What is the Probability - UVa 10056 概率
2015-02-17 17:42
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题意:从1-n轮回掷骰子,每个人每次有p的概率赢,问第m个人赢的概率。
思路:设其为ans,ans=P(前m-1人个都没中)*p+P(n个人都没中)*ans。
整理后为ans=p*pow(q,m-1)/(1-pow(q,n)); q=1-p。
AC代码如下:
思路:设其为ans,ans=P(前m-1人个都没中)*p+P(n个人都没中)*ans。
整理后为ans=p*pow(q,m-1)/(1-pow(q,n)); q=1-p。
AC代码如下:
#include<cstdio> #include<cstring> #include<cmath> using namespace std; int T,t,n,m; double ans,p,q; int main() { int i,j,k; scanf("%d",&T); for(t=1;t<=T;t++) { scanf("%d%lf%d",&n,&p,&m); q=1-p; if(p==0) ans=0; else ans=p*pow(q,m-1)/(1-pow(q,n)); printf("%.4f\n",ans); } }
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