2012Chengdu B （快速组合数）

B - Candy

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 4465

Appoint description:  System Crawler  (2014-10-17)

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

 

http://mikewell.blog.163.com/blog/static/192914172201210180306917/

公式

快速排列组合函数：logC(m,n)，zuhe[i]其实就是i的阶乘，然后给取log值
这样zuhe[m]-zuhe[n]-zuhe[m-n]就是C(m,n)的log值
然后把其余也取对数，然后再求exp就好了！

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define M(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f

using namespace std;

int n;
double p;
double zuhe[400005];

double logC(int m,int n)
{
    return zuhe[m]-zuhe[n]-zuhe[m-n];
}

int main()
{
    int cas = 1;
    zuhe[0] = 0;
    for(int i = 1;i<400005;i++) zuhe[i] = zuhe[i-1]+log(i*1.0);
    while(scanf("%d%lf",&n,&p)==2)
    {
        double res = 0;
        for(int i = 0;i<=n;i++)
        {
            res+=(n-i)*exp(logC(n+i,i)+(n+1)*log(p)+(i)*log(1-p));
            res+=(n-i)*exp(logC(n+i,i)+(n+1)*log(1-p)+(i)*log(p));
        }
        printf("Case %d: %.16f\n",cas++,res);
    }
    return 0;
}