Cows and Cars UVA - 10491——初级概率题目
2017-06-01 21:32
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Think: 1>题意理解:假若有三个门,你如果一开始选了牛,然后可以换的时候直接换,那么最后得到车的概率就是最初选了牛的概率:现在题目扩展,有n头牛,m辆车,打开c个门,这时候问按照换门方法得到车的概率,这时候多了情况变化,那就是你一开始选了车,后面换门的时候还是有可能选到车,还有就是如果你一开始选了牛,不一定后面会选到车,因此实际上三门选车问题就成为了一个特殊条件,展开两种情况,第一种情况选牛再选车,概率为(n/(n+m))*((m)/(n+m-c-1));第二种情况选车再选车,概率为(m/(n+m))*((m-1)/(n-m-c-1));两种情况概率相加即为获得车的概率
以下为Accepted代码
#include <cstdio> #include <cstring> using namespace std; int main(){ int n, m, c; while(scanf("%d %d %d", &n, &m, &c) != EOF){ double f; f = (double)(m*(n+m-1))/(double)((n+m)*(n+m-c-1)); printf("%.5lf\n", f); } return 0; }
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