hdu 5464 Clarke and problem

题意：有n个数，a_0, a_1, ... a_(n-1)，从中选取一部分数(可以一个都不选)，使得其和为p的倍数，问方案数。

解：这题怎么说呢。。。就差在把背包这俩字甩我脸上了。。。but，昨天还是没有做出来。-_-!

#include <iostream>
#include <algorithm>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>

using namespace std;


typedef unsigned int uint;
typedef long long LL;
typedef unsigned long long uLL;

const int N = 1005;
const double PI = acos(-1.0);
const double eps = 1e-6;
const double INF = 1e9+7;
const int MOD = 1000000007;

int f[N][N];
int a[N];
int n, p;

int main() {
    //freopen("data.in", "r", stdin);
    int T;
    scanf("%d", &T);
    while(T--) {
        scanf("%d%d", &n, &p);
        memset(f, 0, sizeof(f));
        //int off = N;
        for(int i = 1; i <= n; ++i) {
            scanf("%d", &a[i]);
            a[i] = (a[i]%p + p)%p;
        }
        f[0][0] = 1;
        for(int i = 1; i <= n; ++i) {
            for(int j = 0; j <= p; ++j) {
                    f[i][j] = (f[i][j] + f[i-1][j])%MOD;
                //if(f[i-1][j + off] != 0) {
                    f[i][(j + a[i])%p] = (f[i][(j + a[i])%p] % MOD + f[i-1][j] % MOD) % MOD;
                //}
            }
        }
        printf("%d\n", f[n][0] % MOD);
    }
    return 0;
}