hdu 5464 Clarke and problem 动态规划

Clarke and problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 844    Accepted Submission(s): 340

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)  

定义dp[i][j]表示前i个数字构成的所有集合中，集合内所有数字相加之和对p取模为j的集合的个数。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;
int dp[2][1001];
#define mod 1000000007
int main(){
    int t,n,p;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&p);
        int w, u = 0, v = 1;
        memset(dp,0,sizeof(dp));
        dp[0][0] = 1;
        for(int i = 0;i < n; i++){
            scanf("%d",&w);
            for(int j = 0;j < p; j++)
                dp[v][j] = dp[u][j];
            w %= p;
            if(w < 0) w+=p;
            for(int j = 0;j < p ;j++){
                dp[v][w] += dp[u][j];
                if(dp[v][w] >= mod) dp[v][w] -= mod;
                w++;
                if(w >= p) w -= p;
            }
            swap(u,v);
        }
        printf("%d\n",dp[u][0]);
    }
    return 0;
}