Candy HDU - 4465 （数学）（概率期望）

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

这题在黑书上是有的，但是我给忘了，这个专题开辟的太晚了，只能希望在比赛的时候不要遇到，但是今天就认真的记录一下吧

那么我们看有多少种情况，假设当停止摸糖时，剩下i颗糖

那么，一共操作了n+n−i

假设这i颗糖在第二个盒子里

则剩下i颗的概率是：Cn2n−ipn+1(1−p)n−i

同理在第一个盒子里剩下：

Cn2n−i(1−p)n+1pn−i

因为精度问题，得取个对数（学到了）:p(i)=elnCn2n−i+(n+1)lnp+(n−i)ln(1−p)+elnCn2n−i+(n+1)ln(1−p)+(n−i)lnp

然后ans=∑i=1np(i)∗i

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;
const int maxn=400040;
double fac[maxn],p,q;
int n;
double cal(int n,int m)
{
return fac
-fac[m]-fac[n-m];
}
int main()
{
int cas=0;
memset(nlog,0,sizeof(nlog));
for(int i=1;i<maxn;i++)
fac[i]=fac[i-1] + log(i);
while(scanf("%d%lf",&n,&p)!=EOF)
{
q=log(1-p);
p=log(p);
double ans=0;
for(int i=1;i<=n;i++)
{
double v1=cal(2*n-i,n)+(n+1)*p+(n-i)*q;
double v2=cal(2*n-i,n)+(n+1)*q+(n-i)*p;
ans+=i*(exp(v1)+exp(v2));
}
printf("Case %d: %.6lf\n",++cas,ans);
}
return 0;
}