[hdu 4586] Play the Dice

Posted 2020-07-10 Ngshily

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

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define maxn 205
 4 double A[maxn],B[maxn];
 5 int main(){
 6     int n,m;
 7     while(scanf("%d",&n)!=EOF){
 8         double all=0;
 9         for(int i=1;i<=n;i++){
10             scanf("%lf",&A[i]);
11             all+=A[i];
12         }
13         scanf("%d",&m);
14         for(int i=1;i<=m;i++)
15             scanf("%lf",&B[i]);
16         if(all==0)printf("%.2lf\\n",0.0);
17         else if(n==m)printf("inf\\n");
18         else printf("%.2lf\\n",all/(n-m));
19     }
20     return 0;
21 }

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