Divisible Group Sums
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
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，，，
1 #include<cstdio>
2 #include<iostream>
3 #include<cstring>
4 #include<cstdlib>
5 #include<algorithm>
6 #include<cmath>
7 #include<queue>
8 #include<deque>
9 #include<set>
10 #include<map>
11 #include<ctime>
12 #define LL long long
13 #define inf 0x7ffffff
14 #define pa pair<LL,int>
15 #define mkp(a,b) make_pair(a,b)
16 using namespace std;
17 inline LL read()
18 {
19 LL x=0,f=1;char ch=getchar();
20 while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
21 while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
22 return x*f;
23 }
24 int n,q;
25 int a[210];
26 int b[210];
27 LL f[210][210];
28 inline void work(int cur)
29 {
30 printf("Case %d:\n",cur);
31 n=read();q=read();
32 for (int i=1;i<=n;i++)a[i]=read();
33 for (int i=1;i<=q;i++)
34 {
35 int mod=read(),m=read();
36 memset(f,0,sizeof(f));f[0][0]=1;
37 for (int j=1;j<=n;j++)b[j]=(a[j]%mod+mod)%mod;
38 for (int j=1;j<=n;j++)
39 {
40 for (int k=m;k>=1;k--)
41 for (int l=m*mod;l>=b[j];l--)
42 f[k][l]+=f[k-1][l-b[j]];
43 }
44 LL sum=0;
45 for (int j=0;j<=m*mod;j+=mod)sum+=f[m][j];
46 printf("%lld\n",sum);
47 }
48 }
49 int main()
50 {
51 int T=read(),tt=0;
52 while (T--)work(++tt);
53 }
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