HDU 4465 Candy【指数表示法】E

Candy

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2775    Accepted Submission(s): 1243
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

解题思路：

这题公式不难推:

E=sigma{ (n-i)*C(n+i,i)*p^n*(1-p)^i+(n-i)*C(n+i,i)*(1-p)^n*p^i}

/

sigma{ C(n+i,i)*p^n*(1-p)^i +C(n+i,i)*(1-p)^n*p^i}

中间值会溢出，用log表示即可。

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<algorithm>#include<map>#include<string>#include<queue>#include<vector>#include<list>#include<bitset>//#pragma comment(linker,"/STACK:1024000000,1024000000")using namespace std;typedef long long ll;#define INF 0x3f3f3f3fint n;double p;int main(){    int tt=0;    while(~scanf("%d%lf",&n,&p))    {        long double ans=0,sum=0;        long double l=(n+1)*log(p),r=(n+1)*log(1-p);        ans=0;        for(int i=0;i<=n;i++)        {            ans+=(n-i)*(exp(l)+exp(r));            sum+=(exp(l)+exp(r));            l=l+log(1-p)+log(n+i+1)-log(i+1);            r=r+log(p)+log(n+i+1)-log(i+1);        }        ans/=sum;        double res=ans;        printf("Case %d: %lf\n",++tt,res);    }   return 0;}