链接：https://ac.nowcoder.com/acm/contest/5203/B
来源：牛客网
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
题目大意：
给出N个数，问我们能否从中选出一些数字，使得其加和为3600的倍数。
思路：
我们不难想到，如果n>3600的话，无论物品的大小，我们肯定能够凑出3600的倍数来。所以问题就可以变成O（3600*3600）的一个暴力背包了。

#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<set>
#include<vector>
#include<map>
#include<queue>
#define ll long long
using namespace std;
bool dp[3605][3605];
int a[150000]; 
int main(){
    int T;
    cin>>T;
    while(T--)
    {
        memset(dp,false,sizeof(dp));
        int n;
        cin >> n;
        for(int i=1;i<=n;i++)
        {
            cin>>a[i];
        }
        if(n>3600){
            cout<<"YES\n";
        }else{
            for(int i=0;i<n;i++){
                dp[i+1][a[i+1]%3600]=true; 
                for(int j=0;j<3600;j++){
                     if(dp[i][j]==true)
                     {
                         dp[i+1][j]=true;
                         dp[i+1][(j+a[i+1])%3600]=true;
                     }
                }
            }
            if(dp[n][0]==true) cout<<"YES\n";
            else cout<<"NO\n";
        }
    }
    return 0;
}

同时也可以使用bitset

#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<set>
#include<vector>
#include<map>
#include<queue>
#include<bitset>
#define ll long long
using namespace std;
bitset<3600>s;
int a[5000000];
int main(){
    int T;
    cin >> T;
    while(T--){
        int n;
        cin>>n;
        s>>=3600;

        for(int i=1;i<=n;i++){
            cin>>a[i];
            a[i]%=3600;
        }
        for(int i=1;i<=n;i++){
            s|=(s<<a[i]|s>>(3600-a[i]));
            s[a[i]]=1;
        }
        if(s[0]==1){
            cout<<"YES\n"; 
        }else{
            cout<<"NO\n";
        }
    }
    return 0;
}