hdu(数值计算 概率）

                                                    Candy

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

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 × 10 ⁵) 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 0.400000
100 0.500000
124 0.432650
325 0.325100
532 0.487520
2276 0.720000
  
  

 

Sample Output

  
  

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

 

本题题意是给你两个罐子，每个里面都有n个糖，现在LazyChild每天要随机从一个里面拿出一个糖，且选择其中一个罐子的概率为p，问当他某天打开罐子时发现里面没有糖了，问另外一个罐子中糖果数的期望。

求出期望值为：

                 （晕啊，发现打出这个式子比求出这个式子还难，汗碎了一键盘啊）

思路：  刚开始的时候想直接求p的n+1次方和q的n+1次方,后来发现测试数组有问题，因为一旦n的值过大后，肯定会出现下溢的，后来发现，反正p 和 q 都是乘进去的，可以在0到n-1循环中一次次的乘进去的，然后还有：

C(n+k,k)=C(n+k-1,k-1)*(n+k)/k;所以C(n+k,k)*p^(n+1)*q^k=C(n+k-1,k-1)*p^(n+1)*q^(k-1)*(n+k-1)/(k-1)*q;

代码：

#include<iostream>
#include<cstdio>
using namespace std;

double sove(int n,double p)
{
    double ans1,ans2,temp=1.000000;
    ans2=n*(1-p);ans1=n*p;//k=0的情况
    for(int k=1;k<=n;k++)//按理说k是小于n的，但这样方便求p和q的n+1次方
    {
        temp = temp*(n+k)*p*(1-p)/k;
        ans1 += temp*(n-k);//而且k=n后ans1和ans2在此处的值也不会改变
        ans2 += temp*(n-k);
        ans1 *= p;
        ans2 *= (1-p);
    }
    return ans1+ans2;
}

int main()
{
    int n;double p;
    int case1=1;
    while(~scanf("%d %lf",&n,&p))
    {
        double ans;
        ans=sove(n,p);
        printf("Case %d: %lf\n",case1++,ans);
    }
    return 0;
}