题目大意：给你两个罐子，里面有糖果每次只能从一个罐子里面取一个糖果，打开A的概率为p，问当一个罐子取完之后，另一个罐子剩糖果的期望是多少。
我们可以知道最少是取第n+1次的时候才会有一个罐子为空，我们可以推出组合公式：
(n-k)*C(n+k, k)*((1-p)^(n+1)*p^k+(1-p)^k*p^(n+k))；0 <= k && k <= n-1。
求一个和就是所有的组合情况了，但是组合数很大我们可以用log来进行优化。
我们已知：C(n,m) = m!/n!/(m-n)! = log(m!)/log(n!)/log(m-n)。
m！= 1*2*……*m = log(1)+log(2)+……+log(m).
先打表在直接求就可以了啊。
C(m,n)=exp(logC(m,n))
Candy
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2229 Accepted Submission(s): 958
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 0.400000 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
Source
2012 Asia Chengdu Regional Contest
#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <time.h>
#include <stack>
#include <map>
#include <set>
#define eps 1e-8
///#define LL long long
#define LL __int64
#define INF 0x3f3f3f
#define PI 3.1415926535898
#define mod 1000000007
using namespace std;
const int maxn = 505000;
double f[maxn];
double logc(int m, int n)///C(n,m) = m!/n!/(m-n)!
{
return f[m]-f[n]-f[m-n];
}
int main()
{
f[0] = 0;
for(int i = 1; i <= 400005; i++) f[i] = f[i-1]+log(i*1.0);
int Case = 1;
int n;
double p;
while(~scanf("%d %lf",&n, &p))
{
double sum = 0.0;
for(int k = 0; k < n; k++)
{
///sum += (n-k)*C(n+k, k)*((1-p)^(n+1)*p^k+(1-p)^k*p^(n+k));
sum += 1.0*(n-k)*(exp(logc(n+k, k)+(n+1)*1.0*log(1.0-p)+k*1.0*log(p*1.0)) + exp(logc(n+k, k)+(n+1)*1.0*log(p*1.0)+k*1.0*log(1.0-p)));
}
printf("Case %d: %.6lf\n",Case++, sum);
}
}