LightOJ1125 Divisible Group Sums

Divisible Group Sums

Given a list of N numbers you will be allowed to choose any M of them. So you can choose in ^NC_M 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

从n个数字里面取m个有C(n,m)种取法，问有多少种取法使得总和是D的倍数

因为询问好多组(m,d)，所以每次都要对不同的d先取模，然后显然就是个背包啦，数字不超过d，只取m个，最大容量只有200

然后特么给的ai还有负数，，，我真是，，，

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<ctime>
#define LL long long
#define inf 0x7ffffff
#define pa pair<LL,int>
#define mkp(a,b) make_pair(a,b)
using namespace std;
inline LL read()
{
    LL x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,q;
int a[210];
int b[210];
LL f[210][210];
inline void work(int cur)
{
    printf("Case %d:
",cur);
    n=read();q=read();
    for (int i=1;i<=n;i++)a[i]=read();
    for (int i=1;i<=q;i++)
    {
        int mod=read(),m=read();
        memset(f,0,sizeof(f));f[0][0]=1;
        for (int j=1;j<=n;j++)b[j]=(a[j]%mod+mod)%mod;
        for (int j=1;j<=n;j++)
        {
            for (int k=m;k>=1;k--)
                for (int l=m*mod;l>=b[j];l--)
                f[k][l]+=f[k-1][l-b[j]];
        }
        LL sum=0;
        for (int j=0;j<=m*mod;j+=mod)sum+=f[m][j];
        printf("%lld
",sum);
    }
}
int main()
{
    int T=read(),tt=0;
    while (T--)work(++tt);
}