首页 > 试题广场 >

Music Problem

[编程题]Music Problem
  • 热度指数:46 时间限制:C/C++ 2秒,其他语言4秒 空间限制:C/C++ 128M,其他语言256M
  • 算法知识视频讲解
Listening to the music is relax, but for obsessive(强迫症), it may be unbearable.
HH is an obsessive, he only start to listen to music at 12:00:00, and he will never stop unless the song he is listening ends at integral points (both minute and second are 0 ), that is, he can stop listen at 13:00:00 or 14:00:00,but he can't stop at 13:01:03 or 13:01:00, since 13:01:03 and 13:01:00 are not an integer hour time.
Now give you the length of some songs, tell HH whether it's possible to choose some songs so he can stop listen at an integral point, or tell him it's impossible.
Every song can be chosen at most once.

输入描述:
The first line contains an positive integer T(1≤T≤60), represents there are T test cases. 
For each test case: 
The first line contains an integer n(1≤n≤105), indicating there are n songs. 
The second line contains n integers a1,a2…an (1≤ai≤109 ), the ith integer ai indicates the ith song lasts ai seconds.


输出描述:
For each test case, output one line "YES" (without quotes) if HH is possible to stop listen at an integral point, and "NO" (without quotes) otherwise.
示例1

输入

3
3
2000 1000 3000
3
2000 3000 1600
2
5400 1800

输出

NO
YES
YES

说明

In the first example it's impossible to stop at an integral point.
In the second example if we choose the first and the third songs, they cost 3600 seconds in total, so HH can stop at 13:00:00
In the third example if we choose the first and the second songs, they cost 7200 seconds in total, so HH can stop at 14:00:00
直接存在性背包会T,所以要模3600后存下模3600后的不同数字各有多少个,然后用多重背包。
发表于 2017-09-22 19:05:16 回复(0)

问题信息

上传者:牛客301599号
难度:
1条回答 1944浏览

热门推荐

通过挑战的用户