Sicily 2682. Die Hard 4

最新推荐文章于 2022-02-20 00:29:31 发布

Nightonke

最新推荐文章于 2022-02-20 00:29:31 发布

阅读量610

点赞数

CC 4.0 BY-SA版权

分类专栏： sicily 文章标签： sicily

本文链接：https://blog.youkuaiyun.com/u012925008/article/details/44280681

sicily 专栏收录该内容

413 篇文章

订阅专栏

在电影《虎胆龙威4》中，恐怖分子提出了一个更复杂的水壶问题。给定不同容量的水壶，目标是在没有秤的情况下精确获取特定体积的水。本文探讨了如何判断这一任务是否可行。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

2682. Die Hard 4

Constraints

Time Limit: 1 secs, Memory Limit: 256 MB

Description

In Die Hard with a Vengeance, Simon's strange game of cat and mouse is played. In the park, a time bomb is about to go off. Detective John McClane and his sidekick Zeus must make exactly four gallons from five and three gallon jugs in five minutes. It’s no so easy for them, but finally they solve it in time:

(1) Fill the five gallon jug. Three gallon jug is empty.

(2) Empty three gallons from the five gallon jug into the three gallon jug.

(3) There remains two gallons in the five gallon jug. Empty the three gallon jug.

(4) Pour the two gallons into the three gallon jug.

(5) Fill the five gallon jug and pour one gallon from it into the three gallon jug - filling the three gallon jug.

(6) Four gallons remain in the five gallon jug.

In Die Hard 4, terrorists start another attack. This time, John faces a similar but more difficult problem. Given n jugs of A1, A2, …, An gallons respectively, no weighing scale, to obtain exactly W gallons. Because terrorists no time to prepare all n jugs, so they just ask whether it is possible.

Input

The input consists of multiple datasets. The end of the input is indicated by a line containing a zero.

The format of each dataset is as follows.

A1 A2 … An

Here, all data items are positive integers. n is the number of jugs (n<=100). A1, A2, …, An (Ai<10^9) are the volume of the jugs. W is the volume they wanted.

Output

For each dataset, output the result of the problem in a separate line: “YES” when the problem is solvable, “NO” otherwise.

Sample Input

Sample Output

YES
NO

Problem Source

系列热身赛5@2011年上半学期算法考试和4+2选拔赛

// Problem#: 2682
// Submission#: 3556152
// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License
// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/
// All Copyright reserved by Informatic Lab of Sun Yat-sen University
#include<cstdio>
int main(){while(1){int N,a,b,t;scanf("%d",&N);if(!N--)break;scanf("%d",&a);while(N--){scanf("%d",&b);while(b){t=a%b;a=b;b=t;}}scanf("%d",&b);printf("%s\n",b%a?"NO":"YES");}return 0;}