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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
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).
For each testcase print a integer, the answer.
1 2 3 1 2
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 }
hdu 5464 Clarke and problem(dp)
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原文地址：http://www.cnblogs.com/UniqueColor/p/4830174.html
