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 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> HDU 5464Clarke and problem(DP)

HDU 5464Clarke and problem(DP)

編輯:C++入門知識

HDU 5464Clarke and problem(DP)


 

Clarke and problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 289 Accepted Submission(s): 131



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.

 

簡單DP題 求n個數中 m個數的和是p的倍數 m~(0-n)

 

#include 
#include 
#include 
#include 
#include 
#include
#include 
#include 
#include 
#include 
using namespace std;

int dp[1010][1010];
int num[1010];
const int Mod=1e9+7;
int main()
{
    int t;
    scanf(%d,&t);
    while(t--)
    {
        int n,p;
        scanf(%d%d,&n,&p);
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
            scanf(%d,&num[i]);
        int sum=1;
        for(int i=1;i<=n;i++)
        {
            num[i]%=p;
            if(num[i]<0) num[i]+=p;
            for(int j=0;j

 

 

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