HDU5464 Clarke and problem

原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=5464

Clarke and problem

Problem Description

Clarke is a patient with multiple personality disorder. One day, Clarke turned into a student and read a book.
Suddenly, a difficult problem appears:
You are given a sequence of number a1,a2,…,an and a number p. Count the number of the way to choose some of number(choose none of them is also a solution) from the sequence that sum of the numbers is a multiple of p(0 is also count as a multiple of p). Since the answer is very large, you only need to output the answer modulo 109+7109+710^9+7

Input

The first line contains one integer T(1≤T≤10) - the number of test cases.
T test cases follow.
The first line contains two positive integers n,p(1≤n,p≤1000)
The second line contains n integers a1,a2,…an(|ai|≤109).

Output

For each testcase print a integer, the answer.

Sample Input

1
2 3
1 2

Sample Output

Hint:

2 choice: choose none and choose all.

题目大意

克拉克是一名人格分裂患者。某一天，克拉克分裂成了一个学生，在做题。
突然一道难题难到了克拉克，这道题是这样的：

给你n个数，要求选一些数（可以不选），把它们加起来，使得和恰好是p的倍数（0也是p的倍数），求方案数。

对于n很小的时候，克拉克是能轻易找到的。然而对于n很大的时候，克拉克没有办法了，所以来求助于你。

题解

我们注意到p⩽1000p⩽1000p\leqslant 1000，但是数据范围是10910910^9，背包显然不可做。但是，考虑到我们只需要凑出p的倍数，那么dp就只跟模p的余数有关系，那么我们读入时就可以将每个数模p缩小数据范围。

接下来，我们用二维数组dp[i][j]dp[i][j]dp[i][j]表示前iii个数可以凑出余数为j" role="presentation" style="position: relative;">jjj的方案数，那么就有状态转移方程：

dp[i][j]=dp[i−1][j]+dp[i−1][(j−x[i]+p) mod p]dp[i][j]=dp[i−1][j]+dp[i−1][(j−x[i]+p)modp]

dp[i][j]=dp[i-1][j]+dp[i-1][(j-x[i]+p)\ mod\ p]

具体来讲，我们将每个数分为两个状态：
1.选了第iii个数,那么方案数就是前i−1" role="presentation" style="position: relative;">i−1i−1i-1个数凑出余数为(j−x[i]+p)(j−x[i]+p)(j-x[i]+p)%p的方案数。
2.不选，那么显然跟dp[i−1][j]dp[i−1][j]dp[i-1][j]一样。

代码

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int M=1005,mod=1e9+7;
int x[M],n,p;
ll dp[M][M];
void in()
{memset(dp,0,sizeof(dp));scanf("%d%d",&n,&p);for(int i=1;i<=n;++i)scanf("%d",&x[i]),x[i]%=p;
}
void ac()
{dp[0][0]=1;for(int i=1;i<=n;++i)for(int j=0;j<=p;++j)dp[i][j]=(dp[i-1][j]+dp[i-1][(j-x[i]+p)%p])%mod;printf("%lld\n",dp[n][0]);
}
int main()
{int T;scanf("%d",&T);for(int i=1;i<=T;++i)in(),ac();return 0;
}