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]

 1 #pragma comment(linker, "/STACK:1024000000,1024000000")
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<math.h>
 7 #include<algorithm>
 8 #include<queue>
 9 #include<set>
10 #include<bitset>
11 #include<map>
12 #include<vector>
13 #include<stdlib.h>
14 #include <stack>
15 using namespace std;
16 int dirx[]={0,0,-1,1};
17 int diry[]={-1,1,0,0};
18 #define PI acos(-1.0)
19 #define max(a,b) (a) > (b) ? (a) : (b)
20 #define min(a,b) (a) < (b) ? (a) : (b)
21 #define ll long long
22 #define eps 1e-10
23 #define MOD 1000000007
24 #define N 1006
25 #define inf 1e12
26 int n,p;
27 int a[N];
28 int dp[N][N];
29 int main()
30 {
31     int t;
32     scanf("%d",&t);
33     while(t--){
34         scanf("%d%d",&n,&p);
35         for(int i=1;i<=n;i++){
36             scanf("%d",&a[i]);
37             a[i]=(a[i]%p+p)%p;
38         }
39
40         memset(dp,0,sizeof(dp));
41         dp[0][0]=1;
42         for(int i=1;i<=n;i++){
43             for(int j=0;j<p;j++){
44                 dp[i][j]+=dp[i-1][j];
45                 dp[i][j]%=MOD;
46
47                 dp[i][(j+a[i])%p]+=(dp[i-1][j]);
48                 dp[i][(j+a[i])%p]%=MOD;
49             }
50         }
51
52         printf("%d\n",dp[n][0]);
53
54     }
55     return 0;
56 }