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;
}