HDU 5776 Sum

题意：问是否存在序列n的某子序列之和是m的倍数

首先求每位%m后的前缀和 【 (a+b)%k == (a%k +b%k)%k】， a[i]=(a[i]+a[i-1])%m，若出现余数为0，则子序列存在

统计每个余数出现的次数，若出现两个相同余数 即a[k] == a[i] k>i, 或n>m（抽屉原理：m个数放进m-1个抽屉  必有两个数在同一个抽屉里）则子序列i+1...k必是m的倍数

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<queue>
#include<stack>
#include<list>
#include<set>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> p;
typedef long double ld;
#define mem(x) memset(x, 0, sizeof(x))
#define me(x) memset(x, -1, sizeof(x))
#define fo(i,n) for(i=0; i<n; i++)
#define sc(x) scanf("%lf", &x)
#define pr(x) printf("%lld
", x)
#define pri(x) printf("%lld ", x)
#define lowbit(x) x&-x
const ll MOD = 1e18 +7;
const ll N = 6e6 +5;
ll a[N], vis[N];
int main()
{
    ll i, j, k, l=0;
    ll n, m, t, x;
    cin>>t;
    while(t--)
    {
        cin>>n>>m;
        mem(vis);
        ll f=0;

        for(i=1; i<=n; i++)
            cin>>a[i], a[i]+=a[i-1];
        if(n>m)
        {
            cout<<"YES"<<endl;
            continue;
        }
        for(i=1; i<=n; i++)
        {
            a[i]%=m;
            vis[a[i]]++;
            if(vis[a[i]]>1 || !a[i])
                f=1;
        }
        if(f) cout<<"YES"<<endl;
        else cout<<"NO"<<endl;
    }
    return 0;
}