Given a list of N numbers you will be allowed to choose any M of them. So you can choose in NCM ways. You will have to determine how many of these chosen groups have a sum, which is divisible by D.
Input
Input starts with an integer T (≤ 20), denoting the number of test cases.
The first line of each case contains two integers N (0 < N ≤ 200) and Q (0 < Q ≤ 10). Here N indicates how many numbers are there and Q is the total number of queries. Each of the next N lines contains one 32 bit signed integer. The queries will have to be answered based on these N numbers. Each of the next Q lines contains two integers D (0 < D ≤ 20) and M (0 < M ≤ 10).
Output
For each case, print the case number in a line. Then for each query, print the number of desired groups in a single line.
Sample Input
2
10 2
1
2
3
4
5
6
7
8
9
10
5 1
5 2
5 1
2
3
4
5
6
6 2
Sample Output
Case 1:
2
9
Case 2:
1
#include<iostream>
#include<stdio.h>
#include<cstring>
#define ll long long
using namespace std;
ll dp[15][20];//dp[j][k]表示选了j个数，余数为k的情况
int num[205];
int main()
{
int t,cnt=0;;
scanf("%d",&t);
while(t--)
{
int n,q;
scanf("%d%d",&n,&q);
for(int i=1;i<=n;i++)
scanf("%d",&num[i]);
int d,m;
printf("Case %d:\n",++cnt);
while(q--)
{
scanf("%d%d",&d,&m);
memset(dp,0,sizeof(dp));
dp[0][0]=1;//取0个数余数为0就只有一种情况
for(int i=1;i<=n;i++)
{
for(int j=m;j>=1;j--)
{
int tmp=num[i]%d;
for(int k=0;k<d;k++)
{
dp[j][k]+=dp[j-1][(k-tmp+d)%d];//要加d，避免出现负数
}
}
}
printf("%lld\n",dp[m][0]);
}
}
return 0;
}