hdu4586 Play the Dice (probability expectation)

Title:

There is an n-sided die, the i-th face has the value a(i),
and m faces have a color.

Now roll the dice once, the score += the value
of the face that was cast . If you cast the dice on the colored face, you can roll it again to
find the desired score.

Data range: n<=200, m<=n, 0<=a(i)<=200,

solution:

n grids, m grids with color, set the weight of the color to be x, and the weight of no color to be y, then E = (xm + E) ∗ mn + yn − m ∗ n − mn The term is: E = x + yn − m = sum {a [i]} n − mn grids, m grids with color, \\ with color grid weight sum as x, no color weight sum Is y,\\ then E=(\frac{x}{m}+E)*\frac{m}{n}+\frac{y}{nm}*\frac{nm}{n}\\ shift Item: \\ E=\frac{x+y}{nm}=\frac{sum\{a[i]\}}{nm} n a grid sub , with a color color of grid sub- m th ,Is provided with a color color of grid sub- weight value and to X ,No color color of right values and to the y- ,That it E=(mx​+E)∗nm​+n−mand∗nn−m​Move items have:E=n−mx+and=n−msum{ a[i]}​

code：

#include <bits/stdc++.h>
using namespace std;

signed main(){
    
    
    int n,m;
    while(scanf("%d",&n)!=EOF){
    
    
        double sum=0;
        for(int i=1;i<=n;i++){
    
    
            int x;scanf("%d",&x);
            sum+=x;
        }
        scanf("%d",&m);
        for(int i=1;i<=m;i++){
    
    
            int x;scanf("%d",&x);
        }
        //要先判sum==0,然后判m==n.
        if(sum==0){
    
    
            puts("0.00");
            continue;
        }
        if(m==n){
    
    
            puts("inf");
            continue;
        }
        //
        double ans=sum*1.0/(n-m);
        printf("%.2f\n",ans);
    }
    return 0;
}

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