买书--计蒜客--dfs减枝，搜索好题

该题就两个状态，买或者不买，跟背包问题的dfs解法大同小异，代码如下：

#include <iostream>
#include <cmath>
using namespace std;
const long long maxn = 35;
long long p[maxn];
bool flag = false;
int cnt = 0;
void dfs(long long index,long long sumW,long long sumC,long long m,long long n,long long k)
{
    if(sumC >k || sumW > m)
        return;
    if(flag)
        return;
    if(index == n)
    {
        if(sumW==m && sumC == k)
        {
            flag = true;
        }
        return;
    }
   
    dfs(index+1,sumW,sumC,m,n,k);
    dfs(index+1,sumW+p[index],sumC+1,m,n,k);
}
int main()
{
    long long m,n,k;
    cin>>m>>n>>k;
    for(int i=0; i<n; i++)
    {
        cin>>p[i];
    }
    dfs(0,0,0,m,n,k);
    if(flag)
        cout<<"Yes"<<endl;
    else
        cout<<"No"<<endl;
    return 0;
}