hdu 5464 Clarke and problem(dp)

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.

 

 

Source

BestCoder Round #56 (div.2)

 

 附上中文题目：

克拉克是一名人格分裂患者。某一天，有两个克拉克（aaa和bbb）在玩一个方格游戏。  
这个方格是一个n∗mn*mn∗m的矩阵，每个格子里有一个数ci,jc_{i, j}c​i,j​​。  
aaa想开挂，想知道如何打败bbb。  
他们要玩qqq次游戏，每一次做一次操作：  
1. 取出当中的一个子矩阵(x1,y1)−(x2,y2)(x_1, y_1)-(x_2, y_2)(x​1​​,y​1​​)−(x​2​​,y​2​​)玩游戏。两个人轮流行动，每一次只能从这个子矩阵中的一个方格ci,jc_{i, j}c​i,j​​中减掉一个的数d(1≤d≤ci,j)d(1 \le d \le c_{i, j})d(1≤d≤c​i,j​​)，当一个格子的数为000时则不能减。如果操作完后另一者无法操作，那么胜利。否则失败。现在aaa作为先手，想知道是否存在一种方案使得自己胜利。  
2. 将ci,jc_{i, j}c​i,j​​的数改成bbb  

 

设d(i,j)表示前i个数，模p为j的方案数，则容易得到  d(0,0)=1；

状态转移：dp[i][j]+=dp[i-1][j];   dp[i][(j+a[i])%p]+=(dp[i-1][j]);

最后的答案为 dp[n][0]

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<bitset>
#include<map>
#include<vector>
#include<stdlib.h>
#include <stack>
using namespace std;
int dirx[]={0,0,-1,1};
int diry[]={-1,1,0,0};
#define PI acos(-1.0)
#define max(a,b) (a) > (b) ? (a) : (b)  
#define min(a,b) (a) < (b) ? (a) : (b)
#define ll long long
#define eps 1e-10
#define MOD 1000000007
#define N 1006
#define inf 1e12
int n,p;
int a[N];
int dp[N][N];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&p);
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            a[i]=(a[i]%p+p)%p;
        }
        
        memset(dp,0,sizeof(dp));
        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][j]%=MOD;

                dp[i][(j+a[i])%p]+=(dp[i-1][j]);
                dp[i][(j+a[i])%p]%=MOD;
            }
        }
        
        printf("%d\n",dp[n][0]);
        
    }
    return 0;
}