Music Problem

文章目录

- 题目描述
- 题意：
- 题解：

传送

时间限制：C/C++ 2秒，其他语言4秒空间限制：C/C++ 131072K，其他语言262144K
64bit IO Format: %lld

题目描述

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

题意：

给你n个数，这些数自由组合能不能凑出3600的倍数

题解：

我一开始想到的是前缀和，后来感觉dp最直接
dp[x]=1表示能组成x这个数
dp = 0表示组不了
cnt是中间数组，暂时存储本轮的数值
因为求能不能组成3600，可以用mod，3600的倍数mod后都是0，直接求dp[0]是否等于1
每读取一个a，就把a与之前所求的值进行相加存在cnt里，然后再给dp[]，cnt就是工具人

#include<bits/stdc++.h>
#define mem(a) memset(a,0,sizeof(a))using namespace std;
const int maxn=1e5+3;
bool dp[maxn],cnt[maxn];
const int mod=3600;
int main()
{int t,n;scanf("%d",&t);while(t--){mem(dp);mem(cnt);scanf("%d",&n);for(int i=1;i<=n;i++){int a;cin>>a;a%=3600;if(!dp[0]){for(int j=0;j<3600;j++){if(dp[j]>0||j==0){cnt[(a+j)%3600]=1;}}for(int j=0;j<3600;j++){if(cnt[j])dp[j]=1;if(cnt[j]==1)cnt[j]=0;}//	mem(cnt);}}if(!dp[0])cout<<"NO"<<endl;else cout<<"YES"<<endl;}return 0;
}