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=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 int n,q;
 int a[];
 int b[];
 LL f[][];
 inline void work(int cur)
 {
     printf("Case %d:\n",cur);
     n=read();q=read();
     for (int i=;i<=n;i++)a[i]=read();
     for (int i=;i<=q;i++)
     {
         int mod=read(),m=read();
         memset(f,,sizeof(f));f[][]=;
         for (int j=;j<=n;j++)b[j]=(a[j]%mod+mod)%mod;
         for (int j=;j<=n;j++)
         {
             for (int k=m;k>=;k--)
                 for (int l=m*mod;l>=b[j];l--)
                 f[k][l]+=f[k-][l-b[j]];
         }
         LL sum=;
         for (int j=;j<=m*mod;j+=mod)sum+=f[m][j];
         printf("%lld\n",sum);
     }
 }
 int main()
 {
     int T=read(),tt=;
     while (T--)work(++tt);
 }

LightOJ 1125