swjtu2383(Music Problem)

题目链接：http://swjtuoj.cn/problem/2383/

当n大于3600时，一定可以，证明：当(presum[i]-presum[j])%3600 == 0时，成立，根据抽屉原理，当n > 3600时，一定有presum[i] == presum[j](mod 3600)

当n<=3600时，dp

f[i+1][j] |= f[i][j];
f[i+1][(j+tt[i])% 3600] |= f[i][j];

bool类型的 DP ，用现在的状态更新将来的状态

#include <iostream>#include <cstdio>#include <memory.h>#include <algorithm>using namespace std;bool f[3700][3700];int tt[100010];int main(){    int t;    cin >> t;    while (t > 0)    {        t--;        int n;        scanf("%d", &n);        for (int i=0; i<n; i++) scanf("%d", &tt[i]);        memset(f,0,sizeof(f));        if (n > 3600)        {            cout << "YES" << endl;            continue;        }        for (int i=0; i<n; i++) f[i+1][tt[i]%3600] = 1;        for (int i=0; i<n; i++)        {            for (int j=0; j<3600; j++)            {                f[i+1][j] |= f[i][j];                f[i+1][(j+tt[i])% 3600] |= f[i][j];            }        }        if (f[n][0])            cout << "YES" << endl;        else            cout << "NO" << endl;    }    return 0;}