Candy

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2912    Accepted Submission(s): 1316

Special Judge

Problem Description

LazyChild is a lazy child who likes candy very much. Despite being very young, he has two large candy boxes, each contains n candies initially. Everyday he chooses one box and open it. He chooses the first box with probability p and the second box with probability (1 - p). For the chosen box, if there are still candies in it, he eats one of them; otherwise, he will be sad and then open the other box.
He has been eating one candy a day for several days. But one day, when opening a box, he finds no candy left. Before opening the other box, he wants to know the expected number of candies left in the other box. Can you help him?

 

Input

There are several test cases.
For each test case, there is a single line containing an integer n (1 ≤ n ≤ 2 × 10⁵) and a real number p (0 ≤ p ≤ 1, with 6 digits after the decimal).
Input is terminated by EOF.

 

Output

For each test case, output one line “Case X: Y” where X is the test case number (starting from 1) and Y is a real number indicating the desired answer.
Any answer with an absolute error less than or equal to 10^-4 would be accepted.

 

Sample Input

10 0.400000100 0.500000124 0.432650325 0.325100532 0.4875202276 0.720000

 

Sample Output

Case 1: 3.528175Case 2: 10.326044Case 3: 28.861945Case 4: 167.965476Case 5: 32.601816Case 6: 1390.500000

 

Source

2012 Asia Chengdu Regional Contest

 

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很巧妙的一道概率题 ，注意题意，最后一次选得的箱子一定是空的

根据题意，我们推出公式：

      

为什么q和p的幂是n+1,是因为题意说最后一次是空的，所以那个盒子被打开n+1次

公    式

C(m,n)=n!/((n-m)!*m!)（m≤n）

性质1

C(m,n)= C(n-m,n)

性质2

C(m,n)=C(m-1,n-1)+C(m-1,n)

数据很大，组合数就不好解决，会造成精度缺失，最后看题解发现一个巧妙的方法

log函数: log(m*n)=log(m)+log(n),  log(m/n)=log(m)-log(n),运用这个性质和组合数公式，对组合数进行打表

代码如下：

代码中f[i]表示log(i的阶乘)

#include<iostream>#include<stdio.h>#include<math.h>#include<string.h>#include<stdlib.h>#include<algorithm>using namespace std;typedef long long ll;const int N=400005;const int mod=1e9+5;double f[N];double logc(int m,int n)  //快速排列组合函数C（n,m）=exp(lggc(n,m));{    return f[m]-(f[n]+f[m-n]);}int main(){    double p,q;    f[0]=0;    for(int i=1;i<N;i++)    {        f[i]=f[i-1]+log(i*1.0);    }    int n,count=1;    while(~scanf("%d%lf",&n,&p))    {        q=1.0-p;        double ans=0.0;         for(int k=1;k<=n;k++)        {            ans+=k*(exp(logc(2*n-k,n-k)+(n+1)*log(p)+(n-k)*log(q))+exp(logc(2*n-k,n-k)+(n+1)*log(q)+(n-k)*log(p)));        }        printf("Case %d: %.6lf\n",count++,ans);    }    return 0;}