dp hdu 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+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

2

Hint:
2 choice: choose none and choose all.

~开心，自己根据写出来的，还特别简洁。

 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <vector>
 #define ll __int64
 #define mod 1000000007
 using namespace std;
 int dp[][],nums[];
 int main(void)
 {
     int t;
     cin>>t;
     while(t--)
     {
         int n,p;
         scanf("%d %d",&n,&p);
         for(int i = ; i <= n; i++)
         {
             scanf("%d",&nums[i]);
             nums[i] %= p;
             if(nums[i] < )
                 nums[i] += p;
         }

         memset(dp,,sizeof(dp));
         dp[][] = ;
         for(int i = ; i <= n; i++)
         {
             for(int j = ; j <= p; j++)
             {
                 dp[i][j] = dp[i-][j] + dp[i-][(j-nums[i]+p)%p];
                 dp[i][j] %= mod;
             }
         }
         printf("%d\n",dp[n][]);
     }
     return ;
 }