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.
简单DP题 求n个数中 m个数的和是p的倍数 m~(0-n)
#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
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<p;j++) dp[i][j]=dp[i-1][j];
dp[i][num[i]]=(dp[i][num[i]]+1)%Mod;
for(int j=0;j<p;j++)
{
dp[i][(num[i]+j)%p]=(dp[i-1][j]+dp[i][(num[i]+j)%p])%Mod;
}
}
printf("%d\n",dp[n][0]+1);
}
}
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