[hdu 4586] Play the Dice

设dp[i]表示掷了i次的期望得分dp[1]=sigma(A[i])/N,dp[i]=dp[i-1]*M/N

可以看出dp数组构成了一个等比数列

ans=sigma(dp[i])=dp[1]*(1-(M/N)^n)/(1-M/N)

当n趋近+oo时(M/N)^n趋近于0,则ans=dp[1]/(1-M/N)=sigma(A[i])/(N-M)

特判当dp[1]!=0且N=M时ans=inf

#include<bits/stdc++.h>
using namespace std;
#define maxn 205
double A[maxn],B[maxn];
int main(){
    int n,m;
    while(scanf("%d",&n)!=EOF){
        double all=0;
        for(int i=1;i<=n;i++){
            scanf("%lf",&A[i]);
            all+=A[i];
        }
        scanf("%d",&m);
        for(int i=1;i<=m;i++)
            scanf("%lf",&B[i]);
        if(all==0)printf("%.2lf
",0.0);
        else if(n==m)printf("inf
");
        else printf("%.2lf
",all/(n-m));
    }
    return 0;
}