标签：term ide alt 背包 lin base span pen ext
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 }
LightOJ1125 Divisible Group Sums
标签：term ide alt 背包 lin base span pen ext
原文地址：http://www.cnblogs.com/zhber/p/7152734.html
