acdream 群赛5 组合数学

Problem B: Dice Dice Dice

Time Limit: 10 Sec Memory Limit:
128 MB

Submit: 81 Solved: 17

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Description

There are 1111 ways in which five 6-sided dice (sides numbered 1 to 6) can be rolled so that the top three numbers sum to 15. Some examples are:
D1,D2,D3,D4,D5 = 4,3,6,3,5
D1,D2,D3,D4,D5 = 4,3,3,5,6
D1,D2,D3,D4,D5 = 3,3,3,6,6
D1,D2,D3,D4,D5 = 6,6,3,3,3
Now we have a extended problem:
In how many ways can n m-sided dice (sides numbered 1 to m) be rolled so that the top k numbers sum to p?

Input

There are multiple test cases. (about 15 groups)
For each test case, there is only four integers n, m, k, p. (1 <= k <= n <= 20, 3 <= m <= 12)

Output

For each test case, output an integer indicating the answer.

Sample Input

5 6 3 156 6 3 15

Sample Output

11117770

HINT

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
 
using namespace std;
#define LL long long
int n,m,k,p;
int re[109];
LL ans;
LL N(int t){
    LL ret = 1;
    for(int i=2;i<=t;i++)
    ret*=i;
    return ret;
}
void dfs(int t,int lim)
{
    if(t<k){
        for(int i=lim;i>0;i--)
        {
            re[t] = i;
            dfs(t+1,i);
        }
    }else if(t==k){
        int s = 0;
        for(int i=0;i<k;i++)
        s+=re[i];
        if(s!=p) return ;
     //   for(int i=0;i<k;i++)
      //  cout<<re[i]<<" ";cout<<endl;
        if(k<n)
        for(int i=lim;i>0;i--)
        {
            re[t] = i;
            dfs(t+1,i);
        }
        else
        {
            re
=lim;
            dfs(t+1,lim);
        }
        return ;
    }else if(t>k&&t<n)
    {
        for(int i=lim;i>0;i--)
        {
            re[t] = i;
            dfs(t+1,i);
        }
        return ;
    }
    if(t>=n){
        LL tmp = N(n);
       // cout<<"con:"<<endl;
       // for(int i=0;i<n;i++)
       // cout<<re[i]<<" ";cout<<endl;
       // cout<<tmp<<endl;
        int c=1;
        re
 = -1;
        for(int i=1;i<=n;i++)
        {
            if(re[i]!=re[i-1])
            {
                tmp/=N(c);
                c=1;
            }
            else
            c++;
        }
        ans+=tmp;
        return ;
    }
}
int main()
{
   // freopen("in.txt","r",stdin);
    while(~scanf("%d%d%d%d",&n,&m,&k,&p))
    {
       // cout<<"<<<<<<<<<<<"<<endl;
        ans = 0;
        dfs(0,m);
        cout<<ans<<endl;
    }
    return 0;
}