Day11 - B - Dice (III) LightOJ - 1248
设dp_i为已经出现了i面,需要的期望次数,dp_n=0
那么dp_i= i/n*dp_i + (n-i)/n*dp_(i+1) + 1
现在已经i面了,i/n的概率再选择一次i面,(n-i)/n的概率选到新的一面,分别乘其期望次数,并且这次丢过,所以+1
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<bits/stdc++.h> using namespace std; #define lowbit(x) ((x)&(-x)) typedef long long LL; typedef pair<int,int> pii; const int maxm = 1e5+5; double dp[maxm]; void run_case() { int n; cin >> n; dp[n] = 0; for(int i = n-1; i >= 0; --i) dp[i] = dp[i+1] + (double)n / (n-i); cout << dp[0] << "\n"; } int main() { ios::sync_with_stdio(false), cin.tie(0); cout.flags(ios::fixed);cout.precision(10); int t; cin >> t; //while(t--) for(int i = 1; i <= t; ++i) { cout << "Case " << i << ": "; run_case(); } cout.flush(); return 0; }