HDU 5464 Clarke and problem(DP 01背包)

Clarke and problem
Time Limit 20001000 MS (JavaOthers)    Memory Limit 6553665536 K (JavaOthers)
Total Submission(s) 400    Accepted Submission(s) 179




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) 

/*01背包DP*/#include <iostream>#include <cstdio>#include <algorithm>#include <cstdlib>#include <string.h>using namespace std;const int N=1000+10;const int mod=1e9+7;int num[N];int p,n;int dp[N][N];int main(){int t,i,j;scanf("%d",&t);while(t--){memset(dp,0,sizeof(dp));dp[0][0]=1;scanf("%d%d",&n,&p);for(i=1;i<=n;i++){scanf("%d",&num[i]);num[i]%=p;//这里被刘大哥指明...  处理负数的  将他变成整数 num[i]=(num[i]+p)%p;   }      for(i=1;i<=n;i++){   //枚举每个数拿还是不拿     for(j=0;j<p;j++){   //枚举余数      //不拿  就是余数为j加上i-1余数为j的数量        dp[i][j]=(dp[i][j]+dp[i-1][j])%mod;   //拿就 是余数为拿后的加上前I-1余数为拿前的 dp[i][(j+num[i])%p]=(dp[i][(j+num[i])%p]+dp[i-1][j])%mod;}}printf("%d\n",dp[n][0]);}    return 0;}/*   暴搜超时 *//*#include <iostream>#include <cstdio>#include <algorithm>#include <cstdlib>using namespace std;const int N=1000+10;const int mod=1000000000+7;int num[N];int p,n;int re;void dfs(int now,int sum){int i;if(sum%p==0)  {  re++;  re%=mod;  return;  }else if(sum>p||now>=n) return;dfs(now+1,sum+num[now]);dfs(now+1,sum);}int main(){int t,i;scanf("%d",&t);while(t--){re=1;scanf("%d%d",&n,&p);for(i=0;i<n;i++)  scanf("%d",&num[i]);sort(num,num+n);dfs(0,0);printf("%d\n",re);}    return 0;}*/