BOJ 1580 Shoot

Shoot

Accept:7     Submit:7

Time Limit:1000MS     Memory Limit:65536KB

Description

Kimi has two types of guns in his gun box. When Kimi use Gun 1 to shoot, there's a probability of p to hit the target. Meanwhile, when he uses Gun 2, the probability of making an accurate shoot becomes q.

One day, Kimi went shooting with totally (a+b) guns, where a for Gun 1 and the other b for Gun 2. He randomly chose a gun in his box and began N times of shootings. But shockingly, he got all his shoots missing! However, when Kimi wanted to took the other type of gun to make the next round of shootings, he found that he didn't remember which type of gun he had taken just then. Now it becomes your task: what's the probability that he'd taken Gun 1 at the first time?

Input Format

An integer T(T≤100) will exist in the first line of input, indicating the number of test cases.

Each test case consists of one line with three integers N,a,b(1≤a,b≤104,1≤N≤10) and two real numbers p,q(0≤p,q≤1).

Output Format

Output the probability that he'd taken Gun 1. Correct the result to 6 decimal places.

Sample Input

1 3 1 2 0.5 0.5

Sample Output

0.333333

 

新生排位赛第四场的D题，一道计算概率的问题

晚上听学长讲貌似是用了一个条件概率的公式，不过还是觉得比赛时自己那个不太专业的做法比较好理解

具体计算的方法参见代码……

 

#include<iostream>
#include<cstdio>

using namespace std;

double power(double a,int b)
{
    double result=1;
    for(int i=1;i<=b;i++)
        result*=a;
    return result;
}

int main()
{
    int t;
    int n;
    long a,b;
    double p,q;

    scanf("%d",&t);

    while(t--)
    {
        double ap,bp;
        scanf("%d %ld %ld %lf %lf",&n,&a,&b,&p,&q);
        ap=a*power(1-p,n)/(a+b);
        bp=b*power(1-q,n)/(a+b);
        printf("%.6lf\n",ap/(ap+bp));
    }

    return 0;
}