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.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, …). Eating the sweet i at the d-th day will cause a sugar penalty of (d ⋅ a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 ≤ m ≤ n ≤ 200\ 000).
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 200\ 000).
Output
You have to output n integers x_1, x_2, …, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 ⋅ a_1 + 1 ⋅ a_4 + 2 ⋅ a_5 + 2 ⋅ a_3 + 3 ⋅ a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30. | instruction | 0 | 96,221 | 9 | 192,442 |
Tags: dp, greedy, math, sortings
Correct Solution:
```
N, M = map(int, input().split())
A = list(map(int, input().split()))
A.sort()
Ans = []
ans = 0
cumA = A[:]
for i in range(N):
j = i - M
if j>=0:
cumA[i] += cumA[j]
for a in cumA:
ans += a
Ans.append(ans)
print(" ".join(map(str, Ans)))
``` | output | 1 | 96,221 | 9 | 192,443 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, …). Eating the sweet i at the d-th day will cause a sugar penalty of (d ⋅ a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 ≤ m ≤ n ≤ 200\ 000).
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 200\ 000).
Output
You have to output n integers x_1, x_2, …, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 ⋅ a_1 + 1 ⋅ a_4 + 2 ⋅ a_5 + 2 ⋅ a_3 + 3 ⋅ a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30. | instruction | 0 | 96,222 | 9 | 192,444 |
Tags: dp, greedy, math, sortings
Correct Solution:
```
n, m = map(int, input().split())
a = [int(i) for i in input().split()]
a.sort()
cum = [0 for _ in range(n)]
dp1 = [0 for _ in range(n)]
dp2 = [0 for _ in range(n)]
cum[0] = a[0]
dp1[0] = a[0]
dp2[0] = a[0]
for i in range(1, n):
cum[i] = a[i] + cum[i - 1]
dp1[i] = cum[i]
if i - m >= 0:
dp1[i] -= cum[i - m]
dp1[i] += dp1[i - m] + dp2[i - m]
dp2[i] = a[i] + dp2[i - 1]
ans = " ".join([str(i) for i in dp1])
print(ans)
``` | output | 1 | 96,222 | 9 | 192,445 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, …). Eating the sweet i at the d-th day will cause a sugar penalty of (d ⋅ a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 ≤ m ≤ n ≤ 200\ 000).
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 200\ 000).
Output
You have to output n integers x_1, x_2, …, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 ⋅ a_1 + 1 ⋅ a_4 + 2 ⋅ a_5 + 2 ⋅ a_3 + 3 ⋅ a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30. | instruction | 0 | 96,223 | 9 | 192,446 |
Tags: dp, greedy, math, sortings
Correct Solution:
```
n,m = map(int,input().split())
a = list(map(int,input().split()))
a.sort()
state = [0]*n
ans = 0
for k in range(n):
state[k%m] += a[k]
ans += state[k%m]
print(ans,end=" ")
``` | output | 1 | 96,223 | 9 | 192,447 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, …). Eating the sweet i at the d-th day will cause a sugar penalty of (d ⋅ a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 ≤ m ≤ n ≤ 200\ 000).
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 200\ 000).
Output
You have to output n integers x_1, x_2, …, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 ⋅ a_1 + 1 ⋅ a_4 + 2 ⋅ a_5 + 2 ⋅ a_3 + 3 ⋅ a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30. | instruction | 0 | 96,224 | 9 | 192,448 |
Tags: dp, greedy, math, sortings
Correct Solution:
```
from collections import *
from functools import reduce
import sys
input = sys.stdin.readline
def factors(n):
return set(reduce(list.__add__,([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
def li():return [int(i) for i in input().rstrip('\n').split(' ')]
def st():return input().rstrip('\n')
def val():return int(input())
n,m = li()
a = li()
a.sort()
# print(a)
helparray = []
for i in range(m):
su = 0
for j in range(i,n,m):
su += a[j]
# print(i+1,su)
helparray.append(su)
helplen = len(helparray)
helparray = helparray[:n%m][::-1] + helparray[n%m:][::-1]
ind = -1
for i in range(m,n,1):
helparray.append(helparray[i-m] - a[ind])
ind-=1
# print(a)
# print(helparray)
helparray = [0] + helparray[::-1]
ans = []
for i in range(1,n+1):
helparray[i] += helparray[i-1]
ans.append(helparray[i])
print(*ans)
``` | output | 1 | 96,224 | 9 | 192,449 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, …). Eating the sweet i at the d-th day will cause a sugar penalty of (d ⋅ a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 ≤ m ≤ n ≤ 200\ 000).
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 200\ 000).
Output
You have to output n integers x_1, x_2, …, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 ⋅ a_1 + 1 ⋅ a_4 + 2 ⋅ a_5 + 2 ⋅ a_3 + 3 ⋅ a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30. | instruction | 0 | 96,225 | 9 | 192,450 |
Tags: dp, greedy, math, sortings
Correct Solution:
```
n, m = map(int, input().split())
a = [int(j) for j in input().split()]
a.sort()
prefix = [0] * n
suffix = [0] * n
# prefix
cur_prefix = 0
for i in range(n):
cur_prefix += a[i]
prefix[i] = cur_prefix
# suffix
suffix[m - 1] = prefix[m - 1]
for i in range(m, n):
suffix[i] = prefix[i] - prefix[i - m]
# print(a)
# print(prefix)
# print(suffix)
ans = []
for k in range(n):
k_look = k - m
if k_look < 0:
ans.append(prefix[k])
else:
ans.append(ans[k_look] + prefix[k_look] + suffix[k])
for i in ans:
print(i)
``` | output | 1 | 96,225 | 9 | 192,451 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, …). Eating the sweet i at the d-th day will cause a sugar penalty of (d ⋅ a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 ≤ m ≤ n ≤ 200\ 000).
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 200\ 000).
Output
You have to output n integers x_1, x_2, …, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 ⋅ a_1 + 1 ⋅ a_4 + 2 ⋅ a_5 + 2 ⋅ a_3 + 3 ⋅ a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30. | instruction | 0 | 96,226 | 9 | 192,452 |
Tags: dp, greedy, math, sortings
Correct Solution:
```
from sys import stdin
from collections import deque
from math import sqrt, floor, ceil, log, log2, log10, pi
def ii(): return int(stdin.readline())
def fi(): return float(stdin.readline())
def mi(): return map(int, stdin.readline().split())
def fmi(): return map(float, stdin.readline().split())
def li(): return list(mi())
def si(): return stdin.readline()
n, k=mi()
a=li()
a.sort()
#p=[]*k
for j in range(k):
for i in range(j+k, n, k):
#s+=a[i]
#p[j].append(s)
a[i]+=a[i-k]
#print(a)
x=[]
s=0
for i in range(n):
s+=a[i]
x.append(s)
print(*x)
``` | output | 1 | 96,226 | 9 | 192,453 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, …). Eating the sweet i at the d-th day will cause a sugar penalty of (d ⋅ a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 ≤ m ≤ n ≤ 200\ 000).
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 200\ 000).
Output
You have to output n integers x_1, x_2, …, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 ⋅ a_1 + 1 ⋅ a_4 + 2 ⋅ a_5 + 2 ⋅ a_3 + 3 ⋅ a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30. | instruction | 0 | 96,227 | 9 | 192,454 |
Tags: dp, greedy, math, sortings
Correct Solution:
```
for _ in range(1):
n,m=list(map(int,input().split()))
a=list(map(int,input().split()))
a.sort()
ans=[]
su=[a[0]]
for i in range(1,m):
su.append(a[i]+su[-1])
for j in range(m,n):
su.append(su[-1]+a[j])
#print(a[i-m])
d=1
#print(su)
for i in range(n):
s=su[i]
d=1
if i-m>=0:
#print(s,ans[i-m],su[i])
s+=ans[i-m]
ans.append(s)
#del(ans[0])
print(*ans)
``` | output | 1 | 96,227 | 9 | 192,455 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, …). Eating the sweet i at the d-th day will cause a sugar penalty of (d ⋅ a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 ≤ m ≤ n ≤ 200\ 000).
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 200\ 000).
Output
You have to output n integers x_1, x_2, …, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 ⋅ a_1 + 1 ⋅ a_4 + 2 ⋅ a_5 + 2 ⋅ a_3 + 3 ⋅ a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30. | instruction | 0 | 96,228 | 9 | 192,456 |
Tags: dp, greedy, math, sortings
Correct Solution:
```
n,m=map(int,input().split())
z=list(map(int,input().split()))
z.sort()
ans=[z[0]]
for i in range(1,len(z)):
ans.append(ans[-1]+z[i])
fin=[]
for i in range(len(z)):
if(i<=m-1):
fin.append(ans[i])
else:
fin.append(ans[i]+fin[i-m])
print(*fin)
``` | output | 1 | 96,228 | 9 | 192,457 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, …). Eating the sweet i at the d-th day will cause a sugar penalty of (d ⋅ a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 ≤ m ≤ n ≤ 200\ 000).
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 200\ 000).
Output
You have to output n integers x_1, x_2, …, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 ⋅ a_1 + 1 ⋅ a_4 + 2 ⋅ a_5 + 2 ⋅ a_3 + 3 ⋅ a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30.
Submitted Solution:
```
n, m = [int(i) for i in input().split()]
d = [int(i) for i in input().split()]
d.sort()
ans = []
tot = 0
psum = [0]*m
for i in range(n):
md = i % m
psum[md] += d[i]
tot += psum[md]
ans.append(tot)
print(' '.join(map(str,ans)))
``` | instruction | 0 | 96,229 | 9 | 192,458 |
Yes | output | 1 | 96,229 | 9 | 192,459 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, …). Eating the sweet i at the d-th day will cause a sugar penalty of (d ⋅ a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 ≤ m ≤ n ≤ 200\ 000).
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 200\ 000).
Output
You have to output n integers x_1, x_2, …, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 ⋅ a_1 + 1 ⋅ a_4 + 2 ⋅ a_5 + 2 ⋅ a_3 + 3 ⋅ a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30.
Submitted Solution:
```
import os
import sys
from bisect import bisect_right
from fractions import Fraction
def main():
n , m = map(int , input().split())
arr=list(map(int , input().split()))
tot=[0]*(n)
cur=0
arr.sort()
for i in range(n):
cur=cur+arr[i]
tot[i]=cur
if(i>=m):
tot[i]+=tot[i-m]
print(*tot)
if __name__ == "__main__":
main()
``` | instruction | 0 | 96,230 | 9 | 192,460 |
Yes | output | 1 | 96,230 | 9 | 192,461 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, …). Eating the sweet i at the d-th day will cause a sugar penalty of (d ⋅ a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 ≤ m ≤ n ≤ 200\ 000).
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 200\ 000).
Output
You have to output n integers x_1, x_2, …, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 ⋅ a_1 + 1 ⋅ a_4 + 2 ⋅ a_5 + 2 ⋅ a_3 + 3 ⋅ a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30.
Submitted Solution:
```
import sys
input = lambda: sys.stdin.readline().strip("\r\n")
n, m = map(int, input().split())
a = sorted(list(map(int, input().split())))
dp = [0] * n
ans = 0
for i in range(n):
ans += a[i]
if i < m:
dp[i] = ans
else:
dp[i] = ans + dp[i-m]
print(*dp)
``` | instruction | 0 | 96,231 | 9 | 192,462 |
Yes | output | 1 | 96,231 | 9 | 192,463 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, …). Eating the sweet i at the d-th day will cause a sugar penalty of (d ⋅ a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 ≤ m ≤ n ≤ 200\ 000).
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 200\ 000).
Output
You have to output n integers x_1, x_2, …, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 ⋅ a_1 + 1 ⋅ a_4 + 2 ⋅ a_5 + 2 ⋅ a_3 + 3 ⋅ a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30.
Submitted Solution:
```
#Ashish_Sagar
n,m=map(int,input().split())
l=list(map(int,input().split()))
l.sort()
ans=[]
prefix=[0]
for i in range(1,n+1):
prefix.append(prefix[-1]+l[i-1])
for i in range(n):
if i<m:
ans.append(prefix[i+1])
else:
ans.append(prefix[i+1]+ans[i-m])
print(*ans)
``` | instruction | 0 | 96,232 | 9 | 192,464 |
Yes | output | 1 | 96,232 | 9 | 192,465 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, …). Eating the sweet i at the d-th day will cause a sugar penalty of (d ⋅ a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 ≤ m ≤ n ≤ 200\ 000).
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 200\ 000).
Output
You have to output n integers x_1, x_2, …, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 ⋅ a_1 + 1 ⋅ a_4 + 2 ⋅ a_5 + 2 ⋅ a_3 + 3 ⋅ a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30.
Submitted Solution:
```
# from sys import stdout
n, m = tuple(map(int, input().split()))
lis = list(map(int, input().split()))
if n==m and m==1:
print(*lis)
exit()
# en = n-1
# di = dict()
# k = 1; eaten = [[2]]
# di[k] = eaten; k = 2; eaten = [[3, 2]]
# di[k] = eaten; k = 3; eaten = [[4, 3], [2]]
# di[k] = eaten; k = 4; eaten = [[4, 4], [3, 2]]
# di[k] = eaten; k = 5; eaten = [[6, 4], [4, 3], [2]]
# di[k] = eaten; k = 6; eaten = [[6, 6], [4, 4], [3, 2]]
# di[k] = eaten; k = 7; eaten = [[7, 6], [6, 4], [4, 3], [2]]
# di[k] = eaten; k = 8; eaten = [[8, 7], [6, 6], [4, 4], [3, 2]]
# di[k] = eaten; k = 9; eaten = [[19, 8], [7, 6], [6, 4], [4, 3], [2]]
# di[k] = eaten
# for k in di.keys():
# #k = [1,2, .., m-1]
# # st = en-k
# li = di[k]
# for i in range(len(li)):
# posi = i+1
# for j in li[i]:
# # print(j,'*',posi, sep='', end = ' + ')
# # print()
# eaten = [[]]
lis.sort(reverse=True)
acc = [sum(lis)]
for i in range(1, n):
acc.append(acc[-1] - lis[i-1])
ans = acc[n-1:n-m-1:-1]
# print('ans =', ans)
# print('lis =', lis)
# print('acc =', acc)
for i in range(m, n):
ans.append(acc[-i-1] + ans[i-m])
# print(acc[-i-1], ans[i-m])
print(*ans)
``` | instruction | 0 | 96,233 | 9 | 192,466 |
No | output | 1 | 96,233 | 9 | 192,467 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, …). Eating the sweet i at the d-th day will cause a sugar penalty of (d ⋅ a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 ≤ m ≤ n ≤ 200\ 000).
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 200\ 000).
Output
You have to output n integers x_1, x_2, …, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 ⋅ a_1 + 1 ⋅ a_4 + 2 ⋅ a_5 + 2 ⋅ a_3 + 3 ⋅ a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30.
Submitted Solution:
```
n, m = map(int, input().split())
l = list(map(int, input().split()))
l.sort()
mass = [0]
for i in range(n):
mass += [mass[-1] + l[i]]
for i in range(1, n + 1):
if i <= m:
print(mass[i])
else:
k = i // m
r = i % m
res = 0
if r > 0:
res += (mass[r]) * (k + 1)
for y in range(k):
res += (mass[y * m + m + r] - mass[y * m + r]) * (k - y)
print(res, i, k, r)
``` | instruction | 0 | 96,234 | 9 | 192,468 |
No | output | 1 | 96,234 | 9 | 192,469 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, …). Eating the sweet i at the d-th day will cause a sugar penalty of (d ⋅ a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 ≤ m ≤ n ≤ 200\ 000).
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 200\ 000).
Output
You have to output n integers x_1, x_2, …, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 ⋅ a_1 + 1 ⋅ a_4 + 2 ⋅ a_5 + 2 ⋅ a_3 + 3 ⋅ a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30.
Submitted Solution:
```
n, m = map(int, input().split())
l1 = list(map(int, input().split()))
l1.sort()
dp = [0] * m
answer = []
for i in range(n):
dp[i % m]=dp[i%m] + l1[i]
answer.append(sum(dp))
for i in range(n):
print(answer[i], end=" ")
``` | instruction | 0 | 96,235 | 9 | 192,470 |
No | output | 1 | 96,235 | 9 | 192,471 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tsumugi brought n delicious sweets to the Light Music Club. They are numbered from 1 to n, where the i-th sweet has a sugar concentration described by an integer a_i.
Yui loves sweets, but she can eat at most m sweets each day for health reasons.
Days are 1-indexed (numbered 1, 2, 3, …). Eating the sweet i at the d-th day will cause a sugar penalty of (d ⋅ a_i), as sweets become more sugary with time. A sweet can be eaten at most once.
The total sugar penalty will be the sum of the individual penalties of each sweet eaten.
Suppose that Yui chooses exactly k sweets, and eats them in any order she wants. What is the minimum total sugar penalty she can get?
Since Yui is an undecided girl, she wants you to answer this question for every value of k between 1 and n.
Input
The first line contains two integers n and m (1 ≤ m ≤ n ≤ 200\ 000).
The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 200\ 000).
Output
You have to output n integers x_1, x_2, …, x_n on a single line, separed by spaces, where x_k is the minimum total sugar penalty Yui can get if she eats exactly k sweets.
Examples
Input
9 2
6 19 3 4 4 2 6 7 8
Output
2 5 11 18 30 43 62 83 121
Input
1 1
7
Output
7
Note
Let's analyze the answer for k = 5 in the first example. Here is one of the possible ways to eat 5 sweets that minimize total sugar penalty:
* Day 1: sweets 1 and 4
* Day 2: sweets 5 and 3
* Day 3 : sweet 6
Total penalty is 1 ⋅ a_1 + 1 ⋅ a_4 + 2 ⋅ a_5 + 2 ⋅ a_3 + 3 ⋅ a_6 = 6 + 4 + 8 + 6 + 6 = 30. We can prove that it's the minimum total sugar penalty Yui can achieve if she eats 5 sweets, hence x_5 = 30.
Submitted Solution:
```
f = lambda: list(map(int, input().split()))
n, max_sweet_per_days = f()
sweets = f()
sweets.sort()
sweet_penalty = []
sweet_penalty.append(sweets[0])
for k in range(2, n + 1):
sweet_penalty.append(sweet_penalty[-1] + ((k // max_sweet_per_days) + 1) * sweets[k - 1])
print(" ".join(list(map(str, sweet_penalty))))
``` | instruction | 0 | 96,236 | 9 | 192,472 |
No | output | 1 | 96,236 | 9 | 192,473 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub helps his grandfather at the farm. Today he must milk the cows. There are n cows sitting in a row, numbered from 1 to n from left to right. Each cow is either facing to the left or facing to the right. When Iahub milks a cow, all the cows that see the current cow get scared and lose one unit of the quantity of milk that they can give. A cow facing left sees all the cows with lower indices than her index, and a cow facing right sees all the cows with higher indices than her index. A cow that got scared once can get scared again (and lose one more unit of milk). A cow that has been milked once cannot get scared and lose any more milk. You can assume that a cow never loses all the milk she can give (a cow gives an infinitely amount of milk).
Iahub can decide the order in which he milks the cows. But he must milk each cow exactly once. Iahub wants to lose as little milk as possible. Print the minimum amount of milk that is lost.
Input
The first line contains an integer n (1 ≤ n ≤ 200000). The second line contains n integers a1, a2, ..., an, where ai is 0 if the cow number i is facing left, and 1 if it is facing right.
Output
Print a single integer, the minimum amount of lost milk.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
4
0 0 1 0
Output
1
Input
5
1 0 1 0 1
Output
3
Note
In the first sample Iahub milks the cows in the following order: cow 3, cow 4, cow 2, cow 1. When he milks cow 3, cow 4 loses 1 unit of milk. After that, no more milk is lost.
Submitted Solution:
```
#in the name of god
#Mr_Rubick
'''n=int(input())
a=list(map(int,input().split()))
cnt,s=0,[0]*n
s[n-1]=1*(a[n-1]==0)
for i in range(n-2,-1,-1): s[i]=s[i+1]+1*(a[i]==0)
for i in range(n): cnt+=s[i]*(a[i]==1)
print(cnt)'''
n=int(input())
a=list(map(int,input().split()))
cnt1,cnt2=0,0
for i in range (n):
if a[i]==0:cnt1+=1
else:cnt2+=1
if abs(cnt1-cnt2)<=1:print(min(cnt1,cnt2))
else:print(min(cnt1,cnt2))
``` | instruction | 0 | 96,478 | 9 | 192,956 |
No | output | 1 | 96,478 | 9 | 192,957 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub helps his grandfather at the farm. Today he must milk the cows. There are n cows sitting in a row, numbered from 1 to n from left to right. Each cow is either facing to the left or facing to the right. When Iahub milks a cow, all the cows that see the current cow get scared and lose one unit of the quantity of milk that they can give. A cow facing left sees all the cows with lower indices than her index, and a cow facing right sees all the cows with higher indices than her index. A cow that got scared once can get scared again (and lose one more unit of milk). A cow that has been milked once cannot get scared and lose any more milk. You can assume that a cow never loses all the milk she can give (a cow gives an infinitely amount of milk).
Iahub can decide the order in which he milks the cows. But he must milk each cow exactly once. Iahub wants to lose as little milk as possible. Print the minimum amount of milk that is lost.
Input
The first line contains an integer n (1 ≤ n ≤ 200000). The second line contains n integers a1, a2, ..., an, where ai is 0 if the cow number i is facing left, and 1 if it is facing right.
Output
Print a single integer, the minimum amount of lost milk.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
4
0 0 1 0
Output
1
Input
5
1 0 1 0 1
Output
3
Note
In the first sample Iahub milks the cows in the following order: cow 3, cow 4, cow 2, cow 1. When he milks cow 3, cow 4 loses 1 unit of milk. After that, no more milk is lost.
Submitted Solution:
```
def cows(n, lst):
zeros, result = 0, 0
for i in range(n - 1, -1, 1):
if lst[i] == 0:
zeros += 1
else:
result += zeros
return result
m = int(input())
a = [int(j) for j in input().split()]
print(cows(m, a))
``` | instruction | 0 | 96,479 | 9 | 192,958 |
No | output | 1 | 96,479 | 9 | 192,959 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub helps his grandfather at the farm. Today he must milk the cows. There are n cows sitting in a row, numbered from 1 to n from left to right. Each cow is either facing to the left or facing to the right. When Iahub milks a cow, all the cows that see the current cow get scared and lose one unit of the quantity of milk that they can give. A cow facing left sees all the cows with lower indices than her index, and a cow facing right sees all the cows with higher indices than her index. A cow that got scared once can get scared again (and lose one more unit of milk). A cow that has been milked once cannot get scared and lose any more milk. You can assume that a cow never loses all the milk she can give (a cow gives an infinitely amount of milk).
Iahub can decide the order in which he milks the cows. But he must milk each cow exactly once. Iahub wants to lose as little milk as possible. Print the minimum amount of milk that is lost.
Input
The first line contains an integer n (1 ≤ n ≤ 200000). The second line contains n integers a1, a2, ..., an, where ai is 0 if the cow number i is facing left, and 1 if it is facing right.
Output
Print a single integer, the minimum amount of lost milk.
Please, do not write the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
4
0 0 1 0
Output
1
Input
5
1 0 1 0 1
Output
3
Note
In the first sample Iahub milks the cows in the following order: cow 3, cow 4, cow 2, cow 1. When he milks cow 3, cow 4 loses 1 unit of milk. After that, no more milk is lost.
Submitted Solution:
```
#in the name of god
#Mr_Rubick
n=int(input())
a=list(map(int,input().split()))
cnt=0
s=[0]*n
if a[n-1]==0:
s[n-1]=1*a[n-1]
for i in range(n-2,-1,-1):
if a[i]==0:
s[i]=s[i+1]+1*a[i]
for i in range(n):
if a[i]==1:
cnt+=s[i]*a[i]
print(cnt)
``` | instruction | 0 | 96,480 | 9 | 192,960 |
No | output | 1 | 96,480 | 9 | 192,961 |
Provide tags and a correct Python 3 solution for this coding contest problem.
[The Duck song](https://www.youtube.com/watch?v=MtN1YnoL46Q)
For simplicity, we'll assume that there are only three types of grapes: green grapes, purple grapes and black grapes.
Andrew, Dmitry and Michal are all grapes' lovers, however their preferences of grapes are different. To make all of them happy, the following should happen:
* Andrew, Dmitry and Michal should eat at least x, y and z grapes, respectively.
* Andrew has an extreme affinity for green grapes, thus he will eat green grapes and green grapes only.
* On the other hand, Dmitry is not a fan of black grapes — any types of grapes except black would do for him. In other words, Dmitry can eat green and purple grapes.
* Michal has a common taste — he enjoys grapes in general and will be pleased with any types of grapes, as long as the quantity is sufficient.
Knowing that his friends are so fond of grapes, Aki decided to host a grape party with them. He has prepared a box with a green grapes, b purple grapes and c black grapes.
However, Aki isn't sure if the box he prepared contains enough grapes to make everyone happy. Can you please find out whether it's possible to distribute grapes so that everyone is happy or Aki has to buy some more grapes?
It is not required to distribute all the grapes, so it's possible that some of them will remain unused.
Input
The first line contains three integers x, y and z (1 ≤ x, y, z ≤ 10^5) — the number of grapes Andrew, Dmitry and Michal want to eat.
The second line contains three integers a, b, c (1 ≤ a, b, c ≤ 10^5) — the number of green, purple and black grapes in the box.
Output
If there is a grape distribution that allows everyone to be happy, print "YES", otherwise print "NO".
Examples
Input
1 6 2
4 3 3
Output
YES
Input
5 1 1
4 3 2
Output
NO
Note
In the first example, there is only one possible distribution:
Andrew should take 1 green grape, Dmitry should take 3 remaining green grapes and 3 purple grapes, and Michal will take 2 out of 3 available black grapes.
In the second test, there is no possible distribution, since Andrew is not be able to eat enough green grapes. :( | instruction | 0 | 96,976 | 9 | 193,952 |
Tags: brute force, greedy, implementation
Correct Solution:
```
if __name__ == "__main__":
i1 = input()
i2 = input()
x,y,z = map(int, i1.split())
a,b,c = map(int, i2.split())
if x > a:
print('NO')
else:
a -= x
if a + b < y:
print('NO')
else:
if a + b + c < y + z:
print('NO')
else:
print('YES')
``` | output | 1 | 96,976 | 9 | 193,953 |
Provide tags and a correct Python 3 solution for this coding contest problem.
[The Duck song](https://www.youtube.com/watch?v=MtN1YnoL46Q)
For simplicity, we'll assume that there are only three types of grapes: green grapes, purple grapes and black grapes.
Andrew, Dmitry and Michal are all grapes' lovers, however their preferences of grapes are different. To make all of them happy, the following should happen:
* Andrew, Dmitry and Michal should eat at least x, y and z grapes, respectively.
* Andrew has an extreme affinity for green grapes, thus he will eat green grapes and green grapes only.
* On the other hand, Dmitry is not a fan of black grapes — any types of grapes except black would do for him. In other words, Dmitry can eat green and purple grapes.
* Michal has a common taste — he enjoys grapes in general and will be pleased with any types of grapes, as long as the quantity is sufficient.
Knowing that his friends are so fond of grapes, Aki decided to host a grape party with them. He has prepared a box with a green grapes, b purple grapes and c black grapes.
However, Aki isn't sure if the box he prepared contains enough grapes to make everyone happy. Can you please find out whether it's possible to distribute grapes so that everyone is happy or Aki has to buy some more grapes?
It is not required to distribute all the grapes, so it's possible that some of them will remain unused.
Input
The first line contains three integers x, y and z (1 ≤ x, y, z ≤ 10^5) — the number of grapes Andrew, Dmitry and Michal want to eat.
The second line contains three integers a, b, c (1 ≤ a, b, c ≤ 10^5) — the number of green, purple and black grapes in the box.
Output
If there is a grape distribution that allows everyone to be happy, print "YES", otherwise print "NO".
Examples
Input
1 6 2
4 3 3
Output
YES
Input
5 1 1
4 3 2
Output
NO
Note
In the first example, there is only one possible distribution:
Andrew should take 1 green grape, Dmitry should take 3 remaining green grapes and 3 purple grapes, and Michal will take 2 out of 3 available black grapes.
In the second test, there is no possible distribution, since Andrew is not be able to eat enough green grapes. :( | instruction | 0 | 96,977 | 9 | 193,954 |
Tags: brute force, greedy, implementation
Correct Solution:
```
x,y,z = map(int,input().split())
a,b,c = map(int,input().split())
if((a>=x) and (b+a-x>=y) and (c+(b+(a-x)-y)>=z)):
print("YES")
else:
print("NO")
``` | output | 1 | 96,977 | 9 | 193,955 |
Provide tags and a correct Python 3 solution for this coding contest problem.
[The Duck song](https://www.youtube.com/watch?v=MtN1YnoL46Q)
For simplicity, we'll assume that there are only three types of grapes: green grapes, purple grapes and black grapes.
Andrew, Dmitry and Michal are all grapes' lovers, however their preferences of grapes are different. To make all of them happy, the following should happen:
* Andrew, Dmitry and Michal should eat at least x, y and z grapes, respectively.
* Andrew has an extreme affinity for green grapes, thus he will eat green grapes and green grapes only.
* On the other hand, Dmitry is not a fan of black grapes — any types of grapes except black would do for him. In other words, Dmitry can eat green and purple grapes.
* Michal has a common taste — he enjoys grapes in general and will be pleased with any types of grapes, as long as the quantity is sufficient.
Knowing that his friends are so fond of grapes, Aki decided to host a grape party with them. He has prepared a box with a green grapes, b purple grapes and c black grapes.
However, Aki isn't sure if the box he prepared contains enough grapes to make everyone happy. Can you please find out whether it's possible to distribute grapes so that everyone is happy or Aki has to buy some more grapes?
It is not required to distribute all the grapes, so it's possible that some of them will remain unused.
Input
The first line contains three integers x, y and z (1 ≤ x, y, z ≤ 10^5) — the number of grapes Andrew, Dmitry and Michal want to eat.
The second line contains three integers a, b, c (1 ≤ a, b, c ≤ 10^5) — the number of green, purple and black grapes in the box.
Output
If there is a grape distribution that allows everyone to be happy, print "YES", otherwise print "NO".
Examples
Input
1 6 2
4 3 3
Output
YES
Input
5 1 1
4 3 2
Output
NO
Note
In the first example, there is only one possible distribution:
Andrew should take 1 green grape, Dmitry should take 3 remaining green grapes and 3 purple grapes, and Michal will take 2 out of 3 available black grapes.
In the second test, there is no possible distribution, since Andrew is not be able to eat enough green grapes. :( | instruction | 0 | 96,978 | 9 | 193,956 |
Tags: brute force, greedy, implementation
Correct Solution:
```
a, d, m = map(int, input().split())
g, p, b = map(int, input().split())
n = g+p+b
if g < a or d > g+p-a or n < a+d+m:
print('NO')
else:
print('YES')
``` | output | 1 | 96,978 | 9 | 193,957 |
Provide tags and a correct Python 3 solution for this coding contest problem.
[The Duck song](https://www.youtube.com/watch?v=MtN1YnoL46Q)
For simplicity, we'll assume that there are only three types of grapes: green grapes, purple grapes and black grapes.
Andrew, Dmitry and Michal are all grapes' lovers, however their preferences of grapes are different. To make all of them happy, the following should happen:
* Andrew, Dmitry and Michal should eat at least x, y and z grapes, respectively.
* Andrew has an extreme affinity for green grapes, thus he will eat green grapes and green grapes only.
* On the other hand, Dmitry is not a fan of black grapes — any types of grapes except black would do for him. In other words, Dmitry can eat green and purple grapes.
* Michal has a common taste — he enjoys grapes in general and will be pleased with any types of grapes, as long as the quantity is sufficient.
Knowing that his friends are so fond of grapes, Aki decided to host a grape party with them. He has prepared a box with a green grapes, b purple grapes and c black grapes.
However, Aki isn't sure if the box he prepared contains enough grapes to make everyone happy. Can you please find out whether it's possible to distribute grapes so that everyone is happy or Aki has to buy some more grapes?
It is not required to distribute all the grapes, so it's possible that some of them will remain unused.
Input
The first line contains three integers x, y and z (1 ≤ x, y, z ≤ 10^5) — the number of grapes Andrew, Dmitry and Michal want to eat.
The second line contains three integers a, b, c (1 ≤ a, b, c ≤ 10^5) — the number of green, purple and black grapes in the box.
Output
If there is a grape distribution that allows everyone to be happy, print "YES", otherwise print "NO".
Examples
Input
1 6 2
4 3 3
Output
YES
Input
5 1 1
4 3 2
Output
NO
Note
In the first example, there is only one possible distribution:
Andrew should take 1 green grape, Dmitry should take 3 remaining green grapes and 3 purple grapes, and Michal will take 2 out of 3 available black grapes.
In the second test, there is no possible distribution, since Andrew is not be able to eat enough green grapes. :( | instruction | 0 | 96,979 | 9 | 193,958 |
Tags: brute force, greedy, implementation
Correct Solution:
```
a,b,c=map(int,input().split())
x,y,z=map(int,input().split())
if x>=a and (x-a)+y>=b and x-a+y-b+z>=c:
print("YES")
else:
print("NO")
``` | output | 1 | 96,979 | 9 | 193,959 |
Provide tags and a correct Python 3 solution for this coding contest problem.
[The Duck song](https://www.youtube.com/watch?v=MtN1YnoL46Q)
For simplicity, we'll assume that there are only three types of grapes: green grapes, purple grapes and black grapes.
Andrew, Dmitry and Michal are all grapes' lovers, however their preferences of grapes are different. To make all of them happy, the following should happen:
* Andrew, Dmitry and Michal should eat at least x, y and z grapes, respectively.
* Andrew has an extreme affinity for green grapes, thus he will eat green grapes and green grapes only.
* On the other hand, Dmitry is not a fan of black grapes — any types of grapes except black would do for him. In other words, Dmitry can eat green and purple grapes.
* Michal has a common taste — he enjoys grapes in general and will be pleased with any types of grapes, as long as the quantity is sufficient.
Knowing that his friends are so fond of grapes, Aki decided to host a grape party with them. He has prepared a box with a green grapes, b purple grapes and c black grapes.
However, Aki isn't sure if the box he prepared contains enough grapes to make everyone happy. Can you please find out whether it's possible to distribute grapes so that everyone is happy or Aki has to buy some more grapes?
It is not required to distribute all the grapes, so it's possible that some of them will remain unused.
Input
The first line contains three integers x, y and z (1 ≤ x, y, z ≤ 10^5) — the number of grapes Andrew, Dmitry and Michal want to eat.
The second line contains three integers a, b, c (1 ≤ a, b, c ≤ 10^5) — the number of green, purple and black grapes in the box.
Output
If there is a grape distribution that allows everyone to be happy, print "YES", otherwise print "NO".
Examples
Input
1 6 2
4 3 3
Output
YES
Input
5 1 1
4 3 2
Output
NO
Note
In the first example, there is only one possible distribution:
Andrew should take 1 green grape, Dmitry should take 3 remaining green grapes and 3 purple grapes, and Michal will take 2 out of 3 available black grapes.
In the second test, there is no possible distribution, since Andrew is not be able to eat enough green grapes. :( | instruction | 0 | 96,980 | 9 | 193,960 |
Tags: brute force, greedy, implementation
Correct Solution:
```
x, y, z = list(map(int, input().split()))
a, b, c = list(map(int, input().split()))
if x <= a and y <= (a-x) + b and z <= a + b + c - x -y:
print('YES')
else:
print('NO')
``` | output | 1 | 96,980 | 9 | 193,961 |
Provide tags and a correct Python 3 solution for this coding contest problem.
[The Duck song](https://www.youtube.com/watch?v=MtN1YnoL46Q)
For simplicity, we'll assume that there are only three types of grapes: green grapes, purple grapes and black grapes.
Andrew, Dmitry and Michal are all grapes' lovers, however their preferences of grapes are different. To make all of them happy, the following should happen:
* Andrew, Dmitry and Michal should eat at least x, y and z grapes, respectively.
* Andrew has an extreme affinity for green grapes, thus he will eat green grapes and green grapes only.
* On the other hand, Dmitry is not a fan of black grapes — any types of grapes except black would do for him. In other words, Dmitry can eat green and purple grapes.
* Michal has a common taste — he enjoys grapes in general and will be pleased with any types of grapes, as long as the quantity is sufficient.
Knowing that his friends are so fond of grapes, Aki decided to host a grape party with them. He has prepared a box with a green grapes, b purple grapes and c black grapes.
However, Aki isn't sure if the box he prepared contains enough grapes to make everyone happy. Can you please find out whether it's possible to distribute grapes so that everyone is happy or Aki has to buy some more grapes?
It is not required to distribute all the grapes, so it's possible that some of them will remain unused.
Input
The first line contains three integers x, y and z (1 ≤ x, y, z ≤ 10^5) — the number of grapes Andrew, Dmitry and Michal want to eat.
The second line contains three integers a, b, c (1 ≤ a, b, c ≤ 10^5) — the number of green, purple and black grapes in the box.
Output
If there is a grape distribution that allows everyone to be happy, print "YES", otherwise print "NO".
Examples
Input
1 6 2
4 3 3
Output
YES
Input
5 1 1
4 3 2
Output
NO
Note
In the first example, there is only one possible distribution:
Andrew should take 1 green grape, Dmitry should take 3 remaining green grapes and 3 purple grapes, and Michal will take 2 out of 3 available black grapes.
In the second test, there is no possible distribution, since Andrew is not be able to eat enough green grapes. :( | instruction | 0 | 96,981 | 9 | 193,962 |
Tags: brute force, greedy, implementation
Correct Solution:
```
x,y,z=map(int,input().split())
a,b,c=map(int,input().split())
bl=True
if(x>a):
bl=False
if(x+y)>(a+b):
bl=False
if(x+y+z)>(a+b+c):
bl=False
if(bl):
print("YES")
else:
print("NO")
``` | output | 1 | 96,981 | 9 | 193,963 |
Provide tags and a correct Python 3 solution for this coding contest problem.
[The Duck song](https://www.youtube.com/watch?v=MtN1YnoL46Q)
For simplicity, we'll assume that there are only three types of grapes: green grapes, purple grapes and black grapes.
Andrew, Dmitry and Michal are all grapes' lovers, however their preferences of grapes are different. To make all of them happy, the following should happen:
* Andrew, Dmitry and Michal should eat at least x, y and z grapes, respectively.
* Andrew has an extreme affinity for green grapes, thus he will eat green grapes and green grapes only.
* On the other hand, Dmitry is not a fan of black grapes — any types of grapes except black would do for him. In other words, Dmitry can eat green and purple grapes.
* Michal has a common taste — he enjoys grapes in general and will be pleased with any types of grapes, as long as the quantity is sufficient.
Knowing that his friends are so fond of grapes, Aki decided to host a grape party with them. He has prepared a box with a green grapes, b purple grapes and c black grapes.
However, Aki isn't sure if the box he prepared contains enough grapes to make everyone happy. Can you please find out whether it's possible to distribute grapes so that everyone is happy or Aki has to buy some more grapes?
It is not required to distribute all the grapes, so it's possible that some of them will remain unused.
Input
The first line contains three integers x, y and z (1 ≤ x, y, z ≤ 10^5) — the number of grapes Andrew, Dmitry and Michal want to eat.
The second line contains three integers a, b, c (1 ≤ a, b, c ≤ 10^5) — the number of green, purple and black grapes in the box.
Output
If there is a grape distribution that allows everyone to be happy, print "YES", otherwise print "NO".
Examples
Input
1 6 2
4 3 3
Output
YES
Input
5 1 1
4 3 2
Output
NO
Note
In the first example, there is only one possible distribution:
Andrew should take 1 green grape, Dmitry should take 3 remaining green grapes and 3 purple grapes, and Michal will take 2 out of 3 available black grapes.
In the second test, there is no possible distribution, since Andrew is not be able to eat enough green grapes. :( | instruction | 0 | 96,982 | 9 | 193,964 |
Tags: brute force, greedy, implementation
Correct Solution:
```
import sys
x, y, z = map(int, input().split())
a, b, c = map(int, input().split())
count = a + b + c
if(x > a):
print("NO")
sys.exit()
a -= x
count -= x
if(y > (a + b)):
print("NO")
sys.exit()
count -= y
if(z > count):
print("NO")
sys.exit()
print("YES")
``` | output | 1 | 96,982 | 9 | 193,965 |
Provide tags and a correct Python 3 solution for this coding contest problem.
[The Duck song](https://www.youtube.com/watch?v=MtN1YnoL46Q)
For simplicity, we'll assume that there are only three types of grapes: green grapes, purple grapes and black grapes.
Andrew, Dmitry and Michal are all grapes' lovers, however their preferences of grapes are different. To make all of them happy, the following should happen:
* Andrew, Dmitry and Michal should eat at least x, y and z grapes, respectively.
* Andrew has an extreme affinity for green grapes, thus he will eat green grapes and green grapes only.
* On the other hand, Dmitry is not a fan of black grapes — any types of grapes except black would do for him. In other words, Dmitry can eat green and purple grapes.
* Michal has a common taste — he enjoys grapes in general and will be pleased with any types of grapes, as long as the quantity is sufficient.
Knowing that his friends are so fond of grapes, Aki decided to host a grape party with them. He has prepared a box with a green grapes, b purple grapes and c black grapes.
However, Aki isn't sure if the box he prepared contains enough grapes to make everyone happy. Can you please find out whether it's possible to distribute grapes so that everyone is happy or Aki has to buy some more grapes?
It is not required to distribute all the grapes, so it's possible that some of them will remain unused.
Input
The first line contains three integers x, y and z (1 ≤ x, y, z ≤ 10^5) — the number of grapes Andrew, Dmitry and Michal want to eat.
The second line contains three integers a, b, c (1 ≤ a, b, c ≤ 10^5) — the number of green, purple and black grapes in the box.
Output
If there is a grape distribution that allows everyone to be happy, print "YES", otherwise print "NO".
Examples
Input
1 6 2
4 3 3
Output
YES
Input
5 1 1
4 3 2
Output
NO
Note
In the first example, there is only one possible distribution:
Andrew should take 1 green grape, Dmitry should take 3 remaining green grapes and 3 purple grapes, and Michal will take 2 out of 3 available black grapes.
In the second test, there is no possible distribution, since Andrew is not be able to eat enough green grapes. :( | instruction | 0 | 96,983 | 9 | 193,966 |
Tags: brute force, greedy, implementation
Correct Solution:
```
x, y, z = map(int, input().split())
a, b, c = map(int, input().split())
if x + y + z > a + b + c:
print("NO")
exit()
if x <= a:
a -= x
x = 0
else:
print("NO")
exit()
if z >= c:
z -= c
c = 0
else:
c -= z
z = 0
if z > 0:
if a + b >= z + y:
print("YES")
else:
print("NO")
exit()
else:
if y <= a + b:
print("YES")
else:
print("NO")
``` | output | 1 | 96,983 | 9 | 193,967 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
[The Duck song](https://www.youtube.com/watch?v=MtN1YnoL46Q)
For simplicity, we'll assume that there are only three types of grapes: green grapes, purple grapes and black grapes.
Andrew, Dmitry and Michal are all grapes' lovers, however their preferences of grapes are different. To make all of them happy, the following should happen:
* Andrew, Dmitry and Michal should eat at least x, y and z grapes, respectively.
* Andrew has an extreme affinity for green grapes, thus he will eat green grapes and green grapes only.
* On the other hand, Dmitry is not a fan of black grapes — any types of grapes except black would do for him. In other words, Dmitry can eat green and purple grapes.
* Michal has a common taste — he enjoys grapes in general and will be pleased with any types of grapes, as long as the quantity is sufficient.
Knowing that his friends are so fond of grapes, Aki decided to host a grape party with them. He has prepared a box with a green grapes, b purple grapes and c black grapes.
However, Aki isn't sure if the box he prepared contains enough grapes to make everyone happy. Can you please find out whether it's possible to distribute grapes so that everyone is happy or Aki has to buy some more grapes?
It is not required to distribute all the grapes, so it's possible that some of them will remain unused.
Input
The first line contains three integers x, y and z (1 ≤ x, y, z ≤ 10^5) — the number of grapes Andrew, Dmitry and Michal want to eat.
The second line contains three integers a, b, c (1 ≤ a, b, c ≤ 10^5) — the number of green, purple and black grapes in the box.
Output
If there is a grape distribution that allows everyone to be happy, print "YES", otherwise print "NO".
Examples
Input
1 6 2
4 3 3
Output
YES
Input
5 1 1
4 3 2
Output
NO
Note
In the first example, there is only one possible distribution:
Andrew should take 1 green grape, Dmitry should take 3 remaining green grapes and 3 purple grapes, and Michal will take 2 out of 3 available black grapes.
In the second test, there is no possible distribution, since Andrew is not be able to eat enough green grapes. :(
Submitted Solution:
```
names=input().split()
grapes=input().split()
x,y,z=int(names[0]),int(names[1]),int(names[2])
a,b,c=int(grapes[0]),int(grapes[1]),int(grapes[2])
flag=0
total=a+b+c
if(x<=total-(c+b)):
a=a-x
total=a+b+c
flag+=1
if(y<=total-c):
temp=(a+b)-y
total=temp+c
flag+=1
if(z<=total):
flag+=1
if(flag==3):
print("YES")
else:
print("No")
``` | instruction | 0 | 96,984 | 9 | 193,968 |
Yes | output | 1 | 96,984 | 9 | 193,969 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
[The Duck song](https://www.youtube.com/watch?v=MtN1YnoL46Q)
For simplicity, we'll assume that there are only three types of grapes: green grapes, purple grapes and black grapes.
Andrew, Dmitry and Michal are all grapes' lovers, however their preferences of grapes are different. To make all of them happy, the following should happen:
* Andrew, Dmitry and Michal should eat at least x, y and z grapes, respectively.
* Andrew has an extreme affinity for green grapes, thus he will eat green grapes and green grapes only.
* On the other hand, Dmitry is not a fan of black grapes — any types of grapes except black would do for him. In other words, Dmitry can eat green and purple grapes.
* Michal has a common taste — he enjoys grapes in general and will be pleased with any types of grapes, as long as the quantity is sufficient.
Knowing that his friends are so fond of grapes, Aki decided to host a grape party with them. He has prepared a box with a green grapes, b purple grapes and c black grapes.
However, Aki isn't sure if the box he prepared contains enough grapes to make everyone happy. Can you please find out whether it's possible to distribute grapes so that everyone is happy or Aki has to buy some more grapes?
It is not required to distribute all the grapes, so it's possible that some of them will remain unused.
Input
The first line contains three integers x, y and z (1 ≤ x, y, z ≤ 10^5) — the number of grapes Andrew, Dmitry and Michal want to eat.
The second line contains three integers a, b, c (1 ≤ a, b, c ≤ 10^5) — the number of green, purple and black grapes in the box.
Output
If there is a grape distribution that allows everyone to be happy, print "YES", otherwise print "NO".
Examples
Input
1 6 2
4 3 3
Output
YES
Input
5 1 1
4 3 2
Output
NO
Note
In the first example, there is only one possible distribution:
Andrew should take 1 green grape, Dmitry should take 3 remaining green grapes and 3 purple grapes, and Michal will take 2 out of 3 available black grapes.
In the second test, there is no possible distribution, since Andrew is not be able to eat enough green grapes. :(
Submitted Solution:
```
x , y , z = list(map(int,input().split()))
g , p , b = list(map(int,input().split()))
if g>=x and (g-x)+p>=y and ((g-x)+p)-y+b>=z:
print("YES")
else:
print("NO")
``` | instruction | 0 | 96,985 | 9 | 193,970 |
Yes | output | 1 | 96,985 | 9 | 193,971 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
[The Duck song](https://www.youtube.com/watch?v=MtN1YnoL46Q)
For simplicity, we'll assume that there are only three types of grapes: green grapes, purple grapes and black grapes.
Andrew, Dmitry and Michal are all grapes' lovers, however their preferences of grapes are different. To make all of them happy, the following should happen:
* Andrew, Dmitry and Michal should eat at least x, y and z grapes, respectively.
* Andrew has an extreme affinity for green grapes, thus he will eat green grapes and green grapes only.
* On the other hand, Dmitry is not a fan of black grapes — any types of grapes except black would do for him. In other words, Dmitry can eat green and purple grapes.
* Michal has a common taste — he enjoys grapes in general and will be pleased with any types of grapes, as long as the quantity is sufficient.
Knowing that his friends are so fond of grapes, Aki decided to host a grape party with them. He has prepared a box with a green grapes, b purple grapes and c black grapes.
However, Aki isn't sure if the box he prepared contains enough grapes to make everyone happy. Can you please find out whether it's possible to distribute grapes so that everyone is happy or Aki has to buy some more grapes?
It is not required to distribute all the grapes, so it's possible that some of them will remain unused.
Input
The first line contains three integers x, y and z (1 ≤ x, y, z ≤ 10^5) — the number of grapes Andrew, Dmitry and Michal want to eat.
The second line contains three integers a, b, c (1 ≤ a, b, c ≤ 10^5) — the number of green, purple and black grapes in the box.
Output
If there is a grape distribution that allows everyone to be happy, print "YES", otherwise print "NO".
Examples
Input
1 6 2
4 3 3
Output
YES
Input
5 1 1
4 3 2
Output
NO
Note
In the first example, there is only one possible distribution:
Andrew should take 1 green grape, Dmitry should take 3 remaining green grapes and 3 purple grapes, and Michal will take 2 out of 3 available black grapes.
In the second test, there is no possible distribution, since Andrew is not be able to eat enough green grapes. :(
Submitted Solution:
```
x,y,z = map(int,input().split(" "))
a,b,c = map(int,input().split(" "))
total_grapes = a + b + c
if a < x:
print("NO")
else:
total_grapes -= x
a -= x
if (a + b) < y:
print("NO")
else:
total_grapes -= y
if total_grapes < z:
print("NO")
else:
print("YES")
``` | instruction | 0 | 96,986 | 9 | 193,972 |
Yes | output | 1 | 96,986 | 9 | 193,973 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
[The Duck song](https://www.youtube.com/watch?v=MtN1YnoL46Q)
For simplicity, we'll assume that there are only three types of grapes: green grapes, purple grapes and black grapes.
Andrew, Dmitry and Michal are all grapes' lovers, however their preferences of grapes are different. To make all of them happy, the following should happen:
* Andrew, Dmitry and Michal should eat at least x, y and z grapes, respectively.
* Andrew has an extreme affinity for green grapes, thus he will eat green grapes and green grapes only.
* On the other hand, Dmitry is not a fan of black grapes — any types of grapes except black would do for him. In other words, Dmitry can eat green and purple grapes.
* Michal has a common taste — he enjoys grapes in general and will be pleased with any types of grapes, as long as the quantity is sufficient.
Knowing that his friends are so fond of grapes, Aki decided to host a grape party with them. He has prepared a box with a green grapes, b purple grapes and c black grapes.
However, Aki isn't sure if the box he prepared contains enough grapes to make everyone happy. Can you please find out whether it's possible to distribute grapes so that everyone is happy or Aki has to buy some more grapes?
It is not required to distribute all the grapes, so it's possible that some of them will remain unused.
Input
The first line contains three integers x, y and z (1 ≤ x, y, z ≤ 10^5) — the number of grapes Andrew, Dmitry and Michal want to eat.
The second line contains three integers a, b, c (1 ≤ a, b, c ≤ 10^5) — the number of green, purple and black grapes in the box.
Output
If there is a grape distribution that allows everyone to be happy, print "YES", otherwise print "NO".
Examples
Input
1 6 2
4 3 3
Output
YES
Input
5 1 1
4 3 2
Output
NO
Note
In the first example, there is only one possible distribution:
Andrew should take 1 green grape, Dmitry should take 3 remaining green grapes and 3 purple grapes, and Michal will take 2 out of 3 available black grapes.
In the second test, there is no possible distribution, since Andrew is not be able to eat enough green grapes. :(
Submitted Solution:
```
# Got any Grapes
def grapes(num, types):
if num[0] > types[0]:
return "NO"
else:
types[0] -= num[0]
if num[1] > (types[0] + types[1]):
return "NO"
else:
if num[1] - types[0] <= 0:
types[0] -= num[1]
else:
num[1] -= types[0]
types[0] = 0
types[1] -= num[1]
if num[2] > sum(types):
return "NO"
return "YES"
num = list(map(int, input().rstrip().split()))
types = list(map(int, input().rstrip().split()))
print(grapes(num, types))
``` | instruction | 0 | 96,987 | 9 | 193,974 |
Yes | output | 1 | 96,987 | 9 | 193,975 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
[The Duck song](https://www.youtube.com/watch?v=MtN1YnoL46Q)
For simplicity, we'll assume that there are only three types of grapes: green grapes, purple grapes and black grapes.
Andrew, Dmitry and Michal are all grapes' lovers, however their preferences of grapes are different. To make all of them happy, the following should happen:
* Andrew, Dmitry and Michal should eat at least x, y and z grapes, respectively.
* Andrew has an extreme affinity for green grapes, thus he will eat green grapes and green grapes only.
* On the other hand, Dmitry is not a fan of black grapes — any types of grapes except black would do for him. In other words, Dmitry can eat green and purple grapes.
* Michal has a common taste — he enjoys grapes in general and will be pleased with any types of grapes, as long as the quantity is sufficient.
Knowing that his friends are so fond of grapes, Aki decided to host a grape party with them. He has prepared a box with a green grapes, b purple grapes and c black grapes.
However, Aki isn't sure if the box he prepared contains enough grapes to make everyone happy. Can you please find out whether it's possible to distribute grapes so that everyone is happy or Aki has to buy some more grapes?
It is not required to distribute all the grapes, so it's possible that some of them will remain unused.
Input
The first line contains three integers x, y and z (1 ≤ x, y, z ≤ 10^5) — the number of grapes Andrew, Dmitry and Michal want to eat.
The second line contains three integers a, b, c (1 ≤ a, b, c ≤ 10^5) — the number of green, purple and black grapes in the box.
Output
If there is a grape distribution that allows everyone to be happy, print "YES", otherwise print "NO".
Examples
Input
1 6 2
4 3 3
Output
YES
Input
5 1 1
4 3 2
Output
NO
Note
In the first example, there is only one possible distribution:
Andrew should take 1 green grape, Dmitry should take 3 remaining green grapes and 3 purple grapes, and Michal will take 2 out of 3 available black grapes.
In the second test, there is no possible distribution, since Andrew is not be able to eat enough green grapes. :(
Submitted Solution:
```
x,y,z=map(int,input().split())
a,b,c=map(int,input().split())
if a<x:
print("NO")
else:
a-=x;
if y>(a+c):
print("NO")
else:
if z>(a+c+b-y):
print("NO")
else:
print("YES")
``` | instruction | 0 | 96,988 | 9 | 193,976 |
No | output | 1 | 96,988 | 9 | 193,977 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
[The Duck song](https://www.youtube.com/watch?v=MtN1YnoL46Q)
For simplicity, we'll assume that there are only three types of grapes: green grapes, purple grapes and black grapes.
Andrew, Dmitry and Michal are all grapes' lovers, however their preferences of grapes are different. To make all of them happy, the following should happen:
* Andrew, Dmitry and Michal should eat at least x, y and z grapes, respectively.
* Andrew has an extreme affinity for green grapes, thus he will eat green grapes and green grapes only.
* On the other hand, Dmitry is not a fan of black grapes — any types of grapes except black would do for him. In other words, Dmitry can eat green and purple grapes.
* Michal has a common taste — he enjoys grapes in general and will be pleased with any types of grapes, as long as the quantity is sufficient.
Knowing that his friends are so fond of grapes, Aki decided to host a grape party with them. He has prepared a box with a green grapes, b purple grapes and c black grapes.
However, Aki isn't sure if the box he prepared contains enough grapes to make everyone happy. Can you please find out whether it's possible to distribute grapes so that everyone is happy or Aki has to buy some more grapes?
It is not required to distribute all the grapes, so it's possible that some of them will remain unused.
Input
The first line contains three integers x, y and z (1 ≤ x, y, z ≤ 10^5) — the number of grapes Andrew, Dmitry and Michal want to eat.
The second line contains three integers a, b, c (1 ≤ a, b, c ≤ 10^5) — the number of green, purple and black grapes in the box.
Output
If there is a grape distribution that allows everyone to be happy, print "YES", otherwise print "NO".
Examples
Input
1 6 2
4 3 3
Output
YES
Input
5 1 1
4 3 2
Output
NO
Note
In the first example, there is only one possible distribution:
Andrew should take 1 green grape, Dmitry should take 3 remaining green grapes and 3 purple grapes, and Michal will take 2 out of 3 available black grapes.
In the second test, there is no possible distribution, since Andrew is not be able to eat enough green grapes. :(
Submitted Solution:
```
a,d,m=map(int,input().split())
x,y,z=map(int,input().split())
count=0
y=x+y
if x>=a:
x-=a;count+=1
if y>=d:
y-=d;count+=1
if y+z>=m:
count+=1
if count==3:
print('YES')
else:
print('NO')
``` | instruction | 0 | 96,989 | 9 | 193,978 |
No | output | 1 | 96,989 | 9 | 193,979 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
[The Duck song](https://www.youtube.com/watch?v=MtN1YnoL46Q)
For simplicity, we'll assume that there are only three types of grapes: green grapes, purple grapes and black grapes.
Andrew, Dmitry and Michal are all grapes' lovers, however their preferences of grapes are different. To make all of them happy, the following should happen:
* Andrew, Dmitry and Michal should eat at least x, y and z grapes, respectively.
* Andrew has an extreme affinity for green grapes, thus he will eat green grapes and green grapes only.
* On the other hand, Dmitry is not a fan of black grapes — any types of grapes except black would do for him. In other words, Dmitry can eat green and purple grapes.
* Michal has a common taste — he enjoys grapes in general and will be pleased with any types of grapes, as long as the quantity is sufficient.
Knowing that his friends are so fond of grapes, Aki decided to host a grape party with them. He has prepared a box with a green grapes, b purple grapes and c black grapes.
However, Aki isn't sure if the box he prepared contains enough grapes to make everyone happy. Can you please find out whether it's possible to distribute grapes so that everyone is happy or Aki has to buy some more grapes?
It is not required to distribute all the grapes, so it's possible that some of them will remain unused.
Input
The first line contains three integers x, y and z (1 ≤ x, y, z ≤ 10^5) — the number of grapes Andrew, Dmitry and Michal want to eat.
The second line contains three integers a, b, c (1 ≤ a, b, c ≤ 10^5) — the number of green, purple and black grapes in the box.
Output
If there is a grape distribution that allows everyone to be happy, print "YES", otherwise print "NO".
Examples
Input
1 6 2
4 3 3
Output
YES
Input
5 1 1
4 3 2
Output
NO
Note
In the first example, there is only one possible distribution:
Andrew should take 1 green grape, Dmitry should take 3 remaining green grapes and 3 purple grapes, and Michal will take 2 out of 3 available black grapes.
In the second test, there is no possible distribution, since Andrew is not be able to eat enough green grapes. :(
Submitted Solution:
```
x,y,z = [int(i) for i in input().split()]
a,b,c = [int(i) for i in input().split()]
andrew = a-x
dmitry = (a-x+b) - y
michal = a+b+c - andrew - dmitry
if andrew >= 0 and dmitry >= 0 and michal >= 0:
print('YES')
else:
print('NO')
``` | instruction | 0 | 96,990 | 9 | 193,980 |
No | output | 1 | 96,990 | 9 | 193,981 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
[The Duck song](https://www.youtube.com/watch?v=MtN1YnoL46Q)
For simplicity, we'll assume that there are only three types of grapes: green grapes, purple grapes and black grapes.
Andrew, Dmitry and Michal are all grapes' lovers, however their preferences of grapes are different. To make all of them happy, the following should happen:
* Andrew, Dmitry and Michal should eat at least x, y and z grapes, respectively.
* Andrew has an extreme affinity for green grapes, thus he will eat green grapes and green grapes only.
* On the other hand, Dmitry is not a fan of black grapes — any types of grapes except black would do for him. In other words, Dmitry can eat green and purple grapes.
* Michal has a common taste — he enjoys grapes in general and will be pleased with any types of grapes, as long as the quantity is sufficient.
Knowing that his friends are so fond of grapes, Aki decided to host a grape party with them. He has prepared a box with a green grapes, b purple grapes and c black grapes.
However, Aki isn't sure if the box he prepared contains enough grapes to make everyone happy. Can you please find out whether it's possible to distribute grapes so that everyone is happy or Aki has to buy some more grapes?
It is not required to distribute all the grapes, so it's possible that some of them will remain unused.
Input
The first line contains three integers x, y and z (1 ≤ x, y, z ≤ 10^5) — the number of grapes Andrew, Dmitry and Michal want to eat.
The second line contains three integers a, b, c (1 ≤ a, b, c ≤ 10^5) — the number of green, purple and black grapes in the box.
Output
If there is a grape distribution that allows everyone to be happy, print "YES", otherwise print "NO".
Examples
Input
1 6 2
4 3 3
Output
YES
Input
5 1 1
4 3 2
Output
NO
Note
In the first example, there is only one possible distribution:
Andrew should take 1 green grape, Dmitry should take 3 remaining green grapes and 3 purple grapes, and Michal will take 2 out of 3 available black grapes.
In the second test, there is no possible distribution, since Andrew is not be able to eat enough green grapes. :(
Submitted Solution:
```
p,q,r=input("").split(" ")
p=int(p)
q=int(q)
r=int(r)
a,b,c=input("").split(" ")
a=int(a)
b=int(b)
c=int(c)
count=0
if p<=a:
count+=1
a-=p
if q<=a+b:
count+=1
d=q-(a+b)
if r<=d+c:
count+=1
if count==3:
print("YES")
else:
print("NO")
``` | instruction | 0 | 96,991 | 9 | 193,982 |
No | output | 1 | 96,991 | 9 | 193,983 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gerald is setting the New Year table. The table has the form of a circle; its radius equals R. Gerald invited many guests and is concerned whether the table has enough space for plates for all those guests. Consider all plates to be round and have the same radii that equal r. Each plate must be completely inside the table and must touch the edge of the table. Of course, the plates must not intersect, but they can touch each other. Help Gerald determine whether the table is large enough for n plates.
Input
The first line contains three integers n, R and r (1 ≤ n ≤ 100, 1 ≤ r, R ≤ 1000) — the number of plates, the radius of the table and the plates' radius.
Output
Print "YES" (without the quotes) if it is possible to place n plates on the table by the rules given above. If it is impossible, print "NO".
Remember, that each plate must touch the edge of the table.
Examples
Input
4 10 4
Output
YES
Input
5 10 4
Output
NO
Input
1 10 10
Output
YES
Note
The possible arrangement of the plates for the first sample is:
<image>
Submitted Solution:
```
# cook your dish here
import math
l = list(map(int,input().split()))
if l[1] == l[2]:
if l[0] == 1:
print("YES")
else:
print("NO")
else:
x = math.asin(l[2]/(l[1]-l[2]))
n = math.pi / x
print(n)
if l[0]-n < 1e-9 or l[0]<=n:
print("YES")
else:
print("NO")
``` | instruction | 0 | 97,182 | 9 | 194,364 |
No | output | 1 | 97,182 | 9 | 194,365 |
Provide a correct Python 3 solution for this coding contest problem.
In "Takahashi-ya", a ramen restaurant, basically they have one menu: "ramen", but N kinds of toppings are also offered. When a customer orders a bowl of ramen, for each kind of topping, he/she can choose whether to put it on top of his/her ramen or not. There is no limit on the number of toppings, and it is allowed to have all kinds of toppings or no topping at all. That is, considering the combination of the toppings, 2^N types of ramen can be ordered.
Akaki entered Takahashi-ya. She is thinking of ordering some bowls of ramen that satisfy both of the following two conditions:
* Do not order multiple bowls of ramen with the exactly same set of toppings.
* Each of the N kinds of toppings is on two or more bowls of ramen ordered.
You are given N and a prime number M. Find the number of the sets of bowls of ramen that satisfy these conditions, disregarding order, modulo M. Since she is in extreme hunger, ordering any number of bowls of ramen is fine.
Constraints
* 2 \leq N \leq 3000
* 10^8 \leq M \leq 10^9 + 9
* N is an integer.
* M is a prime number.
Input
Input is given from Standard Input in the following format:
N M
Output
Print the number of the sets of bowls of ramen that satisfy the conditions, disregarding order, modulo M.
Examples
Input
2 1000000007
Output
2
Input
3 1000000009
Output
118
Input
50 111111113
Output
1456748
Input
3000 123456791
Output
16369789 | instruction | 0 | 98,475 | 9 | 196,950 |
"Correct Solution:
```
N, M = map(int, input().split())
fact = [1]*(N+1)
rfact = [1]*(N+1)
for i in range(1, N+1):
fact[i] = r = (i * fact[i-1]) % M
rfact[i] = pow(r, M-2, M)
memo = {}
def f(N, K):
if (N, K) in memo:
return memo[N, K]
if K == 0:
return N == 0
elif K == 1 or N == K:
return 1
r = memo[N, K] = (f(N-1, K-1) + K*f(N-1, K)) % M
return r
S = [1]
rev2 = pow(2, M-2, M)
base = pow(2, N, M)
ans = 0
S = [1]
for K in range(N+1):
r = (fact[N] * rfact[K] * rfact[N-K]) % M
r = (r * pow(2, pow(2, N - K, M-1), M)) % M
b = 1
v = 0
T = [0]*(K+2)
for L in range(K):
T[L+1] = s = (S[L] + (L+1)*S[L+1]) % M
v += s * b
b = (b * base) % M
v += b
T[K+1] = 1
S = T
r = (r * v) % M
if K % 2:
ans -= r
else:
ans += r
ans %= M
base = (base * rev2) % M
print(ans)
``` | output | 1 | 98,475 | 9 | 196,951 |
Provide a correct Python 3 solution for this coding contest problem.
In "Takahashi-ya", a ramen restaurant, basically they have one menu: "ramen", but N kinds of toppings are also offered. When a customer orders a bowl of ramen, for each kind of topping, he/she can choose whether to put it on top of his/her ramen or not. There is no limit on the number of toppings, and it is allowed to have all kinds of toppings or no topping at all. That is, considering the combination of the toppings, 2^N types of ramen can be ordered.
Akaki entered Takahashi-ya. She is thinking of ordering some bowls of ramen that satisfy both of the following two conditions:
* Do not order multiple bowls of ramen with the exactly same set of toppings.
* Each of the N kinds of toppings is on two or more bowls of ramen ordered.
You are given N and a prime number M. Find the number of the sets of bowls of ramen that satisfy these conditions, disregarding order, modulo M. Since she is in extreme hunger, ordering any number of bowls of ramen is fine.
Constraints
* 2 \leq N \leq 3000
* 10^8 \leq M \leq 10^9 + 9
* N is an integer.
* M is a prime number.
Input
Input is given from Standard Input in the following format:
N M
Output
Print the number of the sets of bowls of ramen that satisfy the conditions, disregarding order, modulo M.
Examples
Input
2 1000000007
Output
2
Input
3 1000000009
Output
118
Input
50 111111113
Output
1456748
Input
3000 123456791
Output
16369789 | instruction | 0 | 98,476 | 9 | 196,952 |
"Correct Solution:
```
from collections import defaultdict
class Combinatorics:
def __init__(self, N, mod):
'''
Preprocess for calculating binomial coefficients nCr (0 <= r <= n, 0 <= n <= N)
over the finite field Z/(mod)Z.
Input:
N (int): maximum n
mod (int): a prime number. The order of the field Z/(mod)Z over which nCr is calculated.
'''
self.mod = mod
self.fact = {i: None for i in range(N+1)} # n!
self.inverse = {i: None for i in range(1, N+1)} # inverse of n in the field Z/(MOD)Z
self.fact_inverse = {i: None for i in range(N+1)} # inverse of n! in the field Z/(MOD)Z
# preprocess
self.fact[0] = self.fact[1] = 1
self.fact_inverse[0] = self.fact_inverse[1] = 1
self.inverse[1] = 1
for i in range(2, N+1):
self.fact[i] = i * self.fact[i-1] % self.mod
q, r = divmod(self.mod, i)
self.inverse[i] = (- (q % self.mod) * self.inverse[r]) % self.mod
self.fact_inverse[i] = self.inverse[i] * self.fact_inverse[i-1] % self.mod
def binom(self, n, r):
'''
Calculate nCr = n! /(r! (n-r)!) % mod
'''
if n < r or n < 0 or r < 0:
return 0
else:
return self.fact[n] * (self.fact_inverse[r] * self.fact_inverse[n-r] % self.mod) % self.mod
N, M = map(int, input().split())
com = Combinatorics(N, M)
ans = 0
# Preprocess
# calculate 2**n and 2**(2**n)
pow2 = [0] * (N*N // 4 + 1)
pow_pow2 = [0] * (N+1)
pow2[0] = 1; pow_pow2[0] = 2
for i in range(1, N*N // 4 + 1):
pow2[i] = (pow2[i-1] * 2) % M
for i in range(1, N+1):
pow_pow2[i] = pow(pow_pow2[i-1], 2, M)
# (#ways to order i ramens, with each of n toppings chosen at most once)
ways2 = {n: defaultdict(int) for n in range(N+1)}
for n in range(N+1):
ways = 0 # (#ways to choose n toppings, with each topping chosen at most once)
temp = 0
if n >= 1:
for i in range(n+1):
ways2[n][i] = (ways2[n][i] + ways2[n-1][i]) % M # n-th topping is not used
if i >= 1: ways2[n][i] = (ways2[n][i] + ways2[n-1][i-1]) % M # only n-th topping is used in an order
ways2[n][i] = (ways2[n][i] + (i*ways2[n-1][i]) % M) % M # n-th topping is chosen with another topping
temp = (temp + (ways2[n][i] * pow2[(N-n)*i]) % M) % M # degree of freedom on each order is 2**(N-n)
else: # n = 0
ways2[n][0] = 1
temp += ways2[n][0]
ways = (temp * pow_pow2[N-n]) % M
if n % 2 == 0:
ans = (ans + (com.binom(N, n) * ways) % M) % M
else:
ans = (ans - (com.binom(N, n) * ways) % M) % M
print(ans)
``` | output | 1 | 98,476 | 9 | 196,953 |
Provide a correct Python 3 solution for this coding contest problem.
In "Takahashi-ya", a ramen restaurant, basically they have one menu: "ramen", but N kinds of toppings are also offered. When a customer orders a bowl of ramen, for each kind of topping, he/she can choose whether to put it on top of his/her ramen or not. There is no limit on the number of toppings, and it is allowed to have all kinds of toppings or no topping at all. That is, considering the combination of the toppings, 2^N types of ramen can be ordered.
Akaki entered Takahashi-ya. She is thinking of ordering some bowls of ramen that satisfy both of the following two conditions:
* Do not order multiple bowls of ramen with the exactly same set of toppings.
* Each of the N kinds of toppings is on two or more bowls of ramen ordered.
You are given N and a prime number M. Find the number of the sets of bowls of ramen that satisfy these conditions, disregarding order, modulo M. Since she is in extreme hunger, ordering any number of bowls of ramen is fine.
Constraints
* 2 \leq N \leq 3000
* 10^8 \leq M \leq 10^9 + 9
* N is an integer.
* M is a prime number.
Input
Input is given from Standard Input in the following format:
N M
Output
Print the number of the sets of bowls of ramen that satisfy the conditions, disregarding order, modulo M.
Examples
Input
2 1000000007
Output
2
Input
3 1000000009
Output
118
Input
50 111111113
Output
1456748
Input
3000 123456791
Output
16369789 | instruction | 0 | 98,477 | 9 | 196,954 |
"Correct Solution:
```
# coding: utf-8
# Your code here!
import sys
sys.setrecursionlimit(10**6)
readline = sys.stdin.readline #文字列入力のときは注意
n,MOD = [int(i) for i in readline().split()]
SIZE=3001; #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
"""
make the table of Sterling numbers of the second kind
Sterling[ball][box]
SIZE = n
Sterling2 = [[0]*(SIZE+1) for _ in range(SIZE+1)]
Sterling2[0][0] = Sterling2[1][1] = 1
for i in range(2,SIZE+1):
for k in range(1,i+1):
Sterling2[i][k] = (Sterling2[i-1][k-1] + k*Sterling2[i-1][k])%MOD
"""
SIZE = n+1
f = [[0]*(SIZE) for _ in range(SIZE)]
f[0][0] = 1
for i in range(1,SIZE):
f[i][0] = 1
for k in range(1,i+1):
f[i][k] = (f[i-1][k] + f[i-1][k-1] + k*f[i-1][k])%MOD
#print(f)
#pow(2,n-s,MOD-1)
pp = [1]*SIZE
for i in range(1,SIZE):
pp[i] = 2*pp[i-1]%(MOD-1)
def size(s):
"""
for b in range(n+1):
res += f[s][b]*pow(2,b*(n-s),MOD)
"""
res = 0
p = 1
e = pow(2,(n-s),MOD)
for b in range(SIZE):
res += f[s][b]*p
p *= e
p %= MOD
return res*pow(2,pp[n-s],MOD)%MOD
#for i in range(n+1):
# print(size(i))
ans = 0
sgn = 1
for j in range(n+1):
ans += sgn*choose(n,j)*size(j)
ans %= MOD
sgn *= -1
print(ans)
``` | output | 1 | 98,477 | 9 | 196,955 |
Provide a correct Python 3 solution for this coding contest problem.
In "Takahashi-ya", a ramen restaurant, basically they have one menu: "ramen", but N kinds of toppings are also offered. When a customer orders a bowl of ramen, for each kind of topping, he/she can choose whether to put it on top of his/her ramen or not. There is no limit on the number of toppings, and it is allowed to have all kinds of toppings or no topping at all. That is, considering the combination of the toppings, 2^N types of ramen can be ordered.
Akaki entered Takahashi-ya. She is thinking of ordering some bowls of ramen that satisfy both of the following two conditions:
* Do not order multiple bowls of ramen with the exactly same set of toppings.
* Each of the N kinds of toppings is on two or more bowls of ramen ordered.
You are given N and a prime number M. Find the number of the sets of bowls of ramen that satisfy these conditions, disregarding order, modulo M. Since she is in extreme hunger, ordering any number of bowls of ramen is fine.
Constraints
* 2 \leq N \leq 3000
* 10^8 \leq M \leq 10^9 + 9
* N is an integer.
* M is a prime number.
Input
Input is given from Standard Input in the following format:
N M
Output
Print the number of the sets of bowls of ramen that satisfy the conditions, disregarding order, modulo M.
Examples
Input
2 1000000007
Output
2
Input
3 1000000009
Output
118
Input
50 111111113
Output
1456748
Input
3000 123456791
Output
16369789 | instruction | 0 | 98,478 | 9 | 196,956 |
"Correct Solution:
```
n, m = map(int, input().split())
MOD = m
list_size = n+2
f_list = [1] * list_size
f_r_list = [1] * list_size
for i in range(list_size - 1):
f_list[i + 1] = int((f_list[i] * (i + 1)) % MOD)
f_r_list = [pow(x, MOD-2, MOD) for x in f_list]
def comb(N, r):
if N < r or r < 0:
return 0
else:
return (((f_list[N] * f_r_list[N - r]) % MOD) * f_r_list[r]) % MOD
st = [[0 for _ in range(n+1)] for _ in range(n+1)]
st[0][0] = 1
for i in range(1, n+1):
st[i][0] = 1
for j in range(1, i+1):
st[i][j] = (st[i-1][j-1] + (j+1) * st[i-1][j]) % MOD
ans = 0
for i in range(n+1):
res = (comb(n, i) * pow(2, pow(2, n-i, MOD-1), MOD)) % MOD
tmp = 0
for j in range(i+1):
tmp += st[i][j] * pow(2, (n-i) * j, MOD)
tmp %= MOD
res *= tmp
res %= MOD
if i%2 == 0:
ans += res
else:
ans -= res
ans %= MOD
print(ans)
``` | output | 1 | 98,478 | 9 | 196,957 |
Provide a correct Python 3 solution for this coding contest problem.
In "Takahashi-ya", a ramen restaurant, basically they have one menu: "ramen", but N kinds of toppings are also offered. When a customer orders a bowl of ramen, for each kind of topping, he/she can choose whether to put it on top of his/her ramen or not. There is no limit on the number of toppings, and it is allowed to have all kinds of toppings or no topping at all. That is, considering the combination of the toppings, 2^N types of ramen can be ordered.
Akaki entered Takahashi-ya. She is thinking of ordering some bowls of ramen that satisfy both of the following two conditions:
* Do not order multiple bowls of ramen with the exactly same set of toppings.
* Each of the N kinds of toppings is on two or more bowls of ramen ordered.
You are given N and a prime number M. Find the number of the sets of bowls of ramen that satisfy these conditions, disregarding order, modulo M. Since she is in extreme hunger, ordering any number of bowls of ramen is fine.
Constraints
* 2 \leq N \leq 3000
* 10^8 \leq M \leq 10^9 + 9
* N is an integer.
* M is a prime number.
Input
Input is given from Standard Input in the following format:
N M
Output
Print the number of the sets of bowls of ramen that satisfy the conditions, disregarding order, modulo M.
Examples
Input
2 1000000007
Output
2
Input
3 1000000009
Output
118
Input
50 111111113
Output
1456748
Input
3000 123456791
Output
16369789 | instruction | 0 | 98,479 | 9 | 196,958 |
"Correct Solution:
```
N, M = map(int, input().split())
fact = [1]*(N+1)
rfact = [1]*(N+1)
for i in range(1, N+1):
fact[i] = r = (i * fact[i-1]) % M
rfact[i] = pow(r, M-2, M)
S = [1]
rev2 = pow(2, M-2, M)
base = pow(2, N, M)
ans = 0
S = [1]
for K in range(N+1):
res = (fact[N] * rfact[K] * rfact[N-K]) % M
res = (res * pow(2, pow(2, N - K, M-1), M)) % M
b = 1
v = 0
T = [0]*(K+2)
for L in range(K):
T[L+1] = s = (S[L] + (L+1)*S[L+1]) % M
v += s * b
b = (b * base) % M
v += b
T[K+1] = 1
S = T
res = (res * v) % M
if K % 2:
ans -= res
else:
ans += res
ans %= M
base = (base * rev2) % M
print(ans)
``` | output | 1 | 98,479 | 9 | 196,959 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In "Takahashi-ya", a ramen restaurant, basically they have one menu: "ramen", but N kinds of toppings are also offered. When a customer orders a bowl of ramen, for each kind of topping, he/she can choose whether to put it on top of his/her ramen or not. There is no limit on the number of toppings, and it is allowed to have all kinds of toppings or no topping at all. That is, considering the combination of the toppings, 2^N types of ramen can be ordered.
Akaki entered Takahashi-ya. She is thinking of ordering some bowls of ramen that satisfy both of the following two conditions:
* Do not order multiple bowls of ramen with the exactly same set of toppings.
* Each of the N kinds of toppings is on two or more bowls of ramen ordered.
You are given N and a prime number M. Find the number of the sets of bowls of ramen that satisfy these conditions, disregarding order, modulo M. Since she is in extreme hunger, ordering any number of bowls of ramen is fine.
Constraints
* 2 \leq N \leq 3000
* 10^8 \leq M \leq 10^9 + 9
* N is an integer.
* M is a prime number.
Input
Input is given from Standard Input in the following format:
N M
Output
Print the number of the sets of bowls of ramen that satisfy the conditions, disregarding order, modulo M.
Examples
Input
2 1000000007
Output
2
Input
3 1000000009
Output
118
Input
50 111111113
Output
1456748
Input
3000 123456791
Output
16369789
Submitted 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
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
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.readline().split()
def S(): return sys.stdin.readline().strip()
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)]
n, mod = 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
sterling = [[0] * (n + 1) for _ in range(n + 1)]
sterling[0][0] = 1
# n個の区別できるものをk個の(n個以下)の区別不可能なグループに分けることに(i, j)が対応
# n >= k
for i in range(1, n + 1):
for j in range(i + 1):
sterling[i][j] = sterling[i - 1][j] * (j + 1)
if j:
sterling[i][j] += sterling[i - 1][j - 1]
sterling[i][j] %= mod
ans = pow(2, pow(2, n, mod), mod)
for l in range(1, n + 1):
ret = 0
for k in range(l + 1):
cumsum = (sterling[l - 1][k - 1] + (k + 1) * sterling[l - 1][k]) % mod
cumsum = cumsum * pow(pow(2, n - l, mod), k, mod) % mod
cumsum = cumsum * pow(2, pow(2, n - l, mod), mod) % mod
cumsum = cumsum * comb(n, l) % mod
ret += cumsum
if l % 2:
ans -= ret
else:
ans += ret
ans %= mod
print(ans)
``` | instruction | 0 | 98,480 | 9 | 196,960 |
No | output | 1 | 98,480 | 9 | 196,961 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In "Takahashi-ya", a ramen restaurant, basically they have one menu: "ramen", but N kinds of toppings are also offered. When a customer orders a bowl of ramen, for each kind of topping, he/she can choose whether to put it on top of his/her ramen or not. There is no limit on the number of toppings, and it is allowed to have all kinds of toppings or no topping at all. That is, considering the combination of the toppings, 2^N types of ramen can be ordered.
Akaki entered Takahashi-ya. She is thinking of ordering some bowls of ramen that satisfy both of the following two conditions:
* Do not order multiple bowls of ramen with the exactly same set of toppings.
* Each of the N kinds of toppings is on two or more bowls of ramen ordered.
You are given N and a prime number M. Find the number of the sets of bowls of ramen that satisfy these conditions, disregarding order, modulo M. Since she is in extreme hunger, ordering any number of bowls of ramen is fine.
Constraints
* 2 \leq N \leq 3000
* 10^8 \leq M \leq 10^9 + 9
* N is an integer.
* M is a prime number.
Input
Input is given from Standard Input in the following format:
N M
Output
Print the number of the sets of bowls of ramen that satisfy the conditions, disregarding order, modulo M.
Examples
Input
2 1000000007
Output
2
Input
3 1000000009
Output
118
Input
50 111111113
Output
1456748
Input
3000 123456791
Output
16369789
Submitted Solution:
```
import sys
sys.setrecursionlimit(10 ** 6)
input = sys.stdin.readline
def main():
def com(com_n, com_r):
return fac[com_n] * inv[com_r] * inv[com_n - com_r] % md
n, md = map(int, input().split())
# combinationの準備
n_max = n+3
fac = [1]
inv = [1] * (n_max + 1)
k_fac_inv = 1
for i in range(1, n_max + 1):
k_fac_inv = k_fac_inv * i % md
fac.append(k_fac_inv)
k_fac_inv = pow(k_fac_inv, md - 2, md)
for i in range(n_max, 1, -1):
inv[i] = k_fac_inv
k_fac_inv = k_fac_inv * i % md
# スターリング数発展の作成
stir = [[1] * (n + 1) for _ in range(n + 1)]
si1 = stir[0]
for i in range(1, n + 1):
si = stir[i]
for j in range(1, i):
si[j] = si1[j] * (j + 1) + si1[j - 1]
si1 = si
# 本体
ans = 0
pm = 1
for m in range(n + 1):
a = 0
e2 = pow(2, n - m)
for k in range(m + 1):
a = (a + stir[m][k] * pow(e2, k, md)) % md
ans = (ans + pm * a * pow(2, e2, md) * com(n, m)) % md
pm *= -1
print(ans)
main()
``` | instruction | 0 | 98,481 | 9 | 196,962 |
No | output | 1 | 98,481 | 9 | 196,963 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In "Takahashi-ya", a ramen restaurant, basically they have one menu: "ramen", but N kinds of toppings are also offered. When a customer orders a bowl of ramen, for each kind of topping, he/she can choose whether to put it on top of his/her ramen or not. There is no limit on the number of toppings, and it is allowed to have all kinds of toppings or no topping at all. That is, considering the combination of the toppings, 2^N types of ramen can be ordered.
Akaki entered Takahashi-ya. She is thinking of ordering some bowls of ramen that satisfy both of the following two conditions:
* Do not order multiple bowls of ramen with the exactly same set of toppings.
* Each of the N kinds of toppings is on two or more bowls of ramen ordered.
You are given N and a prime number M. Find the number of the sets of bowls of ramen that satisfy these conditions, disregarding order, modulo M. Since she is in extreme hunger, ordering any number of bowls of ramen is fine.
Constraints
* 2 \leq N \leq 3000
* 10^8 \leq M \leq 10^9 + 9
* N is an integer.
* M is a prime number.
Input
Input is given from Standard Input in the following format:
N M
Output
Print the number of the sets of bowls of ramen that satisfy the conditions, disregarding order, modulo M.
Examples
Input
2 1000000007
Output
2
Input
3 1000000009
Output
118
Input
50 111111113
Output
1456748
Input
3000 123456791
Output
16369789
Submitted Solution:
```
N,M = list(map(int,input().split()))
table = [1,1]
while len(table) <= N:
temp = 1
for i in range(len(table)-1):
table[i+1] += temp
temp = table[i+1]- temp
table[i+1] = M
table.append(1)
S = [1]
rev2 = pow(2, M-2, M)
base = pow(2, N, M)
ans = 0
S = [1]
for K in range(N+1):
res = table[K] % M
res = (res * pow(2, pow(2, N - K, M-1), M)) % M
b = 1
v = 0
T = [0]*(K+2)
for L in range(K):
T[L+1] = s = (S[L] + (L+1)*S[L+1]) % M
v += s * b
b = (b * base) % M
v += b
T[K+1] = 1
S = T
res = (res * v) % M
if K % 2:
ans -= res
else:
ans += res
ans %= M
base = (base * rev2) % M
print(ans)
``` | instruction | 0 | 98,482 | 9 | 196,964 |
No | output | 1 | 98,482 | 9 | 196,965 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In "Takahashi-ya", a ramen restaurant, basically they have one menu: "ramen", but N kinds of toppings are also offered. When a customer orders a bowl of ramen, for each kind of topping, he/she can choose whether to put it on top of his/her ramen or not. There is no limit on the number of toppings, and it is allowed to have all kinds of toppings or no topping at all. That is, considering the combination of the toppings, 2^N types of ramen can be ordered.
Akaki entered Takahashi-ya. She is thinking of ordering some bowls of ramen that satisfy both of the following two conditions:
* Do not order multiple bowls of ramen with the exactly same set of toppings.
* Each of the N kinds of toppings is on two or more bowls of ramen ordered.
You are given N and a prime number M. Find the number of the sets of bowls of ramen that satisfy these conditions, disregarding order, modulo M. Since she is in extreme hunger, ordering any number of bowls of ramen is fine.
Constraints
* 2 \leq N \leq 3000
* 10^8 \leq M \leq 10^9 + 9
* N is an integer.
* M is a prime number.
Input
Input is given from Standard Input in the following format:
N M
Output
Print the number of the sets of bowls of ramen that satisfy the conditions, disregarding order, modulo M.
Examples
Input
2 1000000007
Output
2
Input
3 1000000009
Output
118
Input
50 111111113
Output
1456748
Input
3000 123456791
Output
16369789
Submitted Solution:
```
import numpy as np
from collections import defaultdict
class Combinatorics:
def __init__(self, N, mod):
'''
Preprocess for calculating binomial coefficients nCr (0 <= r <= n, 0 <= n <= N)
over the finite field Z/(mod)Z.
Input:
N (int): maximum n
mod (int): a prime number. The order of the field Z/(mod)Z over which nCr is calculated.
'''
self.mod = mod
self.fact = {i: None for i in range(N+1)} # n!
self.inverse = {i: None for i in range(1, N+1)} # inverse of n in the field Z/(MOD)Z
self.fact_inverse = {i: None for i in range(N+1)} # inverse of n! in the field Z/(MOD)Z
# preprocess
self.fact[0] = self.fact[1] = 1
self.fact_inverse[0] = self.fact_inverse[1] = 1
self.inverse[1] = 1
for i in range(2, N+1):
self.fact[i] = i * self.fact[i-1] % self.mod
q, r = divmod(self.mod, i)
self.inverse[i] = (- (q % self.mod) * self.inverse[r]) % self.mod
self.fact_inverse[i] = self.inverse[i] * self.fact_inverse[i-1] % self.mod
def binom(self, n, r):
'''
Calculate nCr = n! /(r! (n-r)!) % mod
'''
if n < r or n < 0 or r < 0:
return 0
else:
return self.fact[n] * (self.fact_inverse[r] * self.fact_inverse[n-r] % self.mod) % self.mod
N, M = map(int, input().split())
com = Combinatorics(N, M)
ans = 0
# Preprocess
# calculate 2**n and 2**(2**n)
pow2 = np.power(2, np.array(range(N*N // 4 + 1)))
pow_pow2 = [0] * (N+1)
pow_pow2[0] = 2
for i in range(1, N*N // 4 + 1):
pow2[i] = (pow2[i-1] * 2) % M
for i in range(1, N+1):
pow_pow2[i] = pow(pow_pow2[i-1], 2, M)
# (#ways to order i ramens, with each of n toppings chosen at most once)
ways2 = {n: defaultdict(int) for n in range(N+1)}
for n in range(N+1):
ways = 0 # (#ways to choose n toppings, with each topping chosen at most once)
temp = 0
if n >= 1:
for i in range(n+1):
ways2[n][i] = (ways2[n][i] + ways2[n-1][i]) % M # n-th topping is not used
if i >= 1: ways2[n][i] = (ways2[n][i] + ways2[n-1][i-1]) % M # only n-th topping is used in an order
ways2[n][i] = (ways2[n][i] + (i*ways2[n-1][i]) % M) % M # n-th topping is chosen with another topping
temp = (temp + (ways2[n][i] * pow2[(N-n)*i]) % M) % M # degree of freedom on each order is 2**(N-n)
else: # n = 0
ways2[n][0] = 1
temp += ways2[n][0]
ways = (temp * pow_pow2[N-n]) % M
if n % 2 == 0:
ans = (ans + (com.binom(N, n) * ways) % M) % M
else:
ans = (ans - (com.binom(N, n) * ways) % M) % M
print(ans)
``` | instruction | 0 | 98,483 | 9 | 196,966 |
No | output | 1 | 98,483 | 9 | 196,967 |
Provide a correct Python 3 solution for this coding contest problem.
The restaurant AtCoder serves the following five dishes:
* ABC Don (rice bowl): takes A minutes to serve.
* ARC Curry: takes B minutes to serve.
* AGC Pasta: takes C minutes to serve.
* APC Ramen: takes D minutes to serve.
* ATC Hanbagu (hamburger patty): takes E minutes to serve.
Here, the time to serve a dish is the time between when an order is placed and when the dish is delivered.
This restaurant has the following rules on orders:
* An order can only be placed at a time that is a multiple of 10 (time 0, 10, 20, ...).
* Only one dish can be ordered at a time.
* No new order can be placed when an order is already placed and the dish is still not delivered, but a new order can be placed at the exact time when the dish is delivered.
E869120 arrives at this restaurant at time 0. He will order all five dishes. Find the earliest possible time for the last dish to be delivered.
Here, he can order the dishes in any order he likes, and he can place an order already at time 0.
Constraints
* A, B, C, D and E are integers between 1 and 123 (inclusive).
Input
Input is given from Standard Input in the following format:
A
B
C
D
E
Output
Print the earliest possible time for the last dish to be delivered, as an integer.
Examples
Input
29
20
7
35
120
Output
215
Input
101
86
119
108
57
Output
481
Input
123
123
123
123
123
Output
643 | instruction | 0 | 99,309 | 9 | 198,618 |
"Correct Solution:
```
*a,=map(int,open(0));print(min(~-i%10for i in a)-sum(-i//10*10for i in a)-9)
``` | output | 1 | 99,309 | 9 | 198,619 |
Provide a correct Python 3 solution for this coding contest problem.
The restaurant AtCoder serves the following five dishes:
* ABC Don (rice bowl): takes A minutes to serve.
* ARC Curry: takes B minutes to serve.
* AGC Pasta: takes C minutes to serve.
* APC Ramen: takes D minutes to serve.
* ATC Hanbagu (hamburger patty): takes E minutes to serve.
Here, the time to serve a dish is the time between when an order is placed and when the dish is delivered.
This restaurant has the following rules on orders:
* An order can only be placed at a time that is a multiple of 10 (time 0, 10, 20, ...).
* Only one dish can be ordered at a time.
* No new order can be placed when an order is already placed and the dish is still not delivered, but a new order can be placed at the exact time when the dish is delivered.
E869120 arrives at this restaurant at time 0. He will order all five dishes. Find the earliest possible time for the last dish to be delivered.
Here, he can order the dishes in any order he likes, and he can place an order already at time 0.
Constraints
* A, B, C, D and E are integers between 1 and 123 (inclusive).
Input
Input is given from Standard Input in the following format:
A
B
C
D
E
Output
Print the earliest possible time for the last dish to be delivered, as an integer.
Examples
Input
29
20
7
35
120
Output
215
Input
101
86
119
108
57
Output
481
Input
123
123
123
123
123
Output
643 | instruction | 0 | 99,310 | 9 | 198,620 |
"Correct Solution:
```
dish = [int(input()) for i in range(5)]
reeltime = sorted([10 - (x%10) if x%10 != 0 else (x%10) for x in dish])
print(sum(dish)+sum(reeltime[:-1]))
``` | output | 1 | 99,310 | 9 | 198,621 |
Provide a correct Python 3 solution for this coding contest problem.
The restaurant AtCoder serves the following five dishes:
* ABC Don (rice bowl): takes A minutes to serve.
* ARC Curry: takes B minutes to serve.
* AGC Pasta: takes C minutes to serve.
* APC Ramen: takes D minutes to serve.
* ATC Hanbagu (hamburger patty): takes E minutes to serve.
Here, the time to serve a dish is the time between when an order is placed and when the dish is delivered.
This restaurant has the following rules on orders:
* An order can only be placed at a time that is a multiple of 10 (time 0, 10, 20, ...).
* Only one dish can be ordered at a time.
* No new order can be placed when an order is already placed and the dish is still not delivered, but a new order can be placed at the exact time when the dish is delivered.
E869120 arrives at this restaurant at time 0. He will order all five dishes. Find the earliest possible time for the last dish to be delivered.
Here, he can order the dishes in any order he likes, and he can place an order already at time 0.
Constraints
* A, B, C, D and E are integers between 1 and 123 (inclusive).
Input
Input is given from Standard Input in the following format:
A
B
C
D
E
Output
Print the earliest possible time for the last dish to be delivered, as an integer.
Examples
Input
29
20
7
35
120
Output
215
Input
101
86
119
108
57
Output
481
Input
123
123
123
123
123
Output
643 | instruction | 0 | 99,311 | 9 | 198,622 |
"Correct Solution:
```
abcde = [int(input()) for _ in range(5)]
s = list(map(lambda x: (10-x%10)%10, abcde))
print(sum(abcde)+sum(s)-max(s))
``` | output | 1 | 99,311 | 9 | 198,623 |
Provide a correct Python 3 solution for this coding contest problem.
The restaurant AtCoder serves the following five dishes:
* ABC Don (rice bowl): takes A minutes to serve.
* ARC Curry: takes B minutes to serve.
* AGC Pasta: takes C minutes to serve.
* APC Ramen: takes D minutes to serve.
* ATC Hanbagu (hamburger patty): takes E minutes to serve.
Here, the time to serve a dish is the time between when an order is placed and when the dish is delivered.
This restaurant has the following rules on orders:
* An order can only be placed at a time that is a multiple of 10 (time 0, 10, 20, ...).
* Only one dish can be ordered at a time.
* No new order can be placed when an order is already placed and the dish is still not delivered, but a new order can be placed at the exact time when the dish is delivered.
E869120 arrives at this restaurant at time 0. He will order all five dishes. Find the earliest possible time for the last dish to be delivered.
Here, he can order the dishes in any order he likes, and he can place an order already at time 0.
Constraints
* A, B, C, D and E are integers between 1 and 123 (inclusive).
Input
Input is given from Standard Input in the following format:
A
B
C
D
E
Output
Print the earliest possible time for the last dish to be delivered, as an integer.
Examples
Input
29
20
7
35
120
Output
215
Input
101
86
119
108
57
Output
481
Input
123
123
123
123
123
Output
643 | instruction | 0 | 99,312 | 9 | 198,624 |
"Correct Solution:
```
a=[int(input()) for _ in [0]*5]
b=[-(-i//10)*10-i for i in a]
print(sum(b)+sum(a)-max(b))
``` | output | 1 | 99,312 | 9 | 198,625 |
Provide a correct Python 3 solution for this coding contest problem.
The restaurant AtCoder serves the following five dishes:
* ABC Don (rice bowl): takes A minutes to serve.
* ARC Curry: takes B minutes to serve.
* AGC Pasta: takes C minutes to serve.
* APC Ramen: takes D minutes to serve.
* ATC Hanbagu (hamburger patty): takes E minutes to serve.
Here, the time to serve a dish is the time between when an order is placed and when the dish is delivered.
This restaurant has the following rules on orders:
* An order can only be placed at a time that is a multiple of 10 (time 0, 10, 20, ...).
* Only one dish can be ordered at a time.
* No new order can be placed when an order is already placed and the dish is still not delivered, but a new order can be placed at the exact time when the dish is delivered.
E869120 arrives at this restaurant at time 0. He will order all five dishes. Find the earliest possible time for the last dish to be delivered.
Here, he can order the dishes in any order he likes, and he can place an order already at time 0.
Constraints
* A, B, C, D and E are integers between 1 and 123 (inclusive).
Input
Input is given from Standard Input in the following format:
A
B
C
D
E
Output
Print the earliest possible time for the last dish to be delivered, as an integer.
Examples
Input
29
20
7
35
120
Output
215
Input
101
86
119
108
57
Output
481
Input
123
123
123
123
123
Output
643 | instruction | 0 | 99,313 | 9 | 198,626 |
"Correct Solution:
```
import math
l = [int(input()) for _ in range(5)]
l2 = [math.ceil(i/10)*10 for i in l]
diff = max([i-j for i,j in zip(l2,l)])
print(sum(l2) - diff)
``` | output | 1 | 99,313 | 9 | 198,627 |
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