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Batting Practice

After being all out for 58 and 78 in two matches in the most prestigious tournament in the world, the coach of a certain national cricket team was very upset. He decided to make the batsmen practice a lot. But he was wondering how to make them practice, because the possibility of getting out seems completely random for them. So, he decided to keep them in practice as long as he can and told them to practice in the net until a batsman remains not-out for k₁ consecutive balls. But if the batsman continues to be out for consecutive k₂ balls, then the coach becomes hopeless about the batsman and throws him out of the team. In both cases, the practice session ends for the batsman. Now the coach is wondering how many balls the practice session is expected to take.

For a batsman the probability of being out in a ball is independent and is equal to p. What is the expected number of balls he must face to remain not out for k₁ consecutive balls or become out in consecutive k₂ balls.

Input

Input starts with an integer T (≤ 15000), denoting the number of test cases.

Each case starts with a line containing a real number p (0 ≤ p ≤ 1) and two positive integers k₁ and k₂ (k₁ + k₂ ≤ 50). pwill contain up to three digits after the decimal point.

Output

For each case, print the case number and the expected number of balls the batsman will face. Errors less than 10^-2 will be ignored.

Sample Input

5

0.5 1 1

0.5 1 2

0.5 2 2

0.19 1 3

0.33 2 1

Sample Output

Case 1: 1

Case 2: 1.5

Case 3: 3

Case 4: 1.2261000000

Case 5: 1.67

闲来无事，做做数学题也是极好的，我只是感觉数学很有有趣罢了

独立事件的概率是p，如果连续成功的k1次或者连续失败k2次结束，问需要多少次的期望值

设f(x)为连续成功x次的期望,g(x)为连续失败x次的期望
f(k1)=g(k2)=0

然后根据期望的公式推啊，我要刚开始成功的概率已经是1了，那么期望值肯定是k2，因为我不可能失败，但是我成功的几率接近于0，那么我只要成功k1次就好了

#include <bits/stdc++.h>
using namespace std;
int main(){
int T,ca=1;
scanf("%d",&T);
while(T--){
    double p,q;
    int k1,k2;
    scanf("%lf%d%d",&p,&k1,&k2);
    q=1-p;
    if(p<1e-6)
        printf("Case %d: %d
",ca++,k1);
    else if(q<1e-6)
    printf("Case %d: %d
",ca++,k2);
    else{
    double a=1-pow(q,k1-1),b=1-pow(p,k2-1);
    double x=(a*b/q+a/p)/(1-a*b),y=b*x+b/q;
    printf("Case %d: %f
",ca++,p*y+q*x+1);}
}
return 0;}