) =
# best_result(current_time+1, current_dish_pos) # skip
# or
# abs(dish[current_dish_pos] - current_time) + best_result(current_time+1, current_dish_pos+1) # don't skip
#
#
# # (reverse) fibonacci(n) = fibonacci(n+1)+fibonacci(n+2)
#
# answer = best_result(1,0)
#
# stop at time 2n
# best_result(2n, 0) = 1000000
# best_result(2n, 1) = 1000000
# ...
# best_result(2n, n) = 0
#
# best_result(any x, n) = 0
# if you want recursion to depend on smaller values
# current_time -> time_left
# current_dish_pos -> dishes_left
for _ in range(int(input())):
num_dishes = int(input())
dishes = list(map(int, input().split()))
best_result = [[0] * (num_dishes+1) for i in range(2*num_dishes+1)]
# print(best_result)
dishes.sort()
for i in range(num_dishes):
best_result[2*num_dishes][i] = 1000000
# print()
for t in range(2*num_dishes-1,0,-1):
for d in range(num_dishes-1,-1,-1):
best_result[t][d] = min(best_result[t+1][d], abs(dishes[d] - t) + best_result[t+1][d+1])
# print(best_result)
print(best_result[1][0])
# print()
# print()
``` | output | 1 | 31,943 | 9 | 63,887 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chef Monocarp has just put n dishes into an oven. He knows that the i-th dish has its optimal cooking time equal to t_i minutes.
At any positive integer minute T Monocarp can put no more than one dish out of the oven. If the i-th dish is put out at some minute T, then its unpleasant value is |T - t_i| — the absolute difference between T and t_i. Once the dish is out of the oven, it can't go back in.
Monocarp should put all the dishes out of the oven. What is the minimum total unpleasant value Monocarp can obtain?
Input
The first line contains a single integer q (1 ≤ q ≤ 200) — the number of testcases.
Then q testcases follow.
The first line of the testcase contains a single integer n (1 ≤ n ≤ 200) — the number of dishes in the oven.
The second line of the testcase contains n integers t_1, t_2, ..., t_n (1 ≤ t_i ≤ n) — the optimal cooking time for each dish.
The sum of n over all q testcases doesn't exceed 200.
Output
Print a single integer for each testcase — the minimum total unpleasant value Monocarp can obtain when he puts out all the dishes out of the oven. Remember that Monocarp can only put the dishes out at positive integer minutes and no more than one dish at any minute.
Example
Input
6
6
4 2 4 4 5 2
7
7 7 7 7 7 7 7
1
1
5
5 1 2 4 3
4
1 4 4 4
21
21 8 1 4 1 5 21 1 8 21 11 21 11 3 12 8 19 15 9 11 13
Output
4
12
0
0
2
21
Note
In the first example Monocarp can put out the dishes at minutes 3, 1, 5, 4, 6, 2. That way the total unpleasant value will be |4 - 3| + |2 - 1| + |4 - 5| + |4 - 4| + |6 - 5| + |2 - 2| = 4.
In the second example Monocarp can put out the dishes at minutes 4, 5, 6, 7, 8, 9, 10.
In the third example Monocarp can put out the dish at minute 1.
In the fourth example Monocarp can put out the dishes at minutes 5, 1, 2, 4, 3.
In the fifth example Monocarp can put out the dishes at minutes 1, 3, 4, 5.
Submitted Solution:
```
import sys
input = sys.stdin.readline
for test in range(int(input())):
n = int(input())
t = list(map(int, input().split(" ")))
t.sort()
o = []
u = []
for i in range(n):
if not o or t[i]>o[-1]:
o.append(t[i])
u.append(0)
#print(o)
#print(u)
else:
o.append(o[-1]+1)
u.append(o[-2] - t[i] + 1)
#print(o)
#print(u)
#ualt1 = o[-1] - t[i] + 1
j = len(o)-3
#ualt2p =1
spoj =2
while j>-1 and o[j]==(o[j+1]-1):
spoj +=1
j-=1
#print('j'+str(j))
alt1 = sum([abs(k) for k in u[j+1:len(o)]])
#print('alt1'+str(alt1))
ualt = [abs(k-1) for k in u[j+1:len(o)]]
alt2 = sum(ualt)
#print('alt2'+str(alt2))
if alt2<alt1 and o[j+1]>1:
for h in range(j+1,len(o)):
o[h]-=1
u[h]-=1
#print(o)
#print(u)
print(sum([abs(k) for k in u]))
``` | instruction | 0 | 31,944 | 9 | 63,888 |
Yes | output | 1 | 31,944 | 9 | 63,889 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chef Monocarp has just put n dishes into an oven. He knows that the i-th dish has its optimal cooking time equal to t_i minutes.
At any positive integer minute T Monocarp can put no more than one dish out of the oven. If the i-th dish is put out at some minute T, then its unpleasant value is |T - t_i| — the absolute difference between T and t_i. Once the dish is out of the oven, it can't go back in.
Monocarp should put all the dishes out of the oven. What is the minimum total unpleasant value Monocarp can obtain?
Input
The first line contains a single integer q (1 ≤ q ≤ 200) — the number of testcases.
Then q testcases follow.
The first line of the testcase contains a single integer n (1 ≤ n ≤ 200) — the number of dishes in the oven.
The second line of the testcase contains n integers t_1, t_2, ..., t_n (1 ≤ t_i ≤ n) — the optimal cooking time for each dish.
The sum of n over all q testcases doesn't exceed 200.
Output
Print a single integer for each testcase — the minimum total unpleasant value Monocarp can obtain when he puts out all the dishes out of the oven. Remember that Monocarp can only put the dishes out at positive integer minutes and no more than one dish at any minute.
Example
Input
6
6
4 2 4 4 5 2
7
7 7 7 7 7 7 7
1
1
5
5 1 2 4 3
4
1 4 4 4
21
21 8 1 4 1 5 21 1 8 21 11 21 11 3 12 8 19 15 9 11 13
Output
4
12
0
0
2
21
Note
In the first example Monocarp can put out the dishes at minutes 3, 1, 5, 4, 6, 2. That way the total unpleasant value will be |4 - 3| + |2 - 1| + |4 - 5| + |4 - 4| + |6 - 5| + |2 - 2| = 4.
In the second example Monocarp can put out the dishes at minutes 4, 5, 6, 7, 8, 9, 10.
In the third example Monocarp can put out the dish at minute 1.
In the fourth example Monocarp can put out the dishes at minutes 5, 1, 2, 4, 3.
In the fifth example Monocarp can put out the dishes at minutes 1, 3, 4, 5.
Submitted Solution:
```
def read_int() -> int:
return int(input())
def read_ints():
return map(int, input().split(' '))
INF = 9999999999
def test_case():
n = read_int()
t = sorted(read_ints())
dp = [[(INF if i != 0 else 0) for _ in range(2 * n + 1)] for i in range(n + 1)] # dp[dish][minute]
for dish_index in range(1, n + 1):
for minute in range(1, 2 * n + 1):
dp[dish_index][minute] = min((x for index, x in enumerate(dp[dish_index-1]) if index < minute)) + abs(t[dish_index - 1] - minute)
print(min(dp[n]))
if __name__ == '__main__':
for _ in range(read_int()):
test_case()
``` | instruction | 0 | 31,945 | 9 | 63,890 |
Yes | output | 1 | 31,945 | 9 | 63,891 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chef Monocarp has just put n dishes into an oven. He knows that the i-th dish has its optimal cooking time equal to t_i minutes.
At any positive integer minute T Monocarp can put no more than one dish out of the oven. If the i-th dish is put out at some minute T, then its unpleasant value is |T - t_i| — the absolute difference between T and t_i. Once the dish is out of the oven, it can't go back in.
Monocarp should put all the dishes out of the oven. What is the minimum total unpleasant value Monocarp can obtain?
Input
The first line contains a single integer q (1 ≤ q ≤ 200) — the number of testcases.
Then q testcases follow.
The first line of the testcase contains a single integer n (1 ≤ n ≤ 200) — the number of dishes in the oven.
The second line of the testcase contains n integers t_1, t_2, ..., t_n (1 ≤ t_i ≤ n) — the optimal cooking time for each dish.
The sum of n over all q testcases doesn't exceed 200.
Output
Print a single integer for each testcase — the minimum total unpleasant value Monocarp can obtain when he puts out all the dishes out of the oven. Remember that Monocarp can only put the dishes out at positive integer minutes and no more than one dish at any minute.
Example
Input
6
6
4 2 4 4 5 2
7
7 7 7 7 7 7 7
1
1
5
5 1 2 4 3
4
1 4 4 4
21
21 8 1 4 1 5 21 1 8 21 11 21 11 3 12 8 19 15 9 11 13
Output
4
12
0
0
2
21
Note
In the first example Monocarp can put out the dishes at minutes 3, 1, 5, 4, 6, 2. That way the total unpleasant value will be |4 - 3| + |2 - 1| + |4 - 5| + |4 - 4| + |6 - 5| + |2 - 2| = 4.
In the second example Monocarp can put out the dishes at minutes 4, 5, 6, 7, 8, 9, 10.
In the third example Monocarp can put out the dish at minute 1.
In the fourth example Monocarp can put out the dishes at minutes 5, 1, 2, 4, 3.
In the fifth example Monocarp can put out the dishes at minutes 1, 3, 4, 5.
Submitted Solution:
```
from math import inf, log2
class SegmentTree:
def __init__(self, array, func=max):
self.n = len(array)
self.size = 2**(int(log2(self.n-1))+1) if self.n != 1 else 1
self.func = func
self.default = 0 if self.func != min else inf
self.data = [self.default] * (2 * self.size)
self.process(array)
def process(self, array):
self.data[self.size : self.size+self.n] = array
for i in range(self.size-1, -1, -1):
self.data[i] = self.func(self.data[2*i], self.data[2*i+1])
def query(self, alpha, omega):
"""Returns the result of function over the range (inclusive)!"""
if alpha == omega:
return self.data[alpha + self.size]
res = self.default
alpha += self.size
omega += self.size + 1
while alpha < omega:
if alpha & 1:
res = self.func(res, self.data[alpha])
alpha += 1
if omega & 1:
omega -= 1
res = self.func(res, self.data[omega])
alpha >>= 1
omega >>= 1
return res
def update(self, index, value):
"""Updates the element at index to given value!"""
index += self.size
self.data[index] = value
index >>= 1
while index:
self.data[index] = self.func(self.data[2*index], self.data[2*index+1])
index >>= 1
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
from math import gcd, ceil
def prod(a, mod=10**9+7):
ans = 1
for each in a:
ans = (ans * each) % mod
return ans
def lcm(a, b): return a * b // gcd(a, b)
def binary(x, length=16):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
for _ in range(int(input()) if True else 1):
n = int(input())
#l, r = map(int, input().split())
#a, b = map(int, input().split())
#c, d = map(int, input().split())
a = sorted(list(map(int, input().split())))
#b = list(map(int, input().split()))
dp = [[inf]*(n+1) for __ in range(501)]
dp[0][0] = 0
for j in range(n):
for i in range(500):
dp[i+1][j+1] = min(dp[i+1][j+1], dp[i][j] + abs(a[j] - (i+1)))
dp[i+1][j] = min(dp[i+1][j], dp[i][j])
print(min(x[-1] for x in dp))
``` | instruction | 0 | 31,946 | 9 | 63,892 |
Yes | output | 1 | 31,946 | 9 | 63,893 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chef Monocarp has just put n dishes into an oven. He knows that the i-th dish has its optimal cooking time equal to t_i minutes.
At any positive integer minute T Monocarp can put no more than one dish out of the oven. If the i-th dish is put out at some minute T, then its unpleasant value is |T - t_i| — the absolute difference between T and t_i. Once the dish is out of the oven, it can't go back in.
Monocarp should put all the dishes out of the oven. What is the minimum total unpleasant value Monocarp can obtain?
Input
The first line contains a single integer q (1 ≤ q ≤ 200) — the number of testcases.
Then q testcases follow.
The first line of the testcase contains a single integer n (1 ≤ n ≤ 200) — the number of dishes in the oven.
The second line of the testcase contains n integers t_1, t_2, ..., t_n (1 ≤ t_i ≤ n) — the optimal cooking time for each dish.
The sum of n over all q testcases doesn't exceed 200.
Output
Print a single integer for each testcase — the minimum total unpleasant value Monocarp can obtain when he puts out all the dishes out of the oven. Remember that Monocarp can only put the dishes out at positive integer minutes and no more than one dish at any minute.
Example
Input
6
6
4 2 4 4 5 2
7
7 7 7 7 7 7 7
1
1
5
5 1 2 4 3
4
1 4 4 4
21
21 8 1 4 1 5 21 1 8 21 11 21 11 3 12 8 19 15 9 11 13
Output
4
12
0
0
2
21
Note
In the first example Monocarp can put out the dishes at minutes 3, 1, 5, 4, 6, 2. That way the total unpleasant value will be |4 - 3| + |2 - 1| + |4 - 5| + |4 - 4| + |6 - 5| + |2 - 2| = 4.
In the second example Monocarp can put out the dishes at minutes 4, 5, 6, 7, 8, 9, 10.
In the third example Monocarp can put out the dish at minute 1.
In the fourth example Monocarp can put out the dishes at minutes 5, 1, 2, 4, 3.
In the fifth example Monocarp can put out the dishes at minutes 1, 3, 4, 5.
Submitted Solution:
```
import os, sys
from io import BytesIO, IOBase
from math import log2, ceil, sqrt, gcd
from _collections import deque
import heapq as hp
from bisect import bisect_left, bisect_right
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
mod = 10 ** 9 + 7
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
a.sort()
dp=[[0]*401 for _ in range(n)]
dp[0][0]=a[0]
for i in range(1,401):
dp[0][i]=min(dp[0][i-1],abs(i-a[0]))
for i in range(1,n):
for j in range(i+1,401):
if i+1==j:
dp[i][j]=dp[i-1][j-1]+abs(j-a[i])
continue
dp[i][j]=min(dp[i-1][j-1]+abs(j-a[i]),dp[i][j-1])
print(dp[-1][-1])
``` | instruction | 0 | 31,947 | 9 | 63,894 |
Yes | output | 1 | 31,947 | 9 | 63,895 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chef Monocarp has just put n dishes into an oven. He knows that the i-th dish has its optimal cooking time equal to t_i minutes.
At any positive integer minute T Monocarp can put no more than one dish out of the oven. If the i-th dish is put out at some minute T, then its unpleasant value is |T - t_i| — the absolute difference between T and t_i. Once the dish is out of the oven, it can't go back in.
Monocarp should put all the dishes out of the oven. What is the minimum total unpleasant value Monocarp can obtain?
Input
The first line contains a single integer q (1 ≤ q ≤ 200) — the number of testcases.
Then q testcases follow.
The first line of the testcase contains a single integer n (1 ≤ n ≤ 200) — the number of dishes in the oven.
The second line of the testcase contains n integers t_1, t_2, ..., t_n (1 ≤ t_i ≤ n) — the optimal cooking time for each dish.
The sum of n over all q testcases doesn't exceed 200.
Output
Print a single integer for each testcase — the minimum total unpleasant value Monocarp can obtain when he puts out all the dishes out of the oven. Remember that Monocarp can only put the dishes out at positive integer minutes and no more than one dish at any minute.
Example
Input
6
6
4 2 4 4 5 2
7
7 7 7 7 7 7 7
1
1
5
5 1 2 4 3
4
1 4 4 4
21
21 8 1 4 1 5 21 1 8 21 11 21 11 3 12 8 19 15 9 11 13
Output
4
12
0
0
2
21
Note
In the first example Monocarp can put out the dishes at minutes 3, 1, 5, 4, 6, 2. That way the total unpleasant value will be |4 - 3| + |2 - 1| + |4 - 5| + |4 - 4| + |6 - 5| + |2 - 2| = 4.
In the second example Monocarp can put out the dishes at minutes 4, 5, 6, 7, 8, 9, 10.
In the third example Monocarp can put out the dish at minute 1.
In the fourth example Monocarp can put out the dishes at minutes 5, 1, 2, 4, 3.
In the fifth example Monocarp can put out the dishes at minutes 1, 3, 4, 5.
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
ar = [int(i) for i in input().split()]
s = set()
ar.sort()
ans = 0
for i in ar:
if i not in s:
ans+=0
s.add(i)
else:
k = 1
while True:
if i-k>0 and i-k not in s:
ans+=abs((i-k)-i)
s.add(i-k)
break
elif i+k not in s:
ans+=abs((i+k)-i)
s.add(i+k)
break
k+=1
print(ans)
``` | instruction | 0 | 31,948 | 9 | 63,896 |
No | output | 1 | 31,948 | 9 | 63,897 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chef Monocarp has just put n dishes into an oven. He knows that the i-th dish has its optimal cooking time equal to t_i minutes.
At any positive integer minute T Monocarp can put no more than one dish out of the oven. If the i-th dish is put out at some minute T, then its unpleasant value is |T - t_i| — the absolute difference between T and t_i. Once the dish is out of the oven, it can't go back in.
Monocarp should put all the dishes out of the oven. What is the minimum total unpleasant value Monocarp can obtain?
Input
The first line contains a single integer q (1 ≤ q ≤ 200) — the number of testcases.
Then q testcases follow.
The first line of the testcase contains a single integer n (1 ≤ n ≤ 200) — the number of dishes in the oven.
The second line of the testcase contains n integers t_1, t_2, ..., t_n (1 ≤ t_i ≤ n) — the optimal cooking time for each dish.
The sum of n over all q testcases doesn't exceed 200.
Output
Print a single integer for each testcase — the minimum total unpleasant value Monocarp can obtain when he puts out all the dishes out of the oven. Remember that Monocarp can only put the dishes out at positive integer minutes and no more than one dish at any minute.
Example
Input
6
6
4 2 4 4 5 2
7
7 7 7 7 7 7 7
1
1
5
5 1 2 4 3
4
1 4 4 4
21
21 8 1 4 1 5 21 1 8 21 11 21 11 3 12 8 19 15 9 11 13
Output
4
12
0
0
2
21
Note
In the first example Monocarp can put out the dishes at minutes 3, 1, 5, 4, 6, 2. That way the total unpleasant value will be |4 - 3| + |2 - 1| + |4 - 5| + |4 - 4| + |6 - 5| + |2 - 2| = 4.
In the second example Monocarp can put out the dishes at minutes 4, 5, 6, 7, 8, 9, 10.
In the third example Monocarp can put out the dish at minute 1.
In the fourth example Monocarp can put out the dishes at minutes 5, 1, 2, 4, 3.
In the fifth example Monocarp can put out the dishes at minutes 1, 3, 4, 5.
Submitted Solution:
```
def ab(a):
if a>0:
return a
return -a
for _ in range(int(input())):
N=int(input())
A=[int(i) for i in input().split(" ")]
A.sort()
FT=[0]*(N+2)
for i in range(N):
FT[A[i]]+=1
Truth=[False]*(2*N+1)
Truth[0]=True
for i in range(N):
min=2000
index=0
for j in range(1,2*N+1):
if ab(A[i]-j)<min and not Truth[j]:
min=ab(A[i]-j)
index=j
Truth[index]=True
count=0
counter=1
for i in range(N):
while not Truth[counter]:
counter+=1
count+=ab(A[i]-counter)
counter+=1
print(count)
``` | instruction | 0 | 31,949 | 9 | 63,898 |
No | output | 1 | 31,949 | 9 | 63,899 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chef Monocarp has just put n dishes into an oven. He knows that the i-th dish has its optimal cooking time equal to t_i minutes.
At any positive integer minute T Monocarp can put no more than one dish out of the oven. If the i-th dish is put out at some minute T, then its unpleasant value is |T - t_i| — the absolute difference between T and t_i. Once the dish is out of the oven, it can't go back in.
Monocarp should put all the dishes out of the oven. What is the minimum total unpleasant value Monocarp can obtain?
Input
The first line contains a single integer q (1 ≤ q ≤ 200) — the number of testcases.
Then q testcases follow.
The first line of the testcase contains a single integer n (1 ≤ n ≤ 200) — the number of dishes in the oven.
The second line of the testcase contains n integers t_1, t_2, ..., t_n (1 ≤ t_i ≤ n) — the optimal cooking time for each dish.
The sum of n over all q testcases doesn't exceed 200.
Output
Print a single integer for each testcase — the minimum total unpleasant value Monocarp can obtain when he puts out all the dishes out of the oven. Remember that Monocarp can only put the dishes out at positive integer minutes and no more than one dish at any minute.
Example
Input
6
6
4 2 4 4 5 2
7
7 7 7 7 7 7 7
1
1
5
5 1 2 4 3
4
1 4 4 4
21
21 8 1 4 1 5 21 1 8 21 11 21 11 3 12 8 19 15 9 11 13
Output
4
12
0
0
2
21
Note
In the first example Monocarp can put out the dishes at minutes 3, 1, 5, 4, 6, 2. That way the total unpleasant value will be |4 - 3| + |2 - 1| + |4 - 5| + |4 - 4| + |6 - 5| + |2 - 2| = 4.
In the second example Monocarp can put out the dishes at minutes 4, 5, 6, 7, 8, 9, 10.
In the third example Monocarp can put out the dish at minute 1.
In the fourth example Monocarp can put out the dishes at minutes 5, 1, 2, 4, 3.
In the fifth example Monocarp can put out the dishes at minutes 1, 3, 4, 5.
Submitted Solution:
```
import sys
import math
from functools import reduce
import bisect
def getN():
return int(input())
def getNM():
return map(int, input().split())
def getList():
return list(map(int, input().split()))
def input():
return sys.stdin.readline().rstrip()
def index(a, x):
i = bisect.bisect_left(a, x)
if i != len(a) and a[i] == x:
return i
return False
#############
# MAIN CODE #
#############
for _ in range(getN()):
n = getN()
arr = sorted(getList())
dp = [[float('inf')] * (max(arr) * 2 +1) for _ in range(n + 1)]
dp[0] = [0] * (max(arr) * 2 +1)
for i in range(n):
for j in range(max(arr) * 2 ):
dp[i + 1][j + 1] = min(dp[i][j] + abs(arr[i] - j - 1), dp[i + 1][j])
print(dp[-1][-1])
``` | instruction | 0 | 31,950 | 9 | 63,900 |
No | output | 1 | 31,950 | 9 | 63,901 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chef Monocarp has just put n dishes into an oven. He knows that the i-th dish has its optimal cooking time equal to t_i minutes.
At any positive integer minute T Monocarp can put no more than one dish out of the oven. If the i-th dish is put out at some minute T, then its unpleasant value is |T - t_i| — the absolute difference between T and t_i. Once the dish is out of the oven, it can't go back in.
Monocarp should put all the dishes out of the oven. What is the minimum total unpleasant value Monocarp can obtain?
Input
The first line contains a single integer q (1 ≤ q ≤ 200) — the number of testcases.
Then q testcases follow.
The first line of the testcase contains a single integer n (1 ≤ n ≤ 200) — the number of dishes in the oven.
The second line of the testcase contains n integers t_1, t_2, ..., t_n (1 ≤ t_i ≤ n) — the optimal cooking time for each dish.
The sum of n over all q testcases doesn't exceed 200.
Output
Print a single integer for each testcase — the minimum total unpleasant value Monocarp can obtain when he puts out all the dishes out of the oven. Remember that Monocarp can only put the dishes out at positive integer minutes and no more than one dish at any minute.
Example
Input
6
6
4 2 4 4 5 2
7
7 7 7 7 7 7 7
1
1
5
5 1 2 4 3
4
1 4 4 4
21
21 8 1 4 1 5 21 1 8 21 11 21 11 3 12 8 19 15 9 11 13
Output
4
12
0
0
2
21
Note
In the first example Monocarp can put out the dishes at minutes 3, 1, 5, 4, 6, 2. That way the total unpleasant value will be |4 - 3| + |2 - 1| + |4 - 5| + |4 - 4| + |6 - 5| + |2 - 2| = 4.
In the second example Monocarp can put out the dishes at minutes 4, 5, 6, 7, 8, 9, 10.
In the third example Monocarp can put out the dish at minute 1.
In the fourth example Monocarp can put out the dishes at minutes 5, 1, 2, 4, 3.
In the fifth example Monocarp can put out the dishes at minutes 1, 3, 4, 5.
Submitted Solution:
```
def find_repetition_in_time_list(time_list):
unpleasure=0
repeat_start=0
length=1
i=0
while True:
if i==len(time_list)-1:
break
if time_list[i]!=time_list[i+1]:
if length>1:
unplea,time_list=expand_repetitive(time_list,repeat_start,length)
unpleasure+=unplea
repeat_start+=1
i=repeat_start
length=1
else:
i+=1
length+=1
if length>1:
unplea,time_list=expand_repetitive(time_list,repeat_start,length)
unpleasure+=unplea
return unpleasure
def expand_repetitive(time_list,starter,length):
unpleasure_produced=0
expand_range=1
for _ in range(length-1):
while True:
starter ,has_insert = attempt_insert_left(time_list, expand_range, starter)
if has_insert:
break
starter ,has_insert = attempt_insert_right(time_list, expand_range, starter, length)
if has_insert:
break
expand_range+=1
time_list=sorted(time_list)
unpleasure_produced+=expand_range
return unpleasure_produced,time_list
def attempt_insert_right(time_list, expand_range, starter, length):
has_insert=False
new_minute=time_list[starter]+expand_range
new_index=starter+length+expand_range-1
if can_insert(time_list, new_minute):
new_index=remove_index_exception(new_index)
time_list.insert(new_index,new_minute)
time_list.remove(time_list[starter])
length-=1
has_insert=True
return starter, has_insert
def attempt_insert_left(time_list, expand_range, starter):
has_insert=False
new_minute=time_list[starter]-expand_range
new_index=starter-expand_range+1
if can_insert(time_list, new_minute):
new_index=remove_index_exception(new_index)
time_list.insert(new_index,new_minute)
starter+=1
time_list.remove(time_list[starter])
has_insert=True
return starter, has_insert
def can_insert(time_list, new_minute):
if new_minute not in time_list:#can insert new minute value
if remove_minus_minute_exception(new_minute):
return False
return True
return False
def remove_minus_minute_exception(new_minute):
if new_minute<=0:
return True
return False
def remove_index_exception(new_index):
if new_index<=0:
new_index=0
if new_index>=dish_amount:
new_index=dish_amount-1
return new_index
test_amount=int(input())
for i in range(test_amount):
dish_amount=int(input())
prepare_time_list=list(map(int,input().split()))
prepare_time_list=sorted(prepare_time_list)
#print(prepare_time_list)
print(find_repetition_in_time_list(prepare_time_list))
``` | instruction | 0 | 31,951 | 9 | 63,902 |
No | output | 1 | 31,951 | 9 | 63,903 |
Provide a correct Python 3 solution for this coding contest problem.
You are going to eat X red apples and Y green apples.
You have A red apples of deliciousness p_1,p_2, \dots, p_A, B green apples of deliciousness q_1,q_2, \dots, q_B, and C colorless apples of deliciousness r_1,r_2, \dots, r_C.
Before eating a colorless apple, you can paint it red or green, and it will count as a red or green apple, respectively.
From the apples above, you will choose the apples to eat while making the sum of the deliciousness of the eaten apples as large as possible.
Find the maximum possible sum of the deliciousness of the eaten apples that can be achieved when optimally coloring zero or more colorless apples.
Constraints
* 1 \leq X \leq A \leq 10^5
* 1 \leq Y \leq B \leq 10^5
* 1 \leq C \leq 10^5
* 1 \leq p_i \leq 10^9
* 1 \leq q_i \leq 10^9
* 1 \leq r_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X Y A B C
p_1 p_2 ... p_A
q_1 q_2 ... q_B
r_1 r_2 ... r_C
Output
Print the maximum possible sum of the deliciousness of the eaten apples.
Examples
Input
1 2 2 2 1
2 4
5 1
3
Output
12
Input
2 2 2 2 2
8 6
9 1
2 1
Output
25
Input
2 2 4 4 4
11 12 13 14
21 22 23 24
1 2 3 4
Output
74 | instruction | 0 | 32,255 | 9 | 64,510 |
"Correct Solution:
```
x,y,a,b,c=map(int,input().split()) #x,yだけ見る。
p=list(map(int,input().split()))
q=list(map(int,input().split()))
r=list(map(int,input().split()))
p=sorted(p,reverse=True)[:x]
q=sorted(q,reverse=True)[:y]
print(sum(sorted(p+q+r, reverse=True)[:x+y]))
``` | output | 1 | 32,255 | 9 | 64,511 |
Provide a correct Python 3 solution for this coding contest problem.
You are going to eat X red apples and Y green apples.
You have A red apples of deliciousness p_1,p_2, \dots, p_A, B green apples of deliciousness q_1,q_2, \dots, q_B, and C colorless apples of deliciousness r_1,r_2, \dots, r_C.
Before eating a colorless apple, you can paint it red or green, and it will count as a red or green apple, respectively.
From the apples above, you will choose the apples to eat while making the sum of the deliciousness of the eaten apples as large as possible.
Find the maximum possible sum of the deliciousness of the eaten apples that can be achieved when optimally coloring zero or more colorless apples.
Constraints
* 1 \leq X \leq A \leq 10^5
* 1 \leq Y \leq B \leq 10^5
* 1 \leq C \leq 10^5
* 1 \leq p_i \leq 10^9
* 1 \leq q_i \leq 10^9
* 1 \leq r_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X Y A B C
p_1 p_2 ... p_A
q_1 q_2 ... q_B
r_1 r_2 ... r_C
Output
Print the maximum possible sum of the deliciousness of the eaten apples.
Examples
Input
1 2 2 2 1
2 4
5 1
3
Output
12
Input
2 2 2 2 2
8 6
9 1
2 1
Output
25
Input
2 2 4 4 4
11 12 13 14
21 22 23 24
1 2 3 4
Output
74 | instruction | 0 | 32,256 | 9 | 64,512 |
"Correct Solution:
```
x,y,a,b,c=(int(i) for i in input().split())
p=sorted([int(i) for i in input().split()])
q=sorted([int(i) for i in input().split()])
r=sorted([int(i) for i in input().split()])
l=sorted(p[a-x:]+q[b-y:]+r)
n=len(l)
print(sum(l[n-x-y:]))
``` | output | 1 | 32,256 | 9 | 64,513 |
Provide a correct Python 3 solution for this coding contest problem.
You are going to eat X red apples and Y green apples.
You have A red apples of deliciousness p_1,p_2, \dots, p_A, B green apples of deliciousness q_1,q_2, \dots, q_B, and C colorless apples of deliciousness r_1,r_2, \dots, r_C.
Before eating a colorless apple, you can paint it red or green, and it will count as a red or green apple, respectively.
From the apples above, you will choose the apples to eat while making the sum of the deliciousness of the eaten apples as large as possible.
Find the maximum possible sum of the deliciousness of the eaten apples that can be achieved when optimally coloring zero or more colorless apples.
Constraints
* 1 \leq X \leq A \leq 10^5
* 1 \leq Y \leq B \leq 10^5
* 1 \leq C \leq 10^5
* 1 \leq p_i \leq 10^9
* 1 \leq q_i \leq 10^9
* 1 \leq r_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X Y A B C
p_1 p_2 ... p_A
q_1 q_2 ... q_B
r_1 r_2 ... r_C
Output
Print the maximum possible sum of the deliciousness of the eaten apples.
Examples
Input
1 2 2 2 1
2 4
5 1
3
Output
12
Input
2 2 2 2 2
8 6
9 1
2 1
Output
25
Input
2 2 4 4 4
11 12 13 14
21 22 23 24
1 2 3 4
Output
74 | instruction | 0 | 32,257 | 9 | 64,514 |
"Correct Solution:
```
x,y,a,b,c = map(int,input().split())
p=list(map(int,input().split()))
p=sorted(p)[-x:]
q=list(map(int,input().split()))
q=sorted(q)[-y:]
r=list(map(int,input().split()))
print( sum(sorted(p+q+r)[c:]) )
``` | output | 1 | 32,257 | 9 | 64,515 |
Provide a correct Python 3 solution for this coding contest problem.
You are going to eat X red apples and Y green apples.
You have A red apples of deliciousness p_1,p_2, \dots, p_A, B green apples of deliciousness q_1,q_2, \dots, q_B, and C colorless apples of deliciousness r_1,r_2, \dots, r_C.
Before eating a colorless apple, you can paint it red or green, and it will count as a red or green apple, respectively.
From the apples above, you will choose the apples to eat while making the sum of the deliciousness of the eaten apples as large as possible.
Find the maximum possible sum of the deliciousness of the eaten apples that can be achieved when optimally coloring zero or more colorless apples.
Constraints
* 1 \leq X \leq A \leq 10^5
* 1 \leq Y \leq B \leq 10^5
* 1 \leq C \leq 10^5
* 1 \leq p_i \leq 10^9
* 1 \leq q_i \leq 10^9
* 1 \leq r_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X Y A B C
p_1 p_2 ... p_A
q_1 q_2 ... q_B
r_1 r_2 ... r_C
Output
Print the maximum possible sum of the deliciousness of the eaten apples.
Examples
Input
1 2 2 2 1
2 4
5 1
3
Output
12
Input
2 2 2 2 2
8 6
9 1
2 1
Output
25
Input
2 2 4 4 4
11 12 13 14
21 22 23 24
1 2 3 4
Output
74 | instruction | 0 | 32,258 | 9 | 64,516 |
"Correct Solution:
```
x,y,a,b,c = map(int,input().split())
p= list(map(int,input().split()))
q= list(map(int,input().split()))
r= list(map(int,input().split()))
p = sorted(p)[-x:]
q = sorted(q)[-y:]
r = r+p+q
print(sum(sorted(r)[-(x+y):]))
``` | output | 1 | 32,258 | 9 | 64,517 |
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