HDU - 4336 Card Collector min-max 容斥

公式:$max(S)=\sum_{ T\in S} (-1)^{|T|-1} min(S)$   

code: 

#include <bits/stdc++.h>   
#define ll long long  
#define N 23   
#define setIO(s) freopen(s".in","r",stdin) 
using namespace std; 
int n;     
double perc[N],ans;  
void dfs(int u,double sum,double opt) 
{
    if(u==n) 
    {
        if(sum>1e-9) 
            ans+=opt/sum;   
        return; 
    }
    dfs(u+1,sum+perc[u+1],opt*-1);   
    dfs(u+1,sum,opt);   
}
int main() 
{ 
    // setIO("input");      
    while(scanf("%d",&n)!=EOF) 
    {
        for(int i=1;i<=n;++i) 
            scanf("%lf",&perc[i]);                    
        ans=0,dfs(0,0,-1),printf("%.4f\n",ans);    
    }
    return 0; 
}

  

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转载自www.cnblogs.com/guangheli/p/12820993.html