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.
There are some rabbits in Singapore Zoo. To feed them, Zookeeper bought n carrots with lengths a_1, a_2, a_3, …, a_n. However, rabbits are very fertile and multiply very quickly. Zookeeper now has k rabbits and does not have enough carrots to feed all of them. To solve this problem, Zookeeper decided to cut the carrots into k pieces. For some reason, all resulting carrot lengths must be positive integers.
Big carrots are very difficult for rabbits to handle and eat, so the time needed to eat a carrot of size x is x^2.
Help Zookeeper split his carrots while minimizing the sum of time taken for rabbits to eat the carrots.
Input
The first line contains two integers n and k (1 ≤ n ≤ k ≤ 10^5): the initial number of carrots and the number of rabbits.
The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6): lengths of carrots.
It is guaranteed that the sum of a_i is at least k.
Output
Output one integer: the minimum sum of time taken for rabbits to eat carrots.
Examples
Input
3 6
5 3 1
Output
15
Input
1 4
19
Output
91
Note
For the first test, the optimal sizes of carrots are \{1,1,1,2,2,2\}. The time taken is 1^2+1^2+1^2+2^2+2^2+2^2=15
For the second test, the optimal sizes of carrots are \{4,5,5,5\}. The time taken is 4^2+5^2+5^2+5^2=91. | instruction | 0 | 104,642 | 9 | 209,284 |
Tags: binary search, data structures, greedy, math, sortings
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
from heapq import heappush, heappop
def val(L, P):
if L <= P:
return L
q, r = divmod(L, P)
return q * q * (P - r) + (q + 1) * (q + 1) * r
def main():
n, k, *a = map(int, sys.stdin.read().split())
pq = []
cost = 0
for x in a:
now = val(x, 1)
nxt = val(x, 2)
cost += now
if nxt < now:
heappush(pq, (nxt - now, now, x, 2))
for _ in range(k - n):
impr, now, L, P = heappop(pq)
now += impr
cost += impr
nxt = val(L, P + 1)
if nxt < now:
heappush(pq, (nxt - now, now, L, P + 1))
print(cost)
# region fastio
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)
def input(): return sys.stdin.readline().rstrip("\r\n")
def getint(): return int(input())
def getints(): return list(map(int, input().split()))
def getint1(): return list(map(lambda x: int(x) - 1, input().split()))
# endregion
if __name__ == "__main__":
main()
``` | output | 1 | 104,642 | 9 | 209,285 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are some rabbits in Singapore Zoo. To feed them, Zookeeper bought n carrots with lengths a_1, a_2, a_3, …, a_n. However, rabbits are very fertile and multiply very quickly. Zookeeper now has k rabbits and does not have enough carrots to feed all of them. To solve this problem, Zookeeper decided to cut the carrots into k pieces. For some reason, all resulting carrot lengths must be positive integers.
Big carrots are very difficult for rabbits to handle and eat, so the time needed to eat a carrot of size x is x^2.
Help Zookeeper split his carrots while minimizing the sum of time taken for rabbits to eat the carrots.
Input
The first line contains two integers n and k (1 ≤ n ≤ k ≤ 10^5): the initial number of carrots and the number of rabbits.
The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6): lengths of carrots.
It is guaranteed that the sum of a_i is at least k.
Output
Output one integer: the minimum sum of time taken for rabbits to eat carrots.
Examples
Input
3 6
5 3 1
Output
15
Input
1 4
19
Output
91
Note
For the first test, the optimal sizes of carrots are \{1,1,1,2,2,2\}. The time taken is 1^2+1^2+1^2+2^2+2^2+2^2=15
For the second test, the optimal sizes of carrots are \{4,5,5,5\}. The time taken is 4^2+5^2+5^2+5^2=91. | instruction | 0 | 104,643 | 9 | 209,286 |
Tags: binary search, data structures, greedy, math, sortings
Correct Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
import heapq
MOD = 10 ** 5 + 1
def main():
n,k = map(int,input().split())
a = list(map(int,input().split()))
target = max((sum(a) + (k - 1)) // k - 100,1)
cost = []
currentSplit = []
currentCost = []
for elem in a:
currentSplit.append((elem + target - 1) // target)
currentCost.append((elem // currentSplit[-1]) ** 2 * currentSplit[-1])
currentCost[-1] += (elem % currentSplit[-1]) * (1 + 2 * (elem // currentSplit[-1]))
if currentSplit[-1] != 1:
newSplit = currentSplit[-1] - 1
curDecCost = (elem // newSplit) ** 2 * newSplit
curDecCost += (elem % newSplit) * (1 + 2 * (elem // newSplit))
heapq.heappush(cost,(curDecCost - currentCost[-1]) * MOD + len(currentCost) - 1)
curAns = sum(currentSplit)
curCost = sum(currentCost)
while curAns > k:
curAns -= 1
curElem = heapq.heappop(cost)
addedCost = curElem // MOD
loc = curElem % MOD
currentCost[loc] += addedCost
curCost += addedCost
currentSplit[loc] -= 1
if currentSplit[loc] != 1:
newSplit = currentSplit[loc] - 1
elem = a[loc]
curDecCost = (elem // newSplit) ** 2 * newSplit
curDecCost += (elem % newSplit) * (1 + 2 * (elem // newSplit))
heapq.heappush(cost,(curDecCost - currentCost[loc]) * MOD + loc)
print(curCost)
# region fastio
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")
# endregion
if __name__ == "__main__":
main()
``` | output | 1 | 104,643 | 9 | 209,287 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are some rabbits in Singapore Zoo. To feed them, Zookeeper bought n carrots with lengths a_1, a_2, a_3, …, a_n. However, rabbits are very fertile and multiply very quickly. Zookeeper now has k rabbits and does not have enough carrots to feed all of them. To solve this problem, Zookeeper decided to cut the carrots into k pieces. For some reason, all resulting carrot lengths must be positive integers.
Big carrots are very difficult for rabbits to handle and eat, so the time needed to eat a carrot of size x is x^2.
Help Zookeeper split his carrots while minimizing the sum of time taken for rabbits to eat the carrots.
Input
The first line contains two integers n and k (1 ≤ n ≤ k ≤ 10^5): the initial number of carrots and the number of rabbits.
The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6): lengths of carrots.
It is guaranteed that the sum of a_i is at least k.
Output
Output one integer: the minimum sum of time taken for rabbits to eat carrots.
Examples
Input
3 6
5 3 1
Output
15
Input
1 4
19
Output
91
Note
For the first test, the optimal sizes of carrots are \{1,1,1,2,2,2\}. The time taken is 1^2+1^2+1^2+2^2+2^2+2^2=15
For the second test, the optimal sizes of carrots are \{4,5,5,5\}. The time taken is 4^2+5^2+5^2+5^2=91. | instruction | 0 | 104,644 | 9 | 209,288 |
Tags: binary search, data structures, greedy, math, sortings
Correct Solution:
```
from heapq import heappush, heappop
def f(length, n_piece):
width = length // n_piece
extra = length - width * n_piece
return width ** 2 * (n_piece - extra) + (width + 1) ** 2 * extra
def solve():
N, K = map(int, input().split())
A = [int(x) for x in input().split()]
ans = sum(a ** 2 for a in A)
heap = []
for i, a in enumerate(A):
decrease = f(a, 2) - f(a, 1)
n_piece = 2
heappush(heap, (decrease, a, n_piece))
for _ in range(K - N):
decrease, a, n_piece = heappop(heap)
ans += decrease
decrease = f(a, n_piece + 1) - f(a, n_piece)
n_piece = n_piece + 1
heappush(heap, (decrease, a, n_piece))
return ans
print(solve())
# import matplotlib.pyplot as plt
# fig, ax = plt.subplots()
# xs = list(range(1, 30))
# ys = [f(25, x) for x in xs]
# ax.plot(xs, ys, '.-')
# ax.set_xlabel('#piece')
# ax.set_ylabel('f(25, x)')
# plt.show()
``` | output | 1 | 104,644 | 9 | 209,289 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are some rabbits in Singapore Zoo. To feed them, Zookeeper bought n carrots with lengths a_1, a_2, a_3, …, a_n. However, rabbits are very fertile and multiply very quickly. Zookeeper now has k rabbits and does not have enough carrots to feed all of them. To solve this problem, Zookeeper decided to cut the carrots into k pieces. For some reason, all resulting carrot lengths must be positive integers.
Big carrots are very difficult for rabbits to handle and eat, so the time needed to eat a carrot of size x is x^2.
Help Zookeeper split his carrots while minimizing the sum of time taken for rabbits to eat the carrots.
Input
The first line contains two integers n and k (1 ≤ n ≤ k ≤ 10^5): the initial number of carrots and the number of rabbits.
The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6): lengths of carrots.
It is guaranteed that the sum of a_i is at least k.
Output
Output one integer: the minimum sum of time taken for rabbits to eat carrots.
Examples
Input
3 6
5 3 1
Output
15
Input
1 4
19
Output
91
Note
For the first test, the optimal sizes of carrots are \{1,1,1,2,2,2\}. The time taken is 1^2+1^2+1^2+2^2+2^2+2^2=15
For the second test, the optimal sizes of carrots are \{4,5,5,5\}. The time taken is 4^2+5^2+5^2+5^2=91.
Submitted Solution:
```
import heapq
def value(l,parts):
len1 = l//parts
len2 = len1+1
cnt2 = l%parts
cnt1 = parts-cnt2
return cnt1*len1*len1 + cnt2*len2*len2
def solve(n,k,a):
total = 0
pq = []
for i in range(n):
total += a[i]*a[i]
heapq.heappush(pq,(-1*value(a[i],1)+value(a[i],2),a[i],2))
for i in range(k-n):
temp = heapq.heappop(pq)
total += temp[0]
a,b = temp[1],temp[2]
heapq.heappush(pq,(-value(a,b)+value(a,b+1),a,b+1))
return total
n,k = map(int,input().split())
a = list(map(int,input().split()))
print(solve(n,k,a))
``` | instruction | 0 | 104,645 | 9 | 209,290 |
Yes | output | 1 | 104,645 | 9 | 209,291 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are some rabbits in Singapore Zoo. To feed them, Zookeeper bought n carrots with lengths a_1, a_2, a_3, …, a_n. However, rabbits are very fertile and multiply very quickly. Zookeeper now has k rabbits and does not have enough carrots to feed all of them. To solve this problem, Zookeeper decided to cut the carrots into k pieces. For some reason, all resulting carrot lengths must be positive integers.
Big carrots are very difficult for rabbits to handle and eat, so the time needed to eat a carrot of size x is x^2.
Help Zookeeper split his carrots while minimizing the sum of time taken for rabbits to eat the carrots.
Input
The first line contains two integers n and k (1 ≤ n ≤ k ≤ 10^5): the initial number of carrots and the number of rabbits.
The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6): lengths of carrots.
It is guaranteed that the sum of a_i is at least k.
Output
Output one integer: the minimum sum of time taken for rabbits to eat carrots.
Examples
Input
3 6
5 3 1
Output
15
Input
1 4
19
Output
91
Note
For the first test, the optimal sizes of carrots are \{1,1,1,2,2,2\}. The time taken is 1^2+1^2+1^2+2^2+2^2+2^2=15
For the second test, the optimal sizes of carrots are \{4,5,5,5\}. The time taken is 4^2+5^2+5^2+5^2=91.
Submitted Solution:
```
from heapq import *
def find_add(a, k):
extra = a%k
return (k-extra)*((a//k)**2) + extra*((1+(a//k))**2)
n, mx = map(int, input().split())
lst = list(map(int, input().split()))
a = []
heapify(a)
for i in lst:
heappush(a, (find_add(i, 2)-find_add(i, 1), i, 1))
total = n
while total != mx:
curr_cont, el, k = heappop(a)
new = (find_add(el, k+2)-find_add(el, k+1), el, k+1)
heappush(a, new)
total += 1
sc = 0
while a:
curr_cont, el, k = heappop(a)
sc += find_add(el, k)
print(sc)
``` | instruction | 0 | 104,646 | 9 | 209,292 |
Yes | output | 1 | 104,646 | 9 | 209,293 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are some rabbits in Singapore Zoo. To feed them, Zookeeper bought n carrots with lengths a_1, a_2, a_3, …, a_n. However, rabbits are very fertile and multiply very quickly. Zookeeper now has k rabbits and does not have enough carrots to feed all of them. To solve this problem, Zookeeper decided to cut the carrots into k pieces. For some reason, all resulting carrot lengths must be positive integers.
Big carrots are very difficult for rabbits to handle and eat, so the time needed to eat a carrot of size x is x^2.
Help Zookeeper split his carrots while minimizing the sum of time taken for rabbits to eat the carrots.
Input
The first line contains two integers n and k (1 ≤ n ≤ k ≤ 10^5): the initial number of carrots and the number of rabbits.
The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6): lengths of carrots.
It is guaranteed that the sum of a_i is at least k.
Output
Output one integer: the minimum sum of time taken for rabbits to eat carrots.
Examples
Input
3 6
5 3 1
Output
15
Input
1 4
19
Output
91
Note
For the first test, the optimal sizes of carrots are \{1,1,1,2,2,2\}. The time taken is 1^2+1^2+1^2+2^2+2^2+2^2=15
For the second test, the optimal sizes of carrots are \{4,5,5,5\}. The time taken is 4^2+5^2+5^2+5^2=91.
Submitted Solution:
```
import sys
import bisect as bi
import collections as cc
import heapq as hp
input=sys.stdin.readline
I=lambda:list(map(int,input().split()))
def ch(n,k):
temp=n//k
still=n%k
ans=(k-still)*((temp)**2)+still*((temp+1)**2)
return ans
n,k=I()
l=I()
ans=0
heap=[]
for i in range(n):
ans+=l[i]**2
hp.heappush(heap,(-ch(l[i],1)+ch(l[i],2),l[i],2))
for i in range(abs(n-k)):
now=hp.heappop(heap)
ans+=now[0]
x,y=now[1],now[2]
hp.heappush(heap,(-ch(x,y)+ch(x,y+1),x,y+1))
print(ans)
``` | instruction | 0 | 104,647 | 9 | 209,294 |
Yes | output | 1 | 104,647 | 9 | 209,295 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are some rabbits in Singapore Zoo. To feed them, Zookeeper bought n carrots with lengths a_1, a_2, a_3, …, a_n. However, rabbits are very fertile and multiply very quickly. Zookeeper now has k rabbits and does not have enough carrots to feed all of them. To solve this problem, Zookeeper decided to cut the carrots into k pieces. For some reason, all resulting carrot lengths must be positive integers.
Big carrots are very difficult for rabbits to handle and eat, so the time needed to eat a carrot of size x is x^2.
Help Zookeeper split his carrots while minimizing the sum of time taken for rabbits to eat the carrots.
Input
The first line contains two integers n and k (1 ≤ n ≤ k ≤ 10^5): the initial number of carrots and the number of rabbits.
The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6): lengths of carrots.
It is guaranteed that the sum of a_i is at least k.
Output
Output one integer: the minimum sum of time taken for rabbits to eat carrots.
Examples
Input
3 6
5 3 1
Output
15
Input
1 4
19
Output
91
Note
For the first test, the optimal sizes of carrots are \{1,1,1,2,2,2\}. The time taken is 1^2+1^2+1^2+2^2+2^2+2^2=15
For the second test, the optimal sizes of carrots are \{4,5,5,5\}. The time taken is 4^2+5^2+5^2+5^2=91.
Submitted Solution:
```
import sys
from heapq import *
sys.setrecursionlimit(10 ** 7)
input = sys.stdin.readline
f_inf = float('inf')
mod = 10 ** 9 + 7
def resolve():
def val(l, nums):
unit = l // nums
extra = l - unit * nums
return (nums - extra) * pow(unit, 2) + extra * pow(unit + 1, 2)
n, k = map(int, input().split())
A = list(map(int, input().split()))
pq = []
res = 0
for i in range(n):
res += pow(A[i], 2)
heappush(pq, (-val(A[i], 1) + val(A[i], 2), A[i], 2))
for _ in range(k - n):
temp = heappop(pq)
res += temp[0]
a, b = temp[1], temp[2]
heappush(pq, (-val(a, b) + val(a, b + 1), a, b + 1))
print(res)
if __name__ == '__main__':
resolve()
``` | instruction | 0 | 104,648 | 9 | 209,296 |
Yes | output | 1 | 104,648 | 9 | 209,297 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are some rabbits in Singapore Zoo. To feed them, Zookeeper bought n carrots with lengths a_1, a_2, a_3, …, a_n. However, rabbits are very fertile and multiply very quickly. Zookeeper now has k rabbits and does not have enough carrots to feed all of them. To solve this problem, Zookeeper decided to cut the carrots into k pieces. For some reason, all resulting carrot lengths must be positive integers.
Big carrots are very difficult for rabbits to handle and eat, so the time needed to eat a carrot of size x is x^2.
Help Zookeeper split his carrots while minimizing the sum of time taken for rabbits to eat the carrots.
Input
The first line contains two integers n and k (1 ≤ n ≤ k ≤ 10^5): the initial number of carrots and the number of rabbits.
The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6): lengths of carrots.
It is guaranteed that the sum of a_i is at least k.
Output
Output one integer: the minimum sum of time taken for rabbits to eat carrots.
Examples
Input
3 6
5 3 1
Output
15
Input
1 4
19
Output
91
Note
For the first test, the optimal sizes of carrots are \{1,1,1,2,2,2\}. The time taken is 1^2+1^2+1^2+2^2+2^2+2^2=15
For the second test, the optimal sizes of carrots are \{4,5,5,5\}. The time taken is 4^2+5^2+5^2+5^2=91.
Submitted Solution:
```
n = list(map(int, input().split()))
dem=0
output=0
trungbinh=0
sum=0
la=0
a = list(map(int, input().split()))
a.sort()
for i in range(n[0]):
sum+=a[i]
dem+=1
trungbinh=sum/n[1]
for i in range(n[0]):
for j in range(n[1]):
if (a[la]<=trungbinh):
output+=a[la]*a[la]
sum-=a[la]
la+=1
else:
break
trungbinh=sum/(n[1]-la)
for i in range(n[0]-la):
if(n[1]-la>0):
k=a[n[0]-i-1]/trungbinh
if (k>=k//1+0.5):
k=(k+1)//1
else:
k=k//1
la+=k
m=a[n[0]-i-1]%k
tb=a[n[0]-i-1]//k
output+=tb*tb*(k-m)+(tb+1)*(tb+1)*m
output=int(output)
print(output)
``` | instruction | 0 | 104,649 | 9 | 209,298 |
No | output | 1 | 104,649 | 9 | 209,299 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are some rabbits in Singapore Zoo. To feed them, Zookeeper bought n carrots with lengths a_1, a_2, a_3, …, a_n. However, rabbits are very fertile and multiply very quickly. Zookeeper now has k rabbits and does not have enough carrots to feed all of them. To solve this problem, Zookeeper decided to cut the carrots into k pieces. For some reason, all resulting carrot lengths must be positive integers.
Big carrots are very difficult for rabbits to handle and eat, so the time needed to eat a carrot of size x is x^2.
Help Zookeeper split his carrots while minimizing the sum of time taken for rabbits to eat the carrots.
Input
The first line contains two integers n and k (1 ≤ n ≤ k ≤ 10^5): the initial number of carrots and the number of rabbits.
The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6): lengths of carrots.
It is guaranteed that the sum of a_i is at least k.
Output
Output one integer: the minimum sum of time taken for rabbits to eat carrots.
Examples
Input
3 6
5 3 1
Output
15
Input
1 4
19
Output
91
Note
For the first test, the optimal sizes of carrots are \{1,1,1,2,2,2\}. The time taken is 1^2+1^2+1^2+2^2+2^2+2^2=15
For the second test, the optimal sizes of carrots are \{4,5,5,5\}. The time taken is 4^2+5^2+5^2+5^2=91.
Submitted Solution:
```
for _ in range(1):
# n=(int)(input())
# l=list(map(int,input().split()))
n,k=map(int,input().split())
l=list(map(int,input().split()))
while(len(l)<k):
l.sort(reverse=True)
l.append(l[0]//2)
if l[0]%2==0:
l.append(l[0]//2)
else:
l.append((l[0]//2)+1)
l.pop(0)
# print(*l)
sum=0
# print(l)
for i in range(len(l)):
sum+=(l[i]*l[i])
print(sum)
``` | instruction | 0 | 104,650 | 9 | 209,300 |
No | output | 1 | 104,650 | 9 | 209,301 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are some rabbits in Singapore Zoo. To feed them, Zookeeper bought n carrots with lengths a_1, a_2, a_3, …, a_n. However, rabbits are very fertile and multiply very quickly. Zookeeper now has k rabbits and does not have enough carrots to feed all of them. To solve this problem, Zookeeper decided to cut the carrots into k pieces. For some reason, all resulting carrot lengths must be positive integers.
Big carrots are very difficult for rabbits to handle and eat, so the time needed to eat a carrot of size x is x^2.
Help Zookeeper split his carrots while minimizing the sum of time taken for rabbits to eat the carrots.
Input
The first line contains two integers n and k (1 ≤ n ≤ k ≤ 10^5): the initial number of carrots and the number of rabbits.
The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6): lengths of carrots.
It is guaranteed that the sum of a_i is at least k.
Output
Output one integer: the minimum sum of time taken for rabbits to eat carrots.
Examples
Input
3 6
5 3 1
Output
15
Input
1 4
19
Output
91
Note
For the first test, the optimal sizes of carrots are \{1,1,1,2,2,2\}. The time taken is 1^2+1^2+1^2+2^2+2^2+2^2=15
For the second test, the optimal sizes of carrots are \{4,5,5,5\}. The time taken is 4^2+5^2+5^2+5^2=91.
Submitted Solution:
```
import collections
import heapq as h
n,k=map(int,input().split())
nums=list(map(int,input().split()))
for i in range(n):
nums[i]*=-1
while len(nums)<k:
val=h.heappop(nums)
val*=-1
v1=val//2
v2=val-v1
h.heappush(nums,-v1)
h.heappush(nums,-v2)
out=0
for i in nums:
out+=i**2
print(out)
``` | instruction | 0 | 104,651 | 9 | 209,302 |
No | output | 1 | 104,651 | 9 | 209,303 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are some rabbits in Singapore Zoo. To feed them, Zookeeper bought n carrots with lengths a_1, a_2, a_3, …, a_n. However, rabbits are very fertile and multiply very quickly. Zookeeper now has k rabbits and does not have enough carrots to feed all of them. To solve this problem, Zookeeper decided to cut the carrots into k pieces. For some reason, all resulting carrot lengths must be positive integers.
Big carrots are very difficult for rabbits to handle and eat, so the time needed to eat a carrot of size x is x^2.
Help Zookeeper split his carrots while minimizing the sum of time taken for rabbits to eat the carrots.
Input
The first line contains two integers n and k (1 ≤ n ≤ k ≤ 10^5): the initial number of carrots and the number of rabbits.
The next line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6): lengths of carrots.
It is guaranteed that the sum of a_i is at least k.
Output
Output one integer: the minimum sum of time taken for rabbits to eat carrots.
Examples
Input
3 6
5 3 1
Output
15
Input
1 4
19
Output
91
Note
For the first test, the optimal sizes of carrots are \{1,1,1,2,2,2\}. The time taken is 1^2+1^2+1^2+2^2+2^2+2^2=15
For the second test, the optimal sizes of carrots are \{4,5,5,5\}. The time taken is 4^2+5^2+5^2+5^2=91.
Submitted Solution:
```
from math import modf
n,k = map(int,input().split())
a = [*map(int,input().split())]
s = sum(a)
sumdec = 0
res1 = [0 for i in range(k)]
for i in range(k):
deca = list(i for i in modf(s/k))
res1[i] = int(deca[1])
sumdec += deca[0]
if(n==666):
print(s)
print(sumdec)
sumdec = round(sumdec)
p = sumdec//k
for i in range(k):
res1[i] += p
for i in range(sumdec%k):
res1[i] +=1
res = sum(i**2 for i in res1)
print(res)
#1109481843390
``` | instruction | 0 | 104,652 | 9 | 209,304 |
No | output | 1 | 104,652 | 9 | 209,305 |
Provide a correct Python 3 solution for this coding contest problem.
There are N children, numbered 1,2,\ldots,N. In the next K days, we will give them some cookies. In the i-th day, we will choose a_i children among the N with equal probability, and give one cookie to each child chosen. (We make these K choices independently.)
Let us define the happiness of the children as c_1 \times c_2 \times \ldots \times c_N, where c_i is the number of cookies received by Child i in the K days. Find the expected happiness multiplied by \binom{N}{a_1} \times \binom{N}{a_2} \times \ldots \times \binom{N}{a_K} (we can show that this value is an integer), modulo (10^{9}+7).
Constraints
* 1 \leq N \leq 1000
* 1 \leq K \leq 20
* 1 \leq a_i \leq N
Input
Input is given from Standard Input in the following format:
N K
a_1 a_2 \ldots a_K
Output
Print the answer.
Examples
Input
3 2
3 2
Output
12
Input
856 16
399 263 665 432 206 61 784 548 422 313 848 478 827 26 398 63
Output
337587117 | instruction | 0 | 105,059 | 9 | 210,118 |
"Correct Solution:
```
MOD = 10**9 + 7
list_size = 1001
f_list = [1] * list_size
f_r_list = [1] * list_size
for i in range(list_size-1):
f_list[i+1] = (f_list[i] * (i+1)) % MOD
f_r_list[-1] = pow(f_list[-1], MOD - 2, MOD)
for i in range(list_size-2, -1, -1):
f_r_list[i] = (f_r_list[i+1] * (i+1)) % MOD
def comb(n, r):
if n < r or r < 0:
return 0
elif n == 0 or r == 0 or n == r:
return 1
else:
return (f_list[n] * f_r_list[n-r] * f_r_list[r]) % MOD
n, k = map(int, input().split())
a = list(map(int, input().split()))
dp = [[0 for _ in range(n+1)] for _ in range(k+1)]
dp[0][0] = f_list[n]
for i in range(1, k+1):
for j in range(n+1):
for x in range(j+1):
dp[i][j] += dp[i-1][j-x] * comb(n-x, a[i-1]-x) * f_r_list[x]
dp[i][j] %= MOD
print(dp[k][n])
``` | output | 1 | 105,059 | 9 | 210,119 |
Provide a correct Python 3 solution for this coding contest problem.
There are N children, numbered 1,2,\ldots,N. In the next K days, we will give them some cookies. In the i-th day, we will choose a_i children among the N with equal probability, and give one cookie to each child chosen. (We make these K choices independently.)
Let us define the happiness of the children as c_1 \times c_2 \times \ldots \times c_N, where c_i is the number of cookies received by Child i in the K days. Find the expected happiness multiplied by \binom{N}{a_1} \times \binom{N}{a_2} \times \ldots \times \binom{N}{a_K} (we can show that this value is an integer), modulo (10^{9}+7).
Constraints
* 1 \leq N \leq 1000
* 1 \leq K \leq 20
* 1 \leq a_i \leq N
Input
Input is given from Standard Input in the following format:
N K
a_1 a_2 \ldots a_K
Output
Print the answer.
Examples
Input
3 2
3 2
Output
12
Input
856 16
399 263 665 432 206 61 784 548 422 313 848 478 827 26 398 63
Output
337587117 | instruction | 0 | 105,060 | 9 | 210,120 |
"Correct Solution:
```
import sys
readline = sys.stdin.readline
MOD = 10**9+7
def frac(limit):
frac = [1]*limit
for i in range(2,limit):
frac[i] = i * frac[i-1]%MOD
fraci = [None]*limit
fraci[-1] = pow(frac[-1], MOD -2, MOD)
for i in range(-2, -limit-1, -1):
fraci[i] = fraci[i+1] * (limit + i + 1) % MOD
return frac, fraci
frac, fraci = frac(1398)
def comb(a, b):
if not a >= b >= 0:
return 0
return frac[a]*fraci[b]*fraci[a-b]%MOD
N, K = map(int, readline().split())
A = list(map(int, readline().split()))
dp = [0]*(N+1)
dp[0] = 1
for i in range(K):
a = A[i]
dp2 = [0]*(N+1)
for j in range(N+1):
d = dp[j]
if not d:
continue
for k in range(min(a, N-j)+1):
dp2[j+k] = (dp2[j+k] + d*comb(N-j, k)*comb(N-k, a-k)) %MOD
dp = dp2[:]
print(dp[N])
``` | output | 1 | 105,060 | 9 | 210,121 |
Provide a correct Python 3 solution for this coding contest problem.
There are N children, numbered 1,2,\ldots,N. In the next K days, we will give them some cookies. In the i-th day, we will choose a_i children among the N with equal probability, and give one cookie to each child chosen. (We make these K choices independently.)
Let us define the happiness of the children as c_1 \times c_2 \times \ldots \times c_N, where c_i is the number of cookies received by Child i in the K days. Find the expected happiness multiplied by \binom{N}{a_1} \times \binom{N}{a_2} \times \ldots \times \binom{N}{a_K} (we can show that this value is an integer), modulo (10^{9}+7).
Constraints
* 1 \leq N \leq 1000
* 1 \leq K \leq 20
* 1 \leq a_i \leq N
Input
Input is given from Standard Input in the following format:
N K
a_1 a_2 \ldots a_K
Output
Print the answer.
Examples
Input
3 2
3 2
Output
12
Input
856 16
399 263 665 432 206 61 784 548 422 313 848 478 827 26 398 63
Output
337587117 | instruction | 0 | 105,061 | 9 | 210,122 |
"Correct Solution:
```
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop, heapify
import math
import bisect
import random
from itertools import permutations, accumulate, combinations, product
import sys
from pprint import pprint
from copy import deepcopy
import string
from bisect import bisect_left, bisect_right
from math import factorial, ceil, floor
from operator import mul
from functools import reduce
from pprint import pprint
sys.setrecursionlimit(2147483647)
INF = 10 ** 20
def LI(): return list(map(int, sys.stdin.readline().split()))
def I(): return int(sys.stdin.readline())
def LS(): return sys.stdin.buffer.readline().rstrip().decode('utf-8').split()
def S(): return sys.stdin.buffer.readline().rstrip().decode('utf-8')
def IR(n): return [I() for i in range(n)]
def LIR(n): return [LI() for i in range(n)]
def SR(n): return [S() for i in range(n)]
def LSR(n): return [LS() for i in range(n)]
def SRL(n): return [list(S()) for i in range(n)]
def MSRL(n): return [[int(j) for j in list(S())] for i in range(n)]
mod = 1000000007
n, k = LI()
fac = [1] * (n + 1)
inv = [1] * (n + 1)
for j in range(1, n + 1):
fac[j] = fac[j-1] * j % mod
inv[n] = pow(fac[n], mod-2, mod)
for j in range(n-1, -1, -1):
inv[j] = inv[j+1] * (j+1) % mod
def comb(n, r):
if r > n or n < 0 or r < 0:
return 0
return fac[n] * inv[n - r] * inv[r] % mod
A = LI()
dp = [[0] * (n + 1) for _ in range(k + 1)]
dp[0][0] = 1
for i in range(k):
for j in range(n + 1):
for l in range(min(n - j, A[i]) + 1):
dp[i + 1][j + l] = (dp[i + 1][j + l] + dp[i][j] * comb(n - j, l) % mod
* comb(n - l, A[i] - l)) % mod
print(dp[k][n])
``` | output | 1 | 105,061 | 9 | 210,123 |
Provide a correct Python 3 solution for this coding contest problem.
There are N children, numbered 1,2,\ldots,N. In the next K days, we will give them some cookies. In the i-th day, we will choose a_i children among the N with equal probability, and give one cookie to each child chosen. (We make these K choices independently.)
Let us define the happiness of the children as c_1 \times c_2 \times \ldots \times c_N, where c_i is the number of cookies received by Child i in the K days. Find the expected happiness multiplied by \binom{N}{a_1} \times \binom{N}{a_2} \times \ldots \times \binom{N}{a_K} (we can show that this value is an integer), modulo (10^{9}+7).
Constraints
* 1 \leq N \leq 1000
* 1 \leq K \leq 20
* 1 \leq a_i \leq N
Input
Input is given from Standard Input in the following format:
N K
a_1 a_2 \ldots a_K
Output
Print the answer.
Examples
Input
3 2
3 2
Output
12
Input
856 16
399 263 665 432 206 61 784 548 422 313 848 478 827 26 398 63
Output
337587117 | instruction | 0 | 105,062 | 9 | 210,124 |
"Correct Solution:
```
SIZE=1001; MOD=10**9+7 #998244353 #ここを変更する
SIZE += 1
inv = [0]*SIZE # inv[j] = j^{-1} mod MOD
fac = [0]*SIZE # fac[j] = j! mod MOD
finv = [0]*SIZE # finv[j] = (j!)^{-1} mod MOD
inv[1] = 1
fac[0] = fac[1] = 1
finv[0] = finv[1] = 1
for i in range(2,SIZE):
inv[i] = MOD - (MOD//i)*inv[MOD%i]%MOD
fac[i] = fac[i-1]*i%MOD
finv[i]= finv[i-1]*inv[i]%MOD
def choose(n,r): # nCk mod MOD の計算
if 0 <= r <= n:
return (fac[n]*finv[r]%MOD)*finv[n-r]%MOD
else:
return 0
# coding: utf-8
# Your code here!
import sys
sys.setrecursionlimit(10**6)
readline = sys.stdin.readline
read = sys.stdin.read
n,k = [int(i) for i in readline().split()]
a = [int(i) for i in readline().split()]
def product(p,q):
res = [0]*(n+1)
for i in range(n+1):
for j in range(n-i+1):
res[i+j] += p[i]*q[j]
res[i+j] %= MOD
return res
ans = [1]+[0]*n
for ai in a:
q = [0]*(n+1)
for j in range(n+1):
q[j] = choose(n-j,n-ai)*finv[j]%MOD
ans = product(ans,q)
res = ans[n]
for i in range(1,n+1):
res *= i
res %= MOD
print(res)
``` | output | 1 | 105,062 | 9 | 210,125 |
Provide a correct Python 3 solution for this coding contest problem.
There are N children, numbered 1,2,\ldots,N. In the next K days, we will give them some cookies. In the i-th day, we will choose a_i children among the N with equal probability, and give one cookie to each child chosen. (We make these K choices independently.)
Let us define the happiness of the children as c_1 \times c_2 \times \ldots \times c_N, where c_i is the number of cookies received by Child i in the K days. Find the expected happiness multiplied by \binom{N}{a_1} \times \binom{N}{a_2} \times \ldots \times \binom{N}{a_K} (we can show that this value is an integer), modulo (10^{9}+7).
Constraints
* 1 \leq N \leq 1000
* 1 \leq K \leq 20
* 1 \leq a_i \leq N
Input
Input is given from Standard Input in the following format:
N K
a_1 a_2 \ldots a_K
Output
Print the answer.
Examples
Input
3 2
3 2
Output
12
Input
856 16
399 263 665 432 206 61 784 548 422 313 848 478 827 26 398 63
Output
337587117 | instruction | 0 | 105,063 | 9 | 210,126 |
"Correct Solution:
```
N,K = map(int,input().split())
A = list(map(int,input().split()))
s = sum(A)
if s < N:
print(0)
exit()
MOD = 10**9+7
MAXN = N+5
fac = [1,1] + [0]*MAXN
finv = [1,1] + [0]*MAXN
inv = [0,1] + [0]*MAXN
for i in range(2,MAXN+2):
fac[i] = fac[i-1] * i % MOD
inv[i] = -inv[MOD%i] * (MOD // i) % MOD
finv[i] = finv[i-1] * inv[i] % MOD
def ncr(n,r):
if n < r: return 0
if n < 0 or r < 0: return 0
return fac[n] * (finv[r] * finv[n-r] % MOD) % MOD
dp = [[0]*(N+1) for i in range(K+1)]
dp[0][0] = 1
for i,a in enumerate(A):
for j in range(N+1):
for k in range(j,N+1):
if k-j > a: break
dp[i+1][k] += dp[i][j] * ncr(N-j, k-j) * ncr(N-(k-j), a-(k-j))
dp[i+1][k] %= MOD
print(dp[-1][-1])
``` | output | 1 | 105,063 | 9 | 210,127 |
Provide a correct Python 3 solution for this coding contest problem.
There are N children, numbered 1,2,\ldots,N. In the next K days, we will give them some cookies. In the i-th day, we will choose a_i children among the N with equal probability, and give one cookie to each child chosen. (We make these K choices independently.)
Let us define the happiness of the children as c_1 \times c_2 \times \ldots \times c_N, where c_i is the number of cookies received by Child i in the K days. Find the expected happiness multiplied by \binom{N}{a_1} \times \binom{N}{a_2} \times \ldots \times \binom{N}{a_K} (we can show that this value is an integer), modulo (10^{9}+7).
Constraints
* 1 \leq N \leq 1000
* 1 \leq K \leq 20
* 1 \leq a_i \leq N
Input
Input is given from Standard Input in the following format:
N K
a_1 a_2 \ldots a_K
Output
Print the answer.
Examples
Input
3 2
3 2
Output
12
Input
856 16
399 263 665 432 206 61 784 548 422 313 848 478 827 26 398 63
Output
337587117 | instruction | 0 | 105,064 | 9 | 210,128 |
"Correct Solution:
```
P = 10**9+7
fa = [1]
for i in range(1, 1010): fa.append(fa[-1] * i % P)
fainv = [pow(fa[-1], P-2, P)]
for i in range(1, 1010)[::-1]: fainv.append(fainv[-1] * i % P)
fainv = fainv[::-1]
k = 70
kk = 1<<k
nu = lambda L: int("".join([bin(kk+a)[-k:] for a in L[::-1]]), 2)
st = lambda n: bin(n)[2:] + "0"
li = lambda s: [int(a, 2) % P if len(a) else 0 for a in [s[-(i+1)*k-1:-i*k-1] for i in range(N+1)]]
N, K = map(int, input().split())
A = [int(a) for a in input().split()]
X = [fa[N]] + [0] * N
for a in A:
X = li(st(nu(X) * nu([fainv[i] * fa[N-i] * fainv[a-i] * fainv[N-a] % P if i <= a else 0 for i in range(N+1)])))
print(X[-1])
``` | output | 1 | 105,064 | 9 | 210,129 |
Provide a correct Python 3 solution for this coding contest problem.
There are N children, numbered 1,2,\ldots,N. In the next K days, we will give them some cookies. In the i-th day, we will choose a_i children among the N with equal probability, and give one cookie to each child chosen. (We make these K choices independently.)
Let us define the happiness of the children as c_1 \times c_2 \times \ldots \times c_N, where c_i is the number of cookies received by Child i in the K days. Find the expected happiness multiplied by \binom{N}{a_1} \times \binom{N}{a_2} \times \ldots \times \binom{N}{a_K} (we can show that this value is an integer), modulo (10^{9}+7).
Constraints
* 1 \leq N \leq 1000
* 1 \leq K \leq 20
* 1 \leq a_i \leq N
Input
Input is given from Standard Input in the following format:
N K
a_1 a_2 \ldots a_K
Output
Print the answer.
Examples
Input
3 2
3 2
Output
12
Input
856 16
399 263 665 432 206 61 784 548 422 313 848 478 827 26 398 63
Output
337587117 | instruction | 0 | 105,065 | 9 | 210,130 |
"Correct Solution:
```
# http://drken1215.hatenablog.com/entry/2020/01/12/011700
# 本質的な速度改善の方法は分からない
# dpを二次元配列にするとTLEするので、
# 配列2本で処理した
# 修正点:
# [1]
# enumerate(a) -> a
# 不要なenumerateを消した
# [2]
# counted in range(n + 1) -> enumerate(dp)
# enumerateの方がわずかに速いはず
# [3]
# cmbのif文を消した
def main():
MOD = 10 ** 9 + 7
N, K = map(int, input().split())
*a, = map(int, input().split())
# -+-+-+-prepare combination-+-+-+-
# -+-+-+-START-+-+-+-
fact = [0] * (N + 1)
fact[0] = 1
fact[1] = 1
for x in range(2, N + 1):
fact[x] = x * fact[x - 1] % MOD
invs = [1] * (N + 1)
invs[N] = pow(fact[N], MOD - 2, MOD)
for x in range(N - 1, 0, -1):
invs[x] = invs[x + 1] * (x + 1) % MOD
def cmb(n, r):
return (((invs[r] * invs[n - r]) % MOD) * fact[n]) % MOD
# -+-+-+-END-+-+-+-
dp = [0] * (N + 1)
dp[0] = 1
for ai in a:
ndp = [0] * (N + 1)
for counted, dp_counted in enumerate(dp):
for to_count in range(min(N - counted, ai) + 1):
ndp[counted + to_count] = (
ndp[counted + to_count]
+ dp_counted
* cmb(N - counted, to_count)
* cmb(N - to_count, ai - to_count)
) % MOD
dp = ndp
print(dp[N])
if __name__ == '__main__':
main()
``` | output | 1 | 105,065 | 9 | 210,131 |
Provide a correct Python 3 solution for this coding contest problem.
There are N children, numbered 1,2,\ldots,N. In the next K days, we will give them some cookies. In the i-th day, we will choose a_i children among the N with equal probability, and give one cookie to each child chosen. (We make these K choices independently.)
Let us define the happiness of the children as c_1 \times c_2 \times \ldots \times c_N, where c_i is the number of cookies received by Child i in the K days. Find the expected happiness multiplied by \binom{N}{a_1} \times \binom{N}{a_2} \times \ldots \times \binom{N}{a_K} (we can show that this value is an integer), modulo (10^{9}+7).
Constraints
* 1 \leq N \leq 1000
* 1 \leq K \leq 20
* 1 \leq a_i \leq N
Input
Input is given from Standard Input in the following format:
N K
a_1 a_2 \ldots a_K
Output
Print the answer.
Examples
Input
3 2
3 2
Output
12
Input
856 16
399 263 665 432 206 61 784 548 422 313 848 478 827 26 398 63
Output
337587117 | instruction | 0 | 105,066 | 9 | 210,132 |
"Correct Solution:
```
M=10**9+7;n,k,*A=map(int,open(0).read().split());C=[[0]*(n+1)for _ in"_"*(n+1)]
for i in range(n+1):
for j in range(i+1):C[i][j]=(1,C[i-1][j-1]+C[i-1][j])[j!=0]%M
F=[[0]*(n+1)for _ in"_"*(k+1)];F[0][0]=1
for i in range(k):
for j in range(n+1):
for m in range(j+1):F[i+1][j]+=C[j][m]*F[i][m]*C[n-j+m][A[i]-j+m]
print(F[k][n]%M)
``` | output | 1 | 105,066 | 9 | 210,133 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N children, numbered 1,2,\ldots,N. In the next K days, we will give them some cookies. In the i-th day, we will choose a_i children among the N with equal probability, and give one cookie to each child chosen. (We make these K choices independently.)
Let us define the happiness of the children as c_1 \times c_2 \times \ldots \times c_N, where c_i is the number of cookies received by Child i in the K days. Find the expected happiness multiplied by \binom{N}{a_1} \times \binom{N}{a_2} \times \ldots \times \binom{N}{a_K} (we can show that this value is an integer), modulo (10^{9}+7).
Constraints
* 1 \leq N \leq 1000
* 1 \leq K \leq 20
* 1 \leq a_i \leq N
Input
Input is given from Standard Input in the following format:
N K
a_1 a_2 \ldots a_K
Output
Print the answer.
Examples
Input
3 2
3 2
Output
12
Input
856 16
399 263 665 432 206 61 784 548 422 313 848 478 827 26 398 63
Output
337587117
Submitted Solution:
```
nnnnn = 1010
P = 10 ** 9 + 7
fa = [1]
for i in range(1, nnnnn):
fa.append(fa[-1] * i % P)
fainv = [pow(fa[-1], P-2, P)]
for i in range(1, nnnnn)[::-1]:
fainv.append(fainv[-1] * i % P)
fainv = fainv[::-1]
N, K = map(int, input().split())
A = [int(a) for a in input().split()]
def calc(k):
return [fainv[i] * fa[N-i] * fainv[k-i] * fainv[N-k] % P if i <= k else 0 for i in range(N+1)]
def conv(a, b):
L = [0] * (N+1)
for i in range(N+1):
for j in range(N+1):
if i+j <= N: L[i+j] += a[i] * b[j]
return [l % P for l in L]
X = [1] + [0] * N
for a in A:
X = conv(X, calc(a))
print(X[-1] * fa[N] % P)
``` | instruction | 0 | 105,067 | 9 | 210,134 |
Yes | output | 1 | 105,067 | 9 | 210,135 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N children, numbered 1,2,\ldots,N. In the next K days, we will give them some cookies. In the i-th day, we will choose a_i children among the N with equal probability, and give one cookie to each child chosen. (We make these K choices independently.)
Let us define the happiness of the children as c_1 \times c_2 \times \ldots \times c_N, where c_i is the number of cookies received by Child i in the K days. Find the expected happiness multiplied by \binom{N}{a_1} \times \binom{N}{a_2} \times \ldots \times \binom{N}{a_K} (we can show that this value is an integer), modulo (10^{9}+7).
Constraints
* 1 \leq N \leq 1000
* 1 \leq K \leq 20
* 1 \leq a_i \leq N
Input
Input is given from Standard Input in the following format:
N K
a_1 a_2 \ldots a_K
Output
Print the answer.
Examples
Input
3 2
3 2
Output
12
Input
856 16
399 263 665 432 206 61 784 548 422 313 848 478 827 26 398 63
Output
337587117
Submitted Solution:
```
def inv(a):
m=10**9+7
b=m
(x, lastx) = (0, 1)
(y, lasty) = (1, 0)
while b != 0:
q = a // b
(a, b) = (b, a % b)
(x, lastx) = (lastx - q * x, x)
(y, lasty) = (lasty - q * y, y)
return lastx % m
def main():
MOD=10**9+7
n,k=map(int,input().split())
a=list(map(int,input().split()))
frac=[1]
for i in range(1,n+1):
frac.append(frac[-1]*i%MOD)
fracinv=[inv(frac[i]) for i in range(n+1)]
dp=[[0]*(n+1) for _ in range(k)]
for x in range(a[0]+1):
dp[0][x]=frac[n]*frac[n-x]*fracinv[a[0]-x]%MOD*fracinv[n-a[0]]*fracinv[x]%MOD
for i in range(k-1):
ai=a[i+1]
for j in range(n+1):
for x in range(ai+1):
p=j+x
if p>n: break
tmp=dp[i][j]*frac[n-x]%MOD*fracinv[ai-x]*fracinv[n-ai]%MOD*fracinv[x]
dp[i+1][p]=(dp[i+1][p]+tmp)%MOD
print(dp[k-1][n])
if __name__ == "__main__":
main()
``` | instruction | 0 | 105,068 | 9 | 210,136 |
Yes | output | 1 | 105,068 | 9 | 210,137 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N children, numbered 1,2,\ldots,N. In the next K days, we will give them some cookies. In the i-th day, we will choose a_i children among the N with equal probability, and give one cookie to each child chosen. (We make these K choices independently.)
Let us define the happiness of the children as c_1 \times c_2 \times \ldots \times c_N, where c_i is the number of cookies received by Child i in the K days. Find the expected happiness multiplied by \binom{N}{a_1} \times \binom{N}{a_2} \times \ldots \times \binom{N}{a_K} (we can show that this value is an integer), modulo (10^{9}+7).
Constraints
* 1 \leq N \leq 1000
* 1 \leq K \leq 20
* 1 \leq a_i \leq N
Input
Input is given from Standard Input in the following format:
N K
a_1 a_2 \ldots a_K
Output
Print the answer.
Examples
Input
3 2
3 2
Output
12
Input
856 16
399 263 665 432 206 61 784 548 422 313 848 478 827 26 398 63
Output
337587117
Submitted Solution:
```
import sys
sys.setrecursionlimit(10 ** 6)
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def main():
def com(com_n, com_r):
return fac[com_n] * inv[com_r] * inv[com_n - com_r] % md
n, k = MI()
aa = LI()
# combinationの準備
md = 10 ** 9 + 7
n_max = n + 3
fac = [1] * (n_max + 1)
inv = [1] * (n_max + 1)
for i in range(2, n_max + 1): fac[i] = fac[i - 1] * i % md
inv[n_max] = pow(fac[n_max], md - 2, md)
for i in range(n_max - 1, 1, -1): inv[i] = inv[i + 1] * (i + 1) % md
dp = [0] * (n + 1)
dp[0] = 1
for a in aa:
for j in range(n, -1, -1):
pre = dp[j]
if pre == 0: continue
dp[j] = 0
for dj in range(a + 1):
nj = j + dj
if nj > n or a - dj < 0 or n - j < dj: break
dp[nj] += pre * com(n - j, dj) * com(n - dj, a - dj)
dp[nj] %= md
# print(dp)
print(dp[n])
main()
``` | instruction | 0 | 105,069 | 9 | 210,138 |
Yes | output | 1 | 105,069 | 9 | 210,139 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N children, numbered 1,2,\ldots,N. In the next K days, we will give them some cookies. In the i-th day, we will choose a_i children among the N with equal probability, and give one cookie to each child chosen. (We make these K choices independently.)
Let us define the happiness of the children as c_1 \times c_2 \times \ldots \times c_N, where c_i is the number of cookies received by Child i in the K days. Find the expected happiness multiplied by \binom{N}{a_1} \times \binom{N}{a_2} \times \ldots \times \binom{N}{a_K} (we can show that this value is an integer), modulo (10^{9}+7).
Constraints
* 1 \leq N \leq 1000
* 1 \leq K \leq 20
* 1 \leq a_i \leq N
Input
Input is given from Standard Input in the following format:
N K
a_1 a_2 \ldots a_K
Output
Print the answer.
Examples
Input
3 2
3 2
Output
12
Input
856 16
399 263 665 432 206 61 784 548 422 313 848 478 827 26 398 63
Output
337587117
Submitted Solution:
```
import sys
input = sys.stdin.readline
N,K=map(int,input().split())
A=list(map(int,input().split()))
mod=10**9+7
Combi=[[] for i in range(N+1)]
Combi[0]=[1,0]
for i in range(1,N+1):
Combi[i].append(1)
for j in range(i):
Combi[i].append((Combi[i-1][j]+Combi[i-1][j+1])%mod)
Combi[i].append(0)
DP=[0]*(N+1)
DP[0]=1
for k in range(K):
use=A[k]
NDP=[0]*(N+1)
for i in range(N+1):
ANS=0
for j in range(i+1):
if use-(i-j)>=0:
ANS=(ANS+DP[j]*Combi[N-j][i-j]*Combi[N-(i-j)][use-(i-j)])%mod
NDP[i]=ANS
#print(DP)
DP=NDP
print(DP[N])
``` | instruction | 0 | 105,070 | 9 | 210,140 |
Yes | output | 1 | 105,070 | 9 | 210,141 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N children, numbered 1,2,\ldots,N. In the next K days, we will give them some cookies. In the i-th day, we will choose a_i children among the N with equal probability, and give one cookie to each child chosen. (We make these K choices independently.)
Let us define the happiness of the children as c_1 \times c_2 \times \ldots \times c_N, where c_i is the number of cookies received by Child i in the K days. Find the expected happiness multiplied by \binom{N}{a_1} \times \binom{N}{a_2} \times \ldots \times \binom{N}{a_K} (we can show that this value is an integer), modulo (10^{9}+7).
Constraints
* 1 \leq N \leq 1000
* 1 \leq K \leq 20
* 1 \leq a_i \leq N
Input
Input is given from Standard Input in the following format:
N K
a_1 a_2 \ldots a_K
Output
Print the answer.
Examples
Input
3 2
3 2
Output
12
Input
856 16
399 263 665 432 206 61 784 548 422 313 848 478 827 26 398 63
Output
337587117
Submitted Solution:
```
M=10**9+7;n,k,*A=map(int,open(0).read().split());B=[1];I=[1]*(n+1)
for i in range(1,n+1):B+=[B[-1]*i%M]
for i in range(n+1,0,-1):I[i-1]=pow(B[n],M-2,M)%M if i>n else I[i]*i%M
C=lambda n,k:(B[n]*I[k]*I[n-k],0)[k<0];F=[[0]*(n+1)for _ in"_"*(k+1)];F[0][0]=1
for i in range(k):
for j in range(n+1):
for m in range(j+1):F[i+1][j]+=C(j,m)*F[i][m]*C(n-j+m,A[i]-j+m)%M
print(F[k][n]%M)
``` | instruction | 0 | 105,071 | 9 | 210,142 |
No | output | 1 | 105,071 | 9 | 210,143 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N children, numbered 1,2,\ldots,N. In the next K days, we will give them some cookies. In the i-th day, we will choose a_i children among the N with equal probability, and give one cookie to each child chosen. (We make these K choices independently.)
Let us define the happiness of the children as c_1 \times c_2 \times \ldots \times c_N, where c_i is the number of cookies received by Child i in the K days. Find the expected happiness multiplied by \binom{N}{a_1} \times \binom{N}{a_2} \times \ldots \times \binom{N}{a_K} (we can show that this value is an integer), modulo (10^{9}+7).
Constraints
* 1 \leq N \leq 1000
* 1 \leq K \leq 20
* 1 \leq a_i \leq N
Input
Input is given from Standard Input in the following format:
N K
a_1 a_2 \ldots a_K
Output
Print the answer.
Examples
Input
3 2
3 2
Output
12
Input
856 16
399 263 665 432 206 61 784 548 422 313 848 478 827 26 398 63
Output
337587117
Submitted Solution:
```
import numpy as np
from math import factorial
D = 10**9+7
N, K = [int(n) for n in input().split()]
C = []
CO = {}
E = []
for i in range(1, N+1):
C.append(i)
CO[i] = 0
a = [int(n) for n in input().split()]
for i in range(K):
R = np.random.choice(C, a[i], replace=False)
E.append(factorial(N)//factorial(a[i]) //factorial(N-a[i]))
for r in R:
CO[r] += 1
h = 1
for v in CO.values():
h *= v
ee = 1
for e in E:
ee *= e
print(h*ee%D)
``` | instruction | 0 | 105,072 | 9 | 210,144 |
No | output | 1 | 105,072 | 9 | 210,145 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N children, numbered 1,2,\ldots,N. In the next K days, we will give them some cookies. In the i-th day, we will choose a_i children among the N with equal probability, and give one cookie to each child chosen. (We make these K choices independently.)
Let us define the happiness of the children as c_1 \times c_2 \times \ldots \times c_N, where c_i is the number of cookies received by Child i in the K days. Find the expected happiness multiplied by \binom{N}{a_1} \times \binom{N}{a_2} \times \ldots \times \binom{N}{a_K} (we can show that this value is an integer), modulo (10^{9}+7).
Constraints
* 1 \leq N \leq 1000
* 1 \leq K \leq 20
* 1 \leq a_i \leq N
Input
Input is given from Standard Input in the following format:
N K
a_1 a_2 \ldots a_K
Output
Print the answer.
Examples
Input
3 2
3 2
Output
12
Input
856 16
399 263 665 432 206 61 784 548 422 313 848 478 827 26 398 63
Output
337587117
Submitted Solution:
```
n,k = map(int,input().split())
a = tuple(map(int,input().split()))
mod = 10**9+7
rng = 1001
fctr = [1]
finv = [1]
for i in range(1,rng):
fctr.append(fctr[-1]*i%mod)
for i in range(1,rng):
finv.append(pow(fctr[i],mod-2,mod))
def cmb(n,k):
if n<0 or k<0:
return 0
else:
return fctr[n]*finv[n-k]*finv[k]%mod
dp = [[0 for i in range(n+1)] for j in range(k+1)]
dp[0][0] = 1
ans = 0
for i in range(1,k+1):
x = a[i-1]
for j in range(n+1):
for t in range(min(n-j,x)+1):
dp[i][j+t] = (dp[i][j+t]+dp[i-1][j]*cmb(n-j,t)*cmb(n-t,x-t))%mod
print(dp[k][n])
``` | instruction | 0 | 105,073 | 9 | 210,146 |
No | output | 1 | 105,073 | 9 | 210,147 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N children, numbered 1,2,\ldots,N. In the next K days, we will give them some cookies. In the i-th day, we will choose a_i children among the N with equal probability, and give one cookie to each child chosen. (We make these K choices independently.)
Let us define the happiness of the children as c_1 \times c_2 \times \ldots \times c_N, where c_i is the number of cookies received by Child i in the K days. Find the expected happiness multiplied by \binom{N}{a_1} \times \binom{N}{a_2} \times \ldots \times \binom{N}{a_K} (we can show that this value is an integer), modulo (10^{9}+7).
Constraints
* 1 \leq N \leq 1000
* 1 \leq K \leq 20
* 1 \leq a_i \leq N
Input
Input is given from Standard Input in the following format:
N K
a_1 a_2 \ldots a_K
Output
Print the answer.
Examples
Input
3 2
3 2
Output
12
Input
856 16
399 263 665 432 206 61 784 548 422 313 848 478 827 26 398 63
Output
337587117
Submitted Solution:
```
import sys
input = sys.stdin.readline
N,K=map(int,input().split())
A=list(map(int,input().split()))
mod=10**9+7
FACT=[1]
for i in range(1,2*10**5+1):
FACT.append(FACT[-1]*i%mod)
FACT_INV=[pow(FACT[-1],mod-2,mod)]
for i in range(2*10**5,0,-1):
FACT_INV.append(FACT_INV[-1]*i%mod)
FACT_INV.reverse()
def Combi(a,b):
if 0<=b<=a:
return FACT[a]*FACT_INV[b]*FACT_INV[a-b]%mod
else:
return 0
DP=[0]*(N+1)
DP[0]=1
for k in range(K):
use=A[k]
NDP=[0]*(N+1)
for i in range(N+1):
ANS=0
for j in range(N+1):
ANS=(ANS+DP[j]*Combi(N-j,i-j)*Combi(N-(i-j),use-(i-j)))%mod
NDP[i]=ANS
DP=NDP
print(DP[N])
``` | instruction | 0 | 105,074 | 9 | 210,148 |
No | output | 1 | 105,074 | 9 | 210,149 |
Provide a correct Python 3 solution for this coding contest problem.
"Teishi-zushi", a Japanese restaurant, is a plain restaurant with only one round counter. The outer circumference of the counter is C meters. Customers cannot go inside the counter.
Nakahashi entered Teishi-zushi, and he was guided to the counter. Now, there are N pieces of sushi (vinegared rice with seafood and so on) on the counter. The distance measured clockwise from the point where Nakahashi is standing to the point where the i-th sushi is placed, is x_i meters. Also, the i-th sushi has a nutritive value of v_i kilocalories.
Nakahashi can freely walk around the circumference of the counter. When he reach a point where a sushi is placed, he can eat that sushi and take in its nutrition (naturally, the sushi disappears). However, while walking, he consumes 1 kilocalories per meter.
Whenever he is satisfied, he can leave the restaurant from any place (he does not have to return to the initial place). On balance, at most how much nutrition can he take in before he leaves? That is, what is the maximum possible value of the total nutrition taken in minus the total energy consumed? Assume that there are no other customers, and no new sushi will be added to the counter. Also, since Nakahashi has plenty of nutrition in his body, assume that no matter how much he walks and consumes energy, he never dies from hunger.
Constraints
* 1 ≤ N ≤ 10^5
* 2 ≤ C ≤ 10^{14}
* 1 ≤ x_1 < x_2 < ... < x_N < C
* 1 ≤ v_i ≤ 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N C
x_1 v_1
x_2 v_2
:
x_N v_N
Output
If Nakahashi can take in at most c kilocalories on balance before he leaves the restaurant, print c.
Examples
Input
3 20
2 80
9 120
16 1
Output
191
Input
3 20
2 80
9 1
16 120
Output
192
Input
1 100000000000000
50000000000000 1
Output
0
Input
15 10000000000
400000000 1000000000
800000000 1000000000
1900000000 1000000000
2400000000 1000000000
2900000000 1000000000
3300000000 1000000000
3700000000 1000000000
3800000000 1000000000
4000000000 1000000000
4100000000 1000000000
5200000000 1000000000
6600000000 1000000000
8000000000 1000000000
9300000000 1000000000
9700000000 1000000000
Output
6500000000 | instruction | 0 | 105,100 | 9 | 210,200 |
"Correct Solution:
```
N, C = map(int, input().split())
xv = [[int(i) for i in input().split()] for _ in range(N)]
av = 0
fa1 = 0
bv = 0
fb1 = 0
fa = []
fb = []
ga = []
gb = []
for i in range(N):
ax = xv[i][0]
av += xv[i][1]
bx = C - xv[-(i+1)][0]
bv += xv[-(i+1)][1]
fa1 = max(fa1, av-ax)
fa.append(fa1)
ga.append(av - 2*ax)
fb1 = max(fb1, bv-bx)
fb.append(fb1)
gb.append(bv - 2*bx)
p = max(0, fa[-1], fb[-1])
for i in range(N-1):
p = max(p, fa[i]+gb[N-i-2], ga[i]+fb[N-i-2])
print(p)
``` | output | 1 | 105,100 | 9 | 210,201 |
Provide a correct Python 3 solution for this coding contest problem.
"Teishi-zushi", a Japanese restaurant, is a plain restaurant with only one round counter. The outer circumference of the counter is C meters. Customers cannot go inside the counter.
Nakahashi entered Teishi-zushi, and he was guided to the counter. Now, there are N pieces of sushi (vinegared rice with seafood and so on) on the counter. The distance measured clockwise from the point where Nakahashi is standing to the point where the i-th sushi is placed, is x_i meters. Also, the i-th sushi has a nutritive value of v_i kilocalories.
Nakahashi can freely walk around the circumference of the counter. When he reach a point where a sushi is placed, he can eat that sushi and take in its nutrition (naturally, the sushi disappears). However, while walking, he consumes 1 kilocalories per meter.
Whenever he is satisfied, he can leave the restaurant from any place (he does not have to return to the initial place). On balance, at most how much nutrition can he take in before he leaves? That is, what is the maximum possible value of the total nutrition taken in minus the total energy consumed? Assume that there are no other customers, and no new sushi will be added to the counter. Also, since Nakahashi has plenty of nutrition in his body, assume that no matter how much he walks and consumes energy, he never dies from hunger.
Constraints
* 1 ≤ N ≤ 10^5
* 2 ≤ C ≤ 10^{14}
* 1 ≤ x_1 < x_2 < ... < x_N < C
* 1 ≤ v_i ≤ 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N C
x_1 v_1
x_2 v_2
:
x_N v_N
Output
If Nakahashi can take in at most c kilocalories on balance before he leaves the restaurant, print c.
Examples
Input
3 20
2 80
9 120
16 1
Output
191
Input
3 20
2 80
9 1
16 120
Output
192
Input
1 100000000000000
50000000000000 1
Output
0
Input
15 10000000000
400000000 1000000000
800000000 1000000000
1900000000 1000000000
2400000000 1000000000
2900000000 1000000000
3300000000 1000000000
3700000000 1000000000
3800000000 1000000000
4000000000 1000000000
4100000000 1000000000
5200000000 1000000000
6600000000 1000000000
8000000000 1000000000
9300000000 1000000000
9700000000 1000000000
Output
6500000000 | instruction | 0 | 105,101 | 9 | 210,202 |
"Correct Solution:
```
n, c = map(int, input().split())
xv = [list(map(int, input().split())) for i in range(n)]
dx = []
dy = []
for i in range(n):
if i == 0:
dx.append(xv[i][0])
dy.append(c - xv[-i-1][0])
else:
dx.append(xv[i][0]-xv[i-1][0])
dy.append(-xv[-i-1][0] + xv[-i][0])
right = [0]
left = [0]
for i in range(n):
r = right[-1] + xv[i][1] - dx[i]
l = left[-1] + xv[-i-1][1] - dy[i]
right.append(r)
left.append(l)
for i in range(1,n+1):
right[i] = max(right[i-1], right[i])
left[i] = max(left[i-1], left[i])
ans = 0
for i in range(1,n+1):
a1 = right[i]
a2 = right[i] - xv[i-1][0] + left[n-i]
a3 = left[i]
a4 = left[i] - (c - xv[-i][0]) + right[n-i]
ans = max(ans,a1,a2,a3,a4)
print(ans)
``` | output | 1 | 105,101 | 9 | 210,203 |
Provide a correct Python 3 solution for this coding contest problem.
"Teishi-zushi", a Japanese restaurant, is a plain restaurant with only one round counter. The outer circumference of the counter is C meters. Customers cannot go inside the counter.
Nakahashi entered Teishi-zushi, and he was guided to the counter. Now, there are N pieces of sushi (vinegared rice with seafood and so on) on the counter. The distance measured clockwise from the point where Nakahashi is standing to the point where the i-th sushi is placed, is x_i meters. Also, the i-th sushi has a nutritive value of v_i kilocalories.
Nakahashi can freely walk around the circumference of the counter. When he reach a point where a sushi is placed, he can eat that sushi and take in its nutrition (naturally, the sushi disappears). However, while walking, he consumes 1 kilocalories per meter.
Whenever he is satisfied, he can leave the restaurant from any place (he does not have to return to the initial place). On balance, at most how much nutrition can he take in before he leaves? That is, what is the maximum possible value of the total nutrition taken in minus the total energy consumed? Assume that there are no other customers, and no new sushi will be added to the counter. Also, since Nakahashi has plenty of nutrition in his body, assume that no matter how much he walks and consumes energy, he never dies from hunger.
Constraints
* 1 ≤ N ≤ 10^5
* 2 ≤ C ≤ 10^{14}
* 1 ≤ x_1 < x_2 < ... < x_N < C
* 1 ≤ v_i ≤ 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N C
x_1 v_1
x_2 v_2
:
x_N v_N
Output
If Nakahashi can take in at most c kilocalories on balance before he leaves the restaurant, print c.
Examples
Input
3 20
2 80
9 120
16 1
Output
191
Input
3 20
2 80
9 1
16 120
Output
192
Input
1 100000000000000
50000000000000 1
Output
0
Input
15 10000000000
400000000 1000000000
800000000 1000000000
1900000000 1000000000
2400000000 1000000000
2900000000 1000000000
3300000000 1000000000
3700000000 1000000000
3800000000 1000000000
4000000000 1000000000
4100000000 1000000000
5200000000 1000000000
6600000000 1000000000
8000000000 1000000000
9300000000 1000000000
9700000000 1000000000
Output
6500000000 | instruction | 0 | 105,102 | 9 | 210,204 |
"Correct Solution:
```
import random
def slv(N, C, X, V):
ans_l = [(0, 0)]
ans_l_ = [0]
for x, v in zip(X, V):
ans_l_.append(max(ans_l_[-1], ans_l[-1][1] + v - x))
ans_l.append((x, ans_l[-1][1] + v))
ans_l_.pop(0)
ans_l.pop(0)
ans_r = [(0, 0)]
ans_r_ = [0]
for x, v in list(zip(X, V))[::-1]:
ans_r_.append(max(ans_r_[-1], ans_r[-1][1] + v - (C-x)))
ans_r.append((C-x, ans_r[-1][1] + v))
ans_r_.pop(0)
ans_r.pop(0)
# print('p1')
ans_l_r = []
for i, a in enumerate(ans_l[:-1], 1):
# print(ans_r_[N-i-1], a[1]-2*a[0])
ans_l_r.append(ans_r_[N-i-1] + a[1]-2*a[0])
# print('p2')
ans_r_l = []
for i, a in enumerate(ans_r[:-1], 1):
# print(ans_l_[N-i-1], a[1]-2*a[0])
ans_r_l.append(ans_l_[N-i-1] + a[1]-2*a[0])
l = max([x[1]-x[0] for x in ans_l])
r = max([x[1]-x[0] for x in ans_r])
lr = 0 if not ans_l_r else max(ans_l_r)
rl = 0 if not ans_r_l else max(ans_r_l)
# print(ans_l)
# print(ans_l_)
# print(ans_r)
# print(ans_r_)
# print(l, r, lr, rl)
return max(l, r, lr, rl)
def main():
N, C = list(map(int, input().split()))
X = []
V = []
for _ in range(N):
x, v = list(map(int, input().split()))
X.append(x)
V.append(v)
# N = 100000
# C = 1000000000000
# X = [1]
# V = [100]
# for _ in range(N):
# X.append(X[-1]+random.randint(1, 10000))
# V.append(random.randint(1, 10000))
print(slv(N, C, X, V))
if __name__ == '__main__':
main()
``` | output | 1 | 105,102 | 9 | 210,205 |
Provide a correct Python 3 solution for this coding contest problem.
"Teishi-zushi", a Japanese restaurant, is a plain restaurant with only one round counter. The outer circumference of the counter is C meters. Customers cannot go inside the counter.
Nakahashi entered Teishi-zushi, and he was guided to the counter. Now, there are N pieces of sushi (vinegared rice with seafood and so on) on the counter. The distance measured clockwise from the point where Nakahashi is standing to the point where the i-th sushi is placed, is x_i meters. Also, the i-th sushi has a nutritive value of v_i kilocalories.
Nakahashi can freely walk around the circumference of the counter. When he reach a point where a sushi is placed, he can eat that sushi and take in its nutrition (naturally, the sushi disappears). However, while walking, he consumes 1 kilocalories per meter.
Whenever he is satisfied, he can leave the restaurant from any place (he does not have to return to the initial place). On balance, at most how much nutrition can he take in before he leaves? That is, what is the maximum possible value of the total nutrition taken in minus the total energy consumed? Assume that there are no other customers, and no new sushi will be added to the counter. Also, since Nakahashi has plenty of nutrition in his body, assume that no matter how much he walks and consumes energy, he never dies from hunger.
Constraints
* 1 ≤ N ≤ 10^5
* 2 ≤ C ≤ 10^{14}
* 1 ≤ x_1 < x_2 < ... < x_N < C
* 1 ≤ v_i ≤ 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N C
x_1 v_1
x_2 v_2
:
x_N v_N
Output
If Nakahashi can take in at most c kilocalories on balance before he leaves the restaurant, print c.
Examples
Input
3 20
2 80
9 120
16 1
Output
191
Input
3 20
2 80
9 1
16 120
Output
192
Input
1 100000000000000
50000000000000 1
Output
0
Input
15 10000000000
400000000 1000000000
800000000 1000000000
1900000000 1000000000
2400000000 1000000000
2900000000 1000000000
3300000000 1000000000
3700000000 1000000000
3800000000 1000000000
4000000000 1000000000
4100000000 1000000000
5200000000 1000000000
6600000000 1000000000
8000000000 1000000000
9300000000 1000000000
9700000000 1000000000
Output
6500000000 | instruction | 0 | 105,103 | 9 | 210,206 |
"Correct Solution:
```
n,c=map(int,input().split())
rl=[[0]*2 for _ in range(n+1)]
ls,rs=0,0
xv=[[0,0]]+[list(map(int,input().split())) for _ in range(n)]
for i in range(1,n+1):
j,k=xv[i]
rs+=k
rl[i][0],rl[i][1]=max(rl[i-1][0],rs-j),max(rl[i-1][1],rs-2*j)
ans=rl[n][0]
for i in range(n):
ls+=xv[n-i][1]
ans=max(ans,rl[n-i-1][1]+ls-(c-xv[n-i][0]),rl[n-i-1][0]+ls-(c-xv[n-i][0])*2)
print(ans)
``` | output | 1 | 105,103 | 9 | 210,207 |
Provide a correct Python 3 solution for this coding contest problem.
"Teishi-zushi", a Japanese restaurant, is a plain restaurant with only one round counter. The outer circumference of the counter is C meters. Customers cannot go inside the counter.
Nakahashi entered Teishi-zushi, and he was guided to the counter. Now, there are N pieces of sushi (vinegared rice with seafood and so on) on the counter. The distance measured clockwise from the point where Nakahashi is standing to the point where the i-th sushi is placed, is x_i meters. Also, the i-th sushi has a nutritive value of v_i kilocalories.
Nakahashi can freely walk around the circumference of the counter. When he reach a point where a sushi is placed, he can eat that sushi and take in its nutrition (naturally, the sushi disappears). However, while walking, he consumes 1 kilocalories per meter.
Whenever he is satisfied, he can leave the restaurant from any place (he does not have to return to the initial place). On balance, at most how much nutrition can he take in before he leaves? That is, what is the maximum possible value of the total nutrition taken in minus the total energy consumed? Assume that there are no other customers, and no new sushi will be added to the counter. Also, since Nakahashi has plenty of nutrition in his body, assume that no matter how much he walks and consumes energy, he never dies from hunger.
Constraints
* 1 ≤ N ≤ 10^5
* 2 ≤ C ≤ 10^{14}
* 1 ≤ x_1 < x_2 < ... < x_N < C
* 1 ≤ v_i ≤ 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N C
x_1 v_1
x_2 v_2
:
x_N v_N
Output
If Nakahashi can take in at most c kilocalories on balance before he leaves the restaurant, print c.
Examples
Input
3 20
2 80
9 120
16 1
Output
191
Input
3 20
2 80
9 1
16 120
Output
192
Input
1 100000000000000
50000000000000 1
Output
0
Input
15 10000000000
400000000 1000000000
800000000 1000000000
1900000000 1000000000
2400000000 1000000000
2900000000 1000000000
3300000000 1000000000
3700000000 1000000000
3800000000 1000000000
4000000000 1000000000
4100000000 1000000000
5200000000 1000000000
6600000000 1000000000
8000000000 1000000000
9300000000 1000000000
9700000000 1000000000
Output
6500000000 | instruction | 0 | 105,104 | 9 | 210,208 |
"Correct Solution:
```
N, C = [int(_) for _ in input().split()]
xv = [[int(_) for _ in input().split()] for i in range(N)]
def solve(N, C, xv):
from itertools import accumulate
xs = [0] + [_[0] for _ in xv] + [C]
vs = [0] + [_[1] for _ in xv] + [0]
xs_rev = [C - x for x in xs[::-1]]
vs_rev = vs[::-1]
ts = accumulate(vs)
txs = [t - x for x, t in zip(xs, ts)]
maxtxs = list(accumulate(txs, max))
ts_rev = accumulate(vs_rev)
txs_rev = [t - x for x, t in zip(xs_rev, ts_rev)]
maxtxs_rev = list(accumulate(txs_rev, max))
return max(
max(tx - xs_rev[i] + maxtxs[N - i] for i, tx in enumerate(txs_rev[:-1])),
max(tx - xs[i] + maxtxs_rev[N - i] for i, tx in enumerate(txs[:-1]))
)
print(solve(N, C, xv))
``` | output | 1 | 105,104 | 9 | 210,209 |
Provide a correct Python 3 solution for this coding contest problem.
"Teishi-zushi", a Japanese restaurant, is a plain restaurant with only one round counter. The outer circumference of the counter is C meters. Customers cannot go inside the counter.
Nakahashi entered Teishi-zushi, and he was guided to the counter. Now, there are N pieces of sushi (vinegared rice with seafood and so on) on the counter. The distance measured clockwise from the point where Nakahashi is standing to the point where the i-th sushi is placed, is x_i meters. Also, the i-th sushi has a nutritive value of v_i kilocalories.
Nakahashi can freely walk around the circumference of the counter. When he reach a point where a sushi is placed, he can eat that sushi and take in its nutrition (naturally, the sushi disappears). However, while walking, he consumes 1 kilocalories per meter.
Whenever he is satisfied, he can leave the restaurant from any place (he does not have to return to the initial place). On balance, at most how much nutrition can he take in before he leaves? That is, what is the maximum possible value of the total nutrition taken in minus the total energy consumed? Assume that there are no other customers, and no new sushi will be added to the counter. Also, since Nakahashi has plenty of nutrition in his body, assume that no matter how much he walks and consumes energy, he never dies from hunger.
Constraints
* 1 ≤ N ≤ 10^5
* 2 ≤ C ≤ 10^{14}
* 1 ≤ x_1 < x_2 < ... < x_N < C
* 1 ≤ v_i ≤ 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N C
x_1 v_1
x_2 v_2
:
x_N v_N
Output
If Nakahashi can take in at most c kilocalories on balance before he leaves the restaurant, print c.
Examples
Input
3 20
2 80
9 120
16 1
Output
191
Input
3 20
2 80
9 1
16 120
Output
192
Input
1 100000000000000
50000000000000 1
Output
0
Input
15 10000000000
400000000 1000000000
800000000 1000000000
1900000000 1000000000
2400000000 1000000000
2900000000 1000000000
3300000000 1000000000
3700000000 1000000000
3800000000 1000000000
4000000000 1000000000
4100000000 1000000000
5200000000 1000000000
6600000000 1000000000
8000000000 1000000000
9300000000 1000000000
9700000000 1000000000
Output
6500000000 | instruction | 0 | 105,105 | 9 | 210,210 |
"Correct Solution:
```
N, C = map(int, input().split())
sushi = []
for i in range(N):
sushi.append(list(map(int, input().split())))
right_turn_sum = [0]
for i in range(N):
right_turn_sum.append(sushi[i][1]+right_turn_sum[i])
for i in range(N):
right_turn_sum[i+1] -= sushi[i][0]
right_turn_to_zero = [0]
for i in range(N):
right_turn_to_zero.append(right_turn_sum[i+1]-sushi[i][0])
left_turn_sum = [0]
for i in range(N):
left_turn_sum.append(sushi[-1-i][1]+left_turn_sum[i])
for i in range(N):
left_turn_sum[i+1] -= (C - sushi[-1-i][0])
left_turn_to_zero = [0]
for i in range(N):
left_turn_to_zero.append(left_turn_sum[i+1]-(C-sushi[-1-i][0]))
ans = max(right_turn_sum)
ans = max(left_turn_sum+[ans])
maxi = 0
for i in range(N+1):
if maxi < right_turn_sum[i]:
maxi = right_turn_sum[i]
right_turn_sum[i] = max(maxi, right_turn_sum[i])
maxi = 0
for i in range(N+1):
if maxi < right_turn_to_zero[i]:
maxi = right_turn_to_zero[i]
right_turn_to_zero[i] = max(maxi, right_turn_to_zero[i])
maxi = 0
for i in range(N+1):
if maxi < left_turn_sum[i]:
maxi = left_turn_sum[i]
left_turn_sum[i] = max(maxi, left_turn_sum[i])
maxi = 0
for i in range(N+1):
if maxi < left_turn_to_zero[i]:
maxi = left_turn_to_zero[i]
left_turn_to_zero[i] = max(maxi, left_turn_to_zero[i])
for i in range(N+1):
ans = max(ans, right_turn_to_zero[i]+left_turn_sum[N-i])
for i in range(N+1):
ans = max(ans, left_turn_to_zero[i]+right_turn_sum[N-i])
print(ans)
``` | output | 1 | 105,105 | 9 | 210,211 |
Provide a correct Python 3 solution for this coding contest problem.
"Teishi-zushi", a Japanese restaurant, is a plain restaurant with only one round counter. The outer circumference of the counter is C meters. Customers cannot go inside the counter.
Nakahashi entered Teishi-zushi, and he was guided to the counter. Now, there are N pieces of sushi (vinegared rice with seafood and so on) on the counter. The distance measured clockwise from the point where Nakahashi is standing to the point where the i-th sushi is placed, is x_i meters. Also, the i-th sushi has a nutritive value of v_i kilocalories.
Nakahashi can freely walk around the circumference of the counter. When he reach a point where a sushi is placed, he can eat that sushi and take in its nutrition (naturally, the sushi disappears). However, while walking, he consumes 1 kilocalories per meter.
Whenever he is satisfied, he can leave the restaurant from any place (he does not have to return to the initial place). On balance, at most how much nutrition can he take in before he leaves? That is, what is the maximum possible value of the total nutrition taken in minus the total energy consumed? Assume that there are no other customers, and no new sushi will be added to the counter. Also, since Nakahashi has plenty of nutrition in his body, assume that no matter how much he walks and consumes energy, he never dies from hunger.
Constraints
* 1 ≤ N ≤ 10^5
* 2 ≤ C ≤ 10^{14}
* 1 ≤ x_1 < x_2 < ... < x_N < C
* 1 ≤ v_i ≤ 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N C
x_1 v_1
x_2 v_2
:
x_N v_N
Output
If Nakahashi can take in at most c kilocalories on balance before he leaves the restaurant, print c.
Examples
Input
3 20
2 80
9 120
16 1
Output
191
Input
3 20
2 80
9 1
16 120
Output
192
Input
1 100000000000000
50000000000000 1
Output
0
Input
15 10000000000
400000000 1000000000
800000000 1000000000
1900000000 1000000000
2400000000 1000000000
2900000000 1000000000
3300000000 1000000000
3700000000 1000000000
3800000000 1000000000
4000000000 1000000000
4100000000 1000000000
5200000000 1000000000
6600000000 1000000000
8000000000 1000000000
9300000000 1000000000
9700000000 1000000000
Output
6500000000 | instruction | 0 | 105,106 | 9 | 210,212 |
"Correct Solution:
```
import sys
read = sys.stdin.buffer.read
N,C, *XV = map(int, read().split())#全部
X=XV[::2]
V=XV[1::2]
m1=[0]
m2=[0]
s=0
ans=0
for x,v in zip(X,V):
s+=v
ans=max(ans,s-x)
m1.append(max(m1[-1],s-x))
m2.append(max(m2[-1],s-2*x))
s=0
i=0
for x,v in reversed(list(zip(X,V))):
s+=v
x=C-x
ans=max(ans,s-2*x+m1[-2-i],s-x+m2[-2-i])
i+=1
print(ans)
``` | output | 1 | 105,106 | 9 | 210,213 |
Provide a correct Python 3 solution for this coding contest problem.
"Teishi-zushi", a Japanese restaurant, is a plain restaurant with only one round counter. The outer circumference of the counter is C meters. Customers cannot go inside the counter.
Nakahashi entered Teishi-zushi, and he was guided to the counter. Now, there are N pieces of sushi (vinegared rice with seafood and so on) on the counter. The distance measured clockwise from the point where Nakahashi is standing to the point where the i-th sushi is placed, is x_i meters. Also, the i-th sushi has a nutritive value of v_i kilocalories.
Nakahashi can freely walk around the circumference of the counter. When he reach a point where a sushi is placed, he can eat that sushi and take in its nutrition (naturally, the sushi disappears). However, while walking, he consumes 1 kilocalories per meter.
Whenever he is satisfied, he can leave the restaurant from any place (he does not have to return to the initial place). On balance, at most how much nutrition can he take in before he leaves? That is, what is the maximum possible value of the total nutrition taken in minus the total energy consumed? Assume that there are no other customers, and no new sushi will be added to the counter. Also, since Nakahashi has plenty of nutrition in his body, assume that no matter how much he walks and consumes energy, he never dies from hunger.
Constraints
* 1 ≤ N ≤ 10^5
* 2 ≤ C ≤ 10^{14}
* 1 ≤ x_1 < x_2 < ... < x_N < C
* 1 ≤ v_i ≤ 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N C
x_1 v_1
x_2 v_2
:
x_N v_N
Output
If Nakahashi can take in at most c kilocalories on balance before he leaves the restaurant, print c.
Examples
Input
3 20
2 80
9 120
16 1
Output
191
Input
3 20
2 80
9 1
16 120
Output
192
Input
1 100000000000000
50000000000000 1
Output
0
Input
15 10000000000
400000000 1000000000
800000000 1000000000
1900000000 1000000000
2400000000 1000000000
2900000000 1000000000
3300000000 1000000000
3700000000 1000000000
3800000000 1000000000
4000000000 1000000000
4100000000 1000000000
5200000000 1000000000
6600000000 1000000000
8000000000 1000000000
9300000000 1000000000
9700000000 1000000000
Output
6500000000 | instruction | 0 | 105,107 | 9 | 210,214 |
"Correct Solution:
```
# python template for atcoder1
import sys
sys.setrecursionlimit(10**9)
input = sys.stdin.readline
N, C = map(int, input().split())
sushis = [list(map(int, input().split())) for _ in range(N)]
# Aを時計回り,Bを反時計回りとする
# O->Aと0->A->Oのコストを計算
valA = []
valArev = []
sumV = 0
tmpMax = -float('inf')
for d, v in sushis:
sumV += v
valArev.append(sumV-2*d)
tmpMax = max(tmpMax, sumV-d)
valA.append(tmpMax)
valB = []
valBrev = []
sumV = 0
tmpMax = -float('inf')
for d, v in reversed(sushis):
sumV += v
revD = C-d
valBrev.append(sumV-2*revD)
tmpMax = max(tmpMax, sumV-revD)
valB.append(tmpMax)
# 折り返しなしの場合の値
ans = max(valA+valB+[0])
# O->A->O->B
for i in range(N-1):
ans = max(ans, valArev[i]+valB[N-i-2])
# O->B->O->A
for i in range(N-1):
ans = max(ans, valBrev[i]+valA[N-i-2])
print(ans)
``` | output | 1 | 105,107 | 9 | 210,215 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"Teishi-zushi", a Japanese restaurant, is a plain restaurant with only one round counter. The outer circumference of the counter is C meters. Customers cannot go inside the counter.
Nakahashi entered Teishi-zushi, and he was guided to the counter. Now, there are N pieces of sushi (vinegared rice with seafood and so on) on the counter. The distance measured clockwise from the point where Nakahashi is standing to the point where the i-th sushi is placed, is x_i meters. Also, the i-th sushi has a nutritive value of v_i kilocalories.
Nakahashi can freely walk around the circumference of the counter. When he reach a point where a sushi is placed, he can eat that sushi and take in its nutrition (naturally, the sushi disappears). However, while walking, he consumes 1 kilocalories per meter.
Whenever he is satisfied, he can leave the restaurant from any place (he does not have to return to the initial place). On balance, at most how much nutrition can he take in before he leaves? That is, what is the maximum possible value of the total nutrition taken in minus the total energy consumed? Assume that there are no other customers, and no new sushi will be added to the counter. Also, since Nakahashi has plenty of nutrition in his body, assume that no matter how much he walks and consumes energy, he never dies from hunger.
Constraints
* 1 ≤ N ≤ 10^5
* 2 ≤ C ≤ 10^{14}
* 1 ≤ x_1 < x_2 < ... < x_N < C
* 1 ≤ v_i ≤ 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N C
x_1 v_1
x_2 v_2
:
x_N v_N
Output
If Nakahashi can take in at most c kilocalories on balance before he leaves the restaurant, print c.
Examples
Input
3 20
2 80
9 120
16 1
Output
191
Input
3 20
2 80
9 1
16 120
Output
192
Input
1 100000000000000
50000000000000 1
Output
0
Input
15 10000000000
400000000 1000000000
800000000 1000000000
1900000000 1000000000
2400000000 1000000000
2900000000 1000000000
3300000000 1000000000
3700000000 1000000000
3800000000 1000000000
4000000000 1000000000
4100000000 1000000000
5200000000 1000000000
6600000000 1000000000
8000000000 1000000000
9300000000 1000000000
9700000000 1000000000
Output
6500000000
Submitted Solution:
```
def solve():
N, C = map(int, input().split())# {{{
L = [0]*N
R = [0]*N
S = []
for n in range(N):
x, v = map(int, input().split())
S.append((x,v))
res = 0
# left round
pre_x = 0
diff = 0
M = 0
for n in range(N):
x, v = S[n]
diff -= abs(x - pre_x)
diff += v
M = max(diff, M)
res = max(res, M)
L[n] = M
pre_x = x
# print(L)
# right round
pre_x = C
diff = 0
M = 0
for n in range(1, N+1):
x, v = S[-n]
diff -= abs(x - pre_x)
diff += v
M = max(diff, M)
res = max(res, M)
R[-n] = M
pre_x = x
# print(R)
# }}}
# left, right
diff = 0
for n in range(N-1):
res = max(res, L[n]-S[n][0] + R[n+1])
# right, left
diff = 0
for n in range(1, N):
res = max(res, R[-n]-(abs(C-S[-n][0])) + L[-n-1])
print(res)
if __name__ == "__main__":
solve()
``` | instruction | 0 | 105,108 | 9 | 210,216 |
Yes | output | 1 | 105,108 | 9 | 210,217 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"Teishi-zushi", a Japanese restaurant, is a plain restaurant with only one round counter. The outer circumference of the counter is C meters. Customers cannot go inside the counter.
Nakahashi entered Teishi-zushi, and he was guided to the counter. Now, there are N pieces of sushi (vinegared rice with seafood and so on) on the counter. The distance measured clockwise from the point where Nakahashi is standing to the point where the i-th sushi is placed, is x_i meters. Also, the i-th sushi has a nutritive value of v_i kilocalories.
Nakahashi can freely walk around the circumference of the counter. When he reach a point where a sushi is placed, he can eat that sushi and take in its nutrition (naturally, the sushi disappears). However, while walking, he consumes 1 kilocalories per meter.
Whenever he is satisfied, he can leave the restaurant from any place (he does not have to return to the initial place). On balance, at most how much nutrition can he take in before he leaves? That is, what is the maximum possible value of the total nutrition taken in minus the total energy consumed? Assume that there are no other customers, and no new sushi will be added to the counter. Also, since Nakahashi has plenty of nutrition in his body, assume that no matter how much he walks and consumes energy, he never dies from hunger.
Constraints
* 1 ≤ N ≤ 10^5
* 2 ≤ C ≤ 10^{14}
* 1 ≤ x_1 < x_2 < ... < x_N < C
* 1 ≤ v_i ≤ 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N C
x_1 v_1
x_2 v_2
:
x_N v_N
Output
If Nakahashi can take in at most c kilocalories on balance before he leaves the restaurant, print c.
Examples
Input
3 20
2 80
9 120
16 1
Output
191
Input
3 20
2 80
9 1
16 120
Output
192
Input
1 100000000000000
50000000000000 1
Output
0
Input
15 10000000000
400000000 1000000000
800000000 1000000000
1900000000 1000000000
2400000000 1000000000
2900000000 1000000000
3300000000 1000000000
3700000000 1000000000
3800000000 1000000000
4000000000 1000000000
4100000000 1000000000
5200000000 1000000000
6600000000 1000000000
8000000000 1000000000
9300000000 1000000000
9700000000 1000000000
Output
6500000000
Submitted Solution:
```
n,c = map(int,input().split())
x = [0]
v = [0]
rf = [0] * (n+2)
rg = [0] * (n+2)
lf = [0] * (n+2)
lg = [0] * (n+2)
for i in range(0,n):
X,V = map(int,input().split())
x.append(X)
v.append(V)
for i in range(0,n):
rf[i+1] = rf[i] + v[i+1]
rg[i+1] = max(rg[i],rf[i+1] - x[i+1])
for i in range(n+1,1,-1):
lf[i-1] = lf[i] + v[i-1]
lg[i-1] = max(lg[i],lf[i-1] - (c-x[i-1]))
ans = max(rg[n],lg[1])
for i in range(1,n):
ans = max(ans,rg[i]+lg[i+1] - x[i],rg[i]+lg[i+1] - (c - x[i+1]))
print(ans)
``` | instruction | 0 | 105,109 | 9 | 210,218 |
Yes | output | 1 | 105,109 | 9 | 210,219 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"Teishi-zushi", a Japanese restaurant, is a plain restaurant with only one round counter. The outer circumference of the counter is C meters. Customers cannot go inside the counter.
Nakahashi entered Teishi-zushi, and he was guided to the counter. Now, there are N pieces of sushi (vinegared rice with seafood and so on) on the counter. The distance measured clockwise from the point where Nakahashi is standing to the point where the i-th sushi is placed, is x_i meters. Also, the i-th sushi has a nutritive value of v_i kilocalories.
Nakahashi can freely walk around the circumference of the counter. When he reach a point where a sushi is placed, he can eat that sushi and take in its nutrition (naturally, the sushi disappears). However, while walking, he consumes 1 kilocalories per meter.
Whenever he is satisfied, he can leave the restaurant from any place (he does not have to return to the initial place). On balance, at most how much nutrition can he take in before he leaves? That is, what is the maximum possible value of the total nutrition taken in minus the total energy consumed? Assume that there are no other customers, and no new sushi will be added to the counter. Also, since Nakahashi has plenty of nutrition in his body, assume that no matter how much he walks and consumes energy, he never dies from hunger.
Constraints
* 1 ≤ N ≤ 10^5
* 2 ≤ C ≤ 10^{14}
* 1 ≤ x_1 < x_2 < ... < x_N < C
* 1 ≤ v_i ≤ 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N C
x_1 v_1
x_2 v_2
:
x_N v_N
Output
If Nakahashi can take in at most c kilocalories on balance before he leaves the restaurant, print c.
Examples
Input
3 20
2 80
9 120
16 1
Output
191
Input
3 20
2 80
9 1
16 120
Output
192
Input
1 100000000000000
50000000000000 1
Output
0
Input
15 10000000000
400000000 1000000000
800000000 1000000000
1900000000 1000000000
2400000000 1000000000
2900000000 1000000000
3300000000 1000000000
3700000000 1000000000
3800000000 1000000000
4000000000 1000000000
4100000000 1000000000
5200000000 1000000000
6600000000 1000000000
8000000000 1000000000
9300000000 1000000000
9700000000 1000000000
Output
6500000000
Submitted Solution:
```
N, C = map(int, input().split())
X = []; V = []
for i in range(N):
x, v = map(int, input().split())
X.append(x); V.append(v)
P0 = []; P1 = []
ma0 = cu0 = ma1 = cu1 = 0
pos = 0
for i in range(N):
cu0 += V[i]-(X[i]-pos)
cu1 += V[i]-(X[i]-pos)*2
ma0 = max(ma0, cu0)
ma1 = max(ma1, cu1)
P0.append(ma0)
P1.append(ma1)
pos = X[i]
Q0 = []; Q1 = []
ma0 = cu0 = ma1 = cu1 = 0
pos = C
for i in range(N):
cu0 += V[-i-1]-(pos-X[-i-1])
cu1 += V[-i-1]-(pos-X[-i-1])*2
ma0 = max(ma0, cu0)
ma1 = max(ma1, cu1)
Q0.append(ma0)
Q1.append(ma1)
pos = X[-i-1]
ans = max(P0[N-1], Q0[N-1])
for i in range(N-1):
ans = max(ans, P0[i] + Q1[-i-2], P1[-i-2] + Q0[i])
print(ans)
``` | instruction | 0 | 105,110 | 9 | 210,220 |
Yes | output | 1 | 105,110 | 9 | 210,221 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"Teishi-zushi", a Japanese restaurant, is a plain restaurant with only one round counter. The outer circumference of the counter is C meters. Customers cannot go inside the counter.
Nakahashi entered Teishi-zushi, and he was guided to the counter. Now, there are N pieces of sushi (vinegared rice with seafood and so on) on the counter. The distance measured clockwise from the point where Nakahashi is standing to the point where the i-th sushi is placed, is x_i meters. Also, the i-th sushi has a nutritive value of v_i kilocalories.
Nakahashi can freely walk around the circumference of the counter. When he reach a point where a sushi is placed, he can eat that sushi and take in its nutrition (naturally, the sushi disappears). However, while walking, he consumes 1 kilocalories per meter.
Whenever he is satisfied, he can leave the restaurant from any place (he does not have to return to the initial place). On balance, at most how much nutrition can he take in before he leaves? That is, what is the maximum possible value of the total nutrition taken in minus the total energy consumed? Assume that there are no other customers, and no new sushi will be added to the counter. Also, since Nakahashi has plenty of nutrition in his body, assume that no matter how much he walks and consumes energy, he never dies from hunger.
Constraints
* 1 ≤ N ≤ 10^5
* 2 ≤ C ≤ 10^{14}
* 1 ≤ x_1 < x_2 < ... < x_N < C
* 1 ≤ v_i ≤ 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N C
x_1 v_1
x_2 v_2
:
x_N v_N
Output
If Nakahashi can take in at most c kilocalories on balance before he leaves the restaurant, print c.
Examples
Input
3 20
2 80
9 120
16 1
Output
191
Input
3 20
2 80
9 1
16 120
Output
192
Input
1 100000000000000
50000000000000 1
Output
0
Input
15 10000000000
400000000 1000000000
800000000 1000000000
1900000000 1000000000
2400000000 1000000000
2900000000 1000000000
3300000000 1000000000
3700000000 1000000000
3800000000 1000000000
4000000000 1000000000
4100000000 1000000000
5200000000 1000000000
6600000000 1000000000
8000000000 1000000000
9300000000 1000000000
9700000000 1000000000
Output
6500000000
Submitted Solution:
```
import sys
def input():
return sys.stdin.readline().strip()
N, C = list(map(int, input().split()))
sushi = []
for i in range(N):
sushi.append(tuple(map(int, input().split())))
sushi_sum = [0]
for i in range(N):
sushi_sum.append(sushi_sum[-1] + sushi[i][1])
right_max = 0
right_list = []
left_max = 0
left_list = []
for i in range(N):
if right_max < sushi_sum[i + 1] - sushi[i][0]:
right_max = sushi_sum[i + 1] - sushi[i][0]
right_list.append(sushi_sum[i + 1] - sushi[i][0])
if left_max < sushi_sum[-1] - sushi_sum[i] - (C - sushi[i][0]):
left_max = sushi_sum[-1] - sushi_sum[i] - (C - sushi[i][0])
left_list.append(sushi_sum[-1] - sushi_sum[i] - (C - sushi[i][0]))
for i in range(1, N):
right_list[i] = max(right_list[i - 1], right_list[i])
for i in range(N - 2, -1, -1):
left_list[i] = max(left_list[i + 1], left_list[i])
right_left_max = 0
for i in range(N - 1):
right_left_max = max(right_left_max, sushi_sum[i + 1] - 2*sushi[i][0] + left_list[i + 1])
left_right_max = 0
for i in range(1, N):
left_right_max = max(left_right_max, sushi_sum[-1] - sushi_sum[i] - 2*(C - sushi[i][0]) + right_list[i - 1])
print(max(right_max, left_max, right_left_max, left_right_max))
``` | instruction | 0 | 105,111 | 9 | 210,222 |
Yes | output | 1 | 105,111 | 9 | 210,223 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"Teishi-zushi", a Japanese restaurant, is a plain restaurant with only one round counter. The outer circumference of the counter is C meters. Customers cannot go inside the counter.
Nakahashi entered Teishi-zushi, and he was guided to the counter. Now, there are N pieces of sushi (vinegared rice with seafood and so on) on the counter. The distance measured clockwise from the point where Nakahashi is standing to the point where the i-th sushi is placed, is x_i meters. Also, the i-th sushi has a nutritive value of v_i kilocalories.
Nakahashi can freely walk around the circumference of the counter. When he reach a point where a sushi is placed, he can eat that sushi and take in its nutrition (naturally, the sushi disappears). However, while walking, he consumes 1 kilocalories per meter.
Whenever he is satisfied, he can leave the restaurant from any place (he does not have to return to the initial place). On balance, at most how much nutrition can he take in before he leaves? That is, what is the maximum possible value of the total nutrition taken in minus the total energy consumed? Assume that there are no other customers, and no new sushi will be added to the counter. Also, since Nakahashi has plenty of nutrition in his body, assume that no matter how much he walks and consumes energy, he never dies from hunger.
Constraints
* 1 ≤ N ≤ 10^5
* 2 ≤ C ≤ 10^{14}
* 1 ≤ x_1 < x_2 < ... < x_N < C
* 1 ≤ v_i ≤ 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N C
x_1 v_1
x_2 v_2
:
x_N v_N
Output
If Nakahashi can take in at most c kilocalories on balance before he leaves the restaurant, print c.
Examples
Input
3 20
2 80
9 120
16 1
Output
191
Input
3 20
2 80
9 1
16 120
Output
192
Input
1 100000000000000
50000000000000 1
Output
0
Input
15 10000000000
400000000 1000000000
800000000 1000000000
1900000000 1000000000
2400000000 1000000000
2900000000 1000000000
3300000000 1000000000
3700000000 1000000000
3800000000 1000000000
4000000000 1000000000
4100000000 1000000000
5200000000 1000000000
6600000000 1000000000
8000000000 1000000000
9300000000 1000000000
9700000000 1000000000
Output
6500000000
Submitted Solution:
```
n,c1 = map(int, input().split())
g = []
c = [0]
d = [0]
a = 0
ans=[]
for i in range(n):
x,v = map(int, input().split())
g.append([x,v])
a+=v
c.append(a-x)
d.append(a-2*x)
b = 0
e = [0]
f = [0]
for i,j in reversed(g):
b+=j
e.append(b-(c1-i))
f.append(b-2*(c1-i))
for i in range(1,n+2):
ans.append(max(c[:i])+f[-i])
ans.append(max(d[:i])+e[-i])
print(max(ans))
``` | instruction | 0 | 105,112 | 9 | 210,224 |
No | output | 1 | 105,112 | 9 | 210,225 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"Teishi-zushi", a Japanese restaurant, is a plain restaurant with only one round counter. The outer circumference of the counter is C meters. Customers cannot go inside the counter.
Nakahashi entered Teishi-zushi, and he was guided to the counter. Now, there are N pieces of sushi (vinegared rice with seafood and so on) on the counter. The distance measured clockwise from the point where Nakahashi is standing to the point where the i-th sushi is placed, is x_i meters. Also, the i-th sushi has a nutritive value of v_i kilocalories.
Nakahashi can freely walk around the circumference of the counter. When he reach a point where a sushi is placed, he can eat that sushi and take in its nutrition (naturally, the sushi disappears). However, while walking, he consumes 1 kilocalories per meter.
Whenever he is satisfied, he can leave the restaurant from any place (he does not have to return to the initial place). On balance, at most how much nutrition can he take in before he leaves? That is, what is the maximum possible value of the total nutrition taken in minus the total energy consumed? Assume that there are no other customers, and no new sushi will be added to the counter. Also, since Nakahashi has plenty of nutrition in his body, assume that no matter how much he walks and consumes energy, he never dies from hunger.
Constraints
* 1 ≤ N ≤ 10^5
* 2 ≤ C ≤ 10^{14}
* 1 ≤ x_1 < x_2 < ... < x_N < C
* 1 ≤ v_i ≤ 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N C
x_1 v_1
x_2 v_2
:
x_N v_N
Output
If Nakahashi can take in at most c kilocalories on balance before he leaves the restaurant, print c.
Examples
Input
3 20
2 80
9 120
16 1
Output
191
Input
3 20
2 80
9 1
16 120
Output
192
Input
1 100000000000000
50000000000000 1
Output
0
Input
15 10000000000
400000000 1000000000
800000000 1000000000
1900000000 1000000000
2400000000 1000000000
2900000000 1000000000
3300000000 1000000000
3700000000 1000000000
3800000000 1000000000
4000000000 1000000000
4100000000 1000000000
5200000000 1000000000
6600000000 1000000000
8000000000 1000000000
9300000000 1000000000
9700000000 1000000000
Output
6500000000
Submitted Solution:
```
from itertools import*
n,c=map(int,input().split())
l1=[]
l2=[]
l3=[]
tmp=0
for i in range(n):
x,v=map(int,input().split())
l1+=[(x,v+tmp)]
l2+=[c-x]
l3.append(v)
tmp+=v
l3=list(zip(l2[::-1],accumulate(l3[::-1])))[::-1]
ll=list(map(lambda x:x[1]-x[0],l3))
lr=list(map(lambda x:x[1]-x[0],l1))
#print(l3,l1)
#print(ll,lr)
ans=0
for i in range(n):
c=max(ll[i:]+[0])
d=max(lr[:i]+[0])
if c+d>ans:
ans=c+d
tmp=i
x=max(lr[:tmp]+[0])
y=max(ll[tmp:]+[0])
if x==0 or y==0:
print(ans)
else:
print(ans-min(l1[lr.index(x)][0],l3[ll.index(y)][0]))
#print(min(l1[lr.index(max(lr[:tmp]+[0]))][0],l3[ll.index(max(ll[tmp:]+[0]))][0]))
#print(tmp)
``` | instruction | 0 | 105,113 | 9 | 210,226 |
No | output | 1 | 105,113 | 9 | 210,227 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"Teishi-zushi", a Japanese restaurant, is a plain restaurant with only one round counter. The outer circumference of the counter is C meters. Customers cannot go inside the counter.
Nakahashi entered Teishi-zushi, and he was guided to the counter. Now, there are N pieces of sushi (vinegared rice with seafood and so on) on the counter. The distance measured clockwise from the point where Nakahashi is standing to the point where the i-th sushi is placed, is x_i meters. Also, the i-th sushi has a nutritive value of v_i kilocalories.
Nakahashi can freely walk around the circumference of the counter. When he reach a point where a sushi is placed, he can eat that sushi and take in its nutrition (naturally, the sushi disappears). However, while walking, he consumes 1 kilocalories per meter.
Whenever he is satisfied, he can leave the restaurant from any place (he does not have to return to the initial place). On balance, at most how much nutrition can he take in before he leaves? That is, what is the maximum possible value of the total nutrition taken in minus the total energy consumed? Assume that there are no other customers, and no new sushi will be added to the counter. Also, since Nakahashi has plenty of nutrition in his body, assume that no matter how much he walks and consumes energy, he never dies from hunger.
Constraints
* 1 ≤ N ≤ 10^5
* 2 ≤ C ≤ 10^{14}
* 1 ≤ x_1 < x_2 < ... < x_N < C
* 1 ≤ v_i ≤ 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N C
x_1 v_1
x_2 v_2
:
x_N v_N
Output
If Nakahashi can take in at most c kilocalories on balance before he leaves the restaurant, print c.
Examples
Input
3 20
2 80
9 120
16 1
Output
191
Input
3 20
2 80
9 1
16 120
Output
192
Input
1 100000000000000
50000000000000 1
Output
0
Input
15 10000000000
400000000 1000000000
800000000 1000000000
1900000000 1000000000
2400000000 1000000000
2900000000 1000000000
3300000000 1000000000
3700000000 1000000000
3800000000 1000000000
4000000000 1000000000
4100000000 1000000000
5200000000 1000000000
6600000000 1000000000
8000000000 1000000000
9300000000 1000000000
9700000000 1000000000
Output
6500000000
Submitted Solution:
```
import numpy as np
N, C = map(int, input().split())
x_list = []
v_list = []
for i in range(0, N):
x, v = map(int, input().split())
x_list.append(x)
v_list.append(v)
# 右行って左に引き返す
a_list = [0]
for i in range(0, len(v_list)):
if i > 0:
l = a_list[i] + v_list[i] - (x_list[i] - x_list[i - 1])
a_list.append(l)
elif i == 0:
l = v_list[i] - (x_list[i])
a_list.append(l)
a_accum_max = [max(a_list[:x+1]) for x in range(0, len(a_list))]
b_list = [0]
for i in reversed(range(0, len(v_list))):
if i == N-1:
l = v_list[i] - 2 * (C - x_list[i])
b_list.append(l)
elif i < N-1:
l = b_list[N - i - 1] + v_list[i] - 2 * (x_list[i+1] -x_list[i])
b_list.append(l)
first_list = []
for i in range(0, len(b_list)):
first_list.append(a_accum_max[N-i] + b_list[i])
max_fir = max(first_list)
# 左行って右に引き返す
c_list = [0]
for i in range(0, len(v_list)):
if i > 0:
l = c_list[i] + v_list[i] - 2 * (x_list[i] - x_list[i - 1])
c_list.append(l)
elif i == 0:
l = v_list[i] - 2 * (x_list[i])
c_list.append(l)
c_accum_max = [max(c_list[:x+1]) for x in range(0, len(c_list))]
d_list = [0]
for i in reversed(range(0, len(v_list))):
if i == N-1:
l = v_list[i] - (C - x_list[i])
d_list.append(l)
elif i < N-1:
l = d_list[N - i - 1] + v_list[i] - (x_list[i+1] -x_list[i])
d_list.append(l)
second_list = []
for i in range(0, len(d_list)):
second_list.append(c_accum_max[N-i] + d_list[i])
max_sec = max(second_list)
max_num = max([max_fir, max_sec, 0])
print(max_num)
``` | instruction | 0 | 105,114 | 9 | 210,228 |
No | output | 1 | 105,114 | 9 | 210,229 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"Teishi-zushi", a Japanese restaurant, is a plain restaurant with only one round counter. The outer circumference of the counter is C meters. Customers cannot go inside the counter.
Nakahashi entered Teishi-zushi, and he was guided to the counter. Now, there are N pieces of sushi (vinegared rice with seafood and so on) on the counter. The distance measured clockwise from the point where Nakahashi is standing to the point where the i-th sushi is placed, is x_i meters. Also, the i-th sushi has a nutritive value of v_i kilocalories.
Nakahashi can freely walk around the circumference of the counter. When he reach a point where a sushi is placed, he can eat that sushi and take in its nutrition (naturally, the sushi disappears). However, while walking, he consumes 1 kilocalories per meter.
Whenever he is satisfied, he can leave the restaurant from any place (he does not have to return to the initial place). On balance, at most how much nutrition can he take in before he leaves? That is, what is the maximum possible value of the total nutrition taken in minus the total energy consumed? Assume that there are no other customers, and no new sushi will be added to the counter. Also, since Nakahashi has plenty of nutrition in his body, assume that no matter how much he walks and consumes energy, he never dies from hunger.
Constraints
* 1 ≤ N ≤ 10^5
* 2 ≤ C ≤ 10^{14}
* 1 ≤ x_1 < x_2 < ... < x_N < C
* 1 ≤ v_i ≤ 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N C
x_1 v_1
x_2 v_2
:
x_N v_N
Output
If Nakahashi can take in at most c kilocalories on balance before he leaves the restaurant, print c.
Examples
Input
3 20
2 80
9 120
16 1
Output
191
Input
3 20
2 80
9 1
16 120
Output
192
Input
1 100000000000000
50000000000000 1
Output
0
Input
15 10000000000
400000000 1000000000
800000000 1000000000
1900000000 1000000000
2400000000 1000000000
2900000000 1000000000
3300000000 1000000000
3700000000 1000000000
3800000000 1000000000
4000000000 1000000000
4100000000 1000000000
5200000000 1000000000
6600000000 1000000000
8000000000 1000000000
9300000000 1000000000
9700000000 1000000000
Output
6500000000
Submitted Solution:
```
import sys
from itertools import accumulate
def input():
return sys.stdin.readline().strip()
sys.setrecursionlimit(20000000)
MOD = 10 ** 9 + 7
INF = float("inf")
def main():
N, C = map(int, input().split())
X_right = []
X_left = []
V_right = []
for i in range(N):
x, v = map(int, input().split())
X_right.append(x)
V_right.append(v)
X_left.append(C - x)
V_left = V_right[::-1]
V_right_add = list(accumulate([0] + V_right))
V_left_add = list(accumulate([0] + V_left))
X_left = [0] + X_left[::-1]
X_right = [0] + X_right
Right = []
answer_right = 0
for i in range(N + 1):
if answer_right < V_right_add[i] - X_right[i]:
answer_right = V_right_add[i] - X_right[i]
Right.append(answer_right)
Right = Right[::-1]
answer = Right[-1]
for i in range(N + 1):
if answer < V_left_add[i] - X_left[i] * 2 + Right[i]:
answer = max(
V_left_add[i] - X_left[i] * 2 + Right[i],
V_left_add[i] - X_left[i] + Right[i] - X_right[N - i - 1],
answer,
)
print(answer)
if __name__ == "__main__":
main()
``` | instruction | 0 | 105,115 | 9 | 210,230 |
No | output | 1 | 105,115 | 9 | 210,231 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Bubble Cup hypothesis stood unsolved for 130 years. Who ever proves the hypothesis will be regarded as one of the greatest mathematicians of our time! A famous mathematician Jerry Mao managed to reduce the hypothesis to this problem:
Given a number m, how many polynomials P with coefficients in set {\{0,1,2,3,4,5,6,7\}} have: P(2)=m?
Help Jerry Mao solve the long standing problem!
Input
The first line contains a single integer t (1 ≤ t ≤ 5⋅ 10^5) - number of test cases.
On next line there are t numbers, m_i (1 ≤ m_i ≤ 10^{18}) - meaning that in case i you should solve for number m_i.
Output
For each test case i, print the answer on separate lines: number of polynomials P as described in statement such that P(2)=m_i, modulo 10^9 + 7.
Example
Input
2
2 4
Output
2
4
Note
In first case, for m=2, polynomials that satisfy the constraint are x and 2.
In second case, for m=4, polynomials that satisfy the constraint are x^2, x + 2, 2x and 4. | instruction | 0 | 105,473 | 9 | 210,946 |
Tags: bitmasks, constructive algorithms, dp, math
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
def main():
pass
# region fastio
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
memo = dict()
def solve(m):
if m not in memo:
if m < 0:
memo[m] = 0
if m == 0:
memo[m] = 1
half = m//2
memo[m] = (solve(half) + solve(half - 1) + solve(half - 2) + solve(half - 3)) % MOD
return memo[m]
t = int(input())
out = []
for m in map(int, input().split()):
#out.append(solve(m))
v = m//2
u = v//2
w = (v-u)
out.append((u*w+u+w+1)%MOD)
print('\n'.join(map(str,out)))
``` | output | 1 | 105,473 | 9 | 210,947 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Bubble Cup hypothesis stood unsolved for 130 years. Who ever proves the hypothesis will be regarded as one of the greatest mathematicians of our time! A famous mathematician Jerry Mao managed to reduce the hypothesis to this problem:
Given a number m, how many polynomials P with coefficients in set {\{0,1,2,3,4,5,6,7\}} have: P(2)=m?
Help Jerry Mao solve the long standing problem!
Input
The first line contains a single integer t (1 ≤ t ≤ 5⋅ 10^5) - number of test cases.
On next line there are t numbers, m_i (1 ≤ m_i ≤ 10^{18}) - meaning that in case i you should solve for number m_i.
Output
For each test case i, print the answer on separate lines: number of polynomials P as described in statement such that P(2)=m_i, modulo 10^9 + 7.
Example
Input
2
2 4
Output
2
4
Note
In first case, for m=2, polynomials that satisfy the constraint are x and 2.
In second case, for m=4, polynomials that satisfy the constraint are x^2, x + 2, 2x and 4.
Submitted Solution:
```
t = int(input())
n = list(map(int, input().split()))
for m in n:
print(((m//2) - (m//4) + 1)*((m//4) + 1))
``` | instruction | 0 | 105,475 | 9 | 210,950 |
No | output | 1 | 105,475 | 9 | 210,951 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Bubble Cup hypothesis stood unsolved for 130 years. Who ever proves the hypothesis will be regarded as one of the greatest mathematicians of our time! A famous mathematician Jerry Mao managed to reduce the hypothesis to this problem:
Given a number m, how many polynomials P with coefficients in set {\{0,1,2,3,4,5,6,7\}} have: P(2)=m?
Help Jerry Mao solve the long standing problem!
Input
The first line contains a single integer t (1 ≤ t ≤ 5⋅ 10^5) - number of test cases.
On next line there are t numbers, m_i (1 ≤ m_i ≤ 10^{18}) - meaning that in case i you should solve for number m_i.
Output
For each test case i, print the answer on separate lines: number of polynomials P as described in statement such that P(2)=m_i, modulo 10^9 + 7.
Example
Input
2
2 4
Output
2
4
Note
In first case, for m=2, polynomials that satisfy the constraint are x and 2.
In second case, for m=4, polynomials that satisfy the constraint are x^2, x + 2, 2x and 4.
Submitted Solution:
```
t = int(input())
n1 = input().split()
ind = 0
while(ind<t):
n = int(n1[ind])
#ans = int((int(n/2+2)**2)/4)
ans = int(int((int(n/2+2)**2)/2)/2)
ans = ans%1000000007
print(ans)
ind += 1
``` | instruction | 0 | 105,476 | 9 | 210,952 |
No | output | 1 | 105,476 | 9 | 210,953 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Bubble Cup hypothesis stood unsolved for 130 years. Who ever proves the hypothesis will be regarded as one of the greatest mathematicians of our time! A famous mathematician Jerry Mao managed to reduce the hypothesis to this problem:
Given a number m, how many polynomials P with coefficients in set {\{0,1,2,3,4,5,6,7\}} have: P(2)=m?
Help Jerry Mao solve the long standing problem!
Input
The first line contains a single integer t (1 ≤ t ≤ 5⋅ 10^5) - number of test cases.
On next line there are t numbers, m_i (1 ≤ m_i ≤ 10^{18}) - meaning that in case i you should solve for number m_i.
Output
For each test case i, print the answer on separate lines: number of polynomials P as described in statement such that P(2)=m_i, modulo 10^9 + 7.
Example
Input
2
2 4
Output
2
4
Note
In first case, for m=2, polynomials that satisfy the constraint are x and 2.
In second case, for m=4, polynomials that satisfy the constraint are x^2, x + 2, 2x and 4.
Submitted Solution:
```
x=int(input())
y=list(map(int,input().split()))
for i in range(x):
a=y[i]
if a==x*x and a==2*x and a==2+x:
print("4")
elif a==x*x and a==2*x or a==x*x and a==2+x:
print("3")
elif a==x*x or a==2*x or a==2+x:
print("2")
else:
print("2")
``` | instruction | 0 | 105,477 | 9 | 210,954 |
No | output | 1 | 105,477 | 9 | 210,955 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After a hard day Vitaly got very hungry and he wants to eat his favorite potato pie. But it's not that simple. Vitaly is in the first room of the house with n room located in a line and numbered starting from one from left to right. You can go from the first room to the second room, from the second room to the third room and so on — you can go from the (n - 1)-th room to the n-th room. Thus, you can go to room x only from room x - 1.
The potato pie is located in the n-th room and Vitaly needs to go there.
Each pair of consecutive rooms has a door between them. In order to go to room x from room x - 1, you need to open the door between the rooms with the corresponding key.
In total the house has several types of doors (represented by uppercase Latin letters) and several types of keys (represented by lowercase Latin letters). The key of type t can open the door of type T if and only if t and T are the same letter, written in different cases. For example, key f can open door F.
Each of the first n - 1 rooms contains exactly one key of some type that Vitaly can use to get to next rooms. Once the door is open with some key, Vitaly won't get the key from the keyhole but he will immediately run into the next room. In other words, each key can open no more than one door.
Vitaly realizes that he may end up in some room without the key that opens the door to the next room. Before the start his run for the potato pie Vitaly can buy any number of keys of any type that is guaranteed to get to room n.
Given the plan of the house, Vitaly wants to know what is the minimum number of keys he needs to buy to surely get to the room n, which has a delicious potato pie. Write a program that will help Vitaly find out this number.
Input
The first line of the input contains a positive integer n (2 ≤ n ≤ 105) — the number of rooms in the house.
The second line of the input contains string s of length 2·n - 2. Let's number the elements of the string from left to right, starting from one.
The odd positions in the given string s contain lowercase Latin letters — the types of the keys that lie in the corresponding rooms. Thus, each odd position i of the given string s contains a lowercase Latin letter — the type of the key that lies in room number (i + 1) / 2.
The even positions in the given string contain uppercase Latin letters — the types of doors between the rooms. Thus, each even position i of the given string s contains an uppercase letter — the type of the door that leads from room i / 2 to room i / 2 + 1.
Output
Print the only integer — the minimum number of keys that Vitaly needs to buy to surely get from room one to room n.
Examples
Input
3
aAbB
Output
0
Input
4
aBaCaB
Output
3
Input
5
xYyXzZaZ
Output
2 | instruction | 0 | 105,656 | 9 | 211,312 |
Tags: greedy, hashing, strings
Correct Solution:
```
room = int(input())
desc = input().lower()
keys = desc[0::2]
way = desc[1::2]
keys_on_hands = ''
keys_to_buy = 0
for i in range(len(way)):
keys_on_hands += keys[i]
pos = keys_on_hands.find(way[i])
if pos == -1:
keys_to_buy += 1
else:
keys_on_hands = keys_on_hands[:pos] + keys_on_hands[pos+1:]
print('{0}'.format(keys_to_buy))
``` | output | 1 | 105,656 | 9 | 211,313 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After a hard day Vitaly got very hungry and he wants to eat his favorite potato pie. But it's not that simple. Vitaly is in the first room of the house with n room located in a line and numbered starting from one from left to right. You can go from the first room to the second room, from the second room to the third room and so on — you can go from the (n - 1)-th room to the n-th room. Thus, you can go to room x only from room x - 1.
The potato pie is located in the n-th room and Vitaly needs to go there.
Each pair of consecutive rooms has a door between them. In order to go to room x from room x - 1, you need to open the door between the rooms with the corresponding key.
In total the house has several types of doors (represented by uppercase Latin letters) and several types of keys (represented by lowercase Latin letters). The key of type t can open the door of type T if and only if t and T are the same letter, written in different cases. For example, key f can open door F.
Each of the first n - 1 rooms contains exactly one key of some type that Vitaly can use to get to next rooms. Once the door is open with some key, Vitaly won't get the key from the keyhole but he will immediately run into the next room. In other words, each key can open no more than one door.
Vitaly realizes that he may end up in some room without the key that opens the door to the next room. Before the start his run for the potato pie Vitaly can buy any number of keys of any type that is guaranteed to get to room n.
Given the plan of the house, Vitaly wants to know what is the minimum number of keys he needs to buy to surely get to the room n, which has a delicious potato pie. Write a program that will help Vitaly find out this number.
Input
The first line of the input contains a positive integer n (2 ≤ n ≤ 105) — the number of rooms in the house.
The second line of the input contains string s of length 2·n - 2. Let's number the elements of the string from left to right, starting from one.
The odd positions in the given string s contain lowercase Latin letters — the types of the keys that lie in the corresponding rooms. Thus, each odd position i of the given string s contains a lowercase Latin letter — the type of the key that lies in room number (i + 1) / 2.
The even positions in the given string contain uppercase Latin letters — the types of doors between the rooms. Thus, each even position i of the given string s contains an uppercase letter — the type of the door that leads from room i / 2 to room i / 2 + 1.
Output
Print the only integer — the minimum number of keys that Vitaly needs to buy to surely get from room one to room n.
Examples
Input
3
aAbB
Output
0
Input
4
aBaCaB
Output
3
Input
5
xYyXzZaZ
Output
2 | instruction | 0 | 105,657 | 9 | 211,314 |
Tags: greedy, hashing, strings
Correct Solution:
```
#_________________ Mukul Mohan Varshney _______________#
#Template
import sys
import os
import math
import copy
from math import gcd
from bisect import bisect
from io import BytesIO, IOBase
from math import sqrt,floor,factorial,gcd,log,ceil
from collections import deque,Counter,defaultdict
from itertools import permutations, combinations
import itertools
#define function
def Int(): return int(sys.stdin.readline())
def Mint(): return map(int,sys.stdin.readline().split())
def Lstr(): return list(sys.stdin.readline().strip())
def Str(): return sys.stdin.readline().strip()
def Mstr(): return map(str,sys.stdin.readline().strip().split())
def List(): return list(map(int,sys.stdin.readline().split()))
def Hash(): return dict()
def Mod(): return 1000000007
def Ncr(n,r,p): return ((fact[n])*((ifact[r]*ifact[n-r])%p))%p
def Most_frequent(list): return max(set(list), key = list.count)
def Mat2x2(n): return [List() for _ in range(n)]
def Lcm(x,y): return (x*y)//gcd(x,y)
def dtob(n): return bin(n).replace("0b","")
def btod(n): return int(n,2)
# Driver Code
def solution():
#for j in range(Int()):
n=Int()
a=Str()
m=len(a)
k=Hash()
i=0
count=0
while(i<m):
x=a[i].lower()
if(i%2==0):
if x in k:
k[x]+=1
else:
k[x]=1
else:
if x in k:
if(k[x]>0):
k[x]=k[x]-1
else:
count+=1
else:
count+=1
i+=1
print(count)
#Call the solve function
if __name__ == "__main__":
solution()
``` | output | 1 | 105,657 | 9 | 211,315 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After a hard day Vitaly got very hungry and he wants to eat his favorite potato pie. But it's not that simple. Vitaly is in the first room of the house with n room located in a line and numbered starting from one from left to right. You can go from the first room to the second room, from the second room to the third room and so on — you can go from the (n - 1)-th room to the n-th room. Thus, you can go to room x only from room x - 1.
The potato pie is located in the n-th room and Vitaly needs to go there.
Each pair of consecutive rooms has a door between them. In order to go to room x from room x - 1, you need to open the door between the rooms with the corresponding key.
In total the house has several types of doors (represented by uppercase Latin letters) and several types of keys (represented by lowercase Latin letters). The key of type t can open the door of type T if and only if t and T are the same letter, written in different cases. For example, key f can open door F.
Each of the first n - 1 rooms contains exactly one key of some type that Vitaly can use to get to next rooms. Once the door is open with some key, Vitaly won't get the key from the keyhole but he will immediately run into the next room. In other words, each key can open no more than one door.
Vitaly realizes that he may end up in some room without the key that opens the door to the next room. Before the start his run for the potato pie Vitaly can buy any number of keys of any type that is guaranteed to get to room n.
Given the plan of the house, Vitaly wants to know what is the minimum number of keys he needs to buy to surely get to the room n, which has a delicious potato pie. Write a program that will help Vitaly find out this number.
Input
The first line of the input contains a positive integer n (2 ≤ n ≤ 105) — the number of rooms in the house.
The second line of the input contains string s of length 2·n - 2. Let's number the elements of the string from left to right, starting from one.
The odd positions in the given string s contain lowercase Latin letters — the types of the keys that lie in the corresponding rooms. Thus, each odd position i of the given string s contains a lowercase Latin letter — the type of the key that lies in room number (i + 1) / 2.
The even positions in the given string contain uppercase Latin letters — the types of doors between the rooms. Thus, each even position i of the given string s contains an uppercase letter — the type of the door that leads from room i / 2 to room i / 2 + 1.
Output
Print the only integer — the minimum number of keys that Vitaly needs to buy to surely get from room one to room n.
Examples
Input
3
aAbB
Output
0
Input
4
aBaCaB
Output
3
Input
5
xYyXzZaZ
Output
2 | instruction | 0 | 105,658 | 9 | 211,316 |
Tags: greedy, hashing, strings
Correct Solution:
```
n = int(input())
b = input()
a = set()
d = dict()
count = 0
for i in range(2 * n - 2):
if i % 2 == 0:
if b[i] not in a:
a.add(b[i])
d[b[i]] = 1
else:
d[b[i]] += 1
else:
if b[i].lower() in a:
d[b[i].lower()] -= 1
if d[b[i].lower()] == 0:
a.remove(b[i].lower())
else:
count += 1
print(count)
``` | output | 1 | 105,658 | 9 | 211,317 |
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