HDU 4465 (candy)(期望+log优化）

Candy

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2890    Accepted Submission(s): 1305
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

 

Source

2012 Asia Chengdu Regional Contest

分析：有2个大小为n的盒子，里面放满了糖果，有个小孩一天吃一个，然后直到一天有一个盒子吃完了，算一下另外一个盒子的期望。

简单暴力的期望题目，但是呢不会做，因为没看过类似期望的东西，训练赛后推公式补题。

第一个盒子的概率是P  第二个盒子的概率是（1-p)

得到公式：C(n+i,i)*(p^(n+1)*(1-p)^i+p^i*(1-p)^(n+1))*(n-i)

第n+i+1次是代表取空了某一个盒子，所以就不再C（）里面了，所以只需要考虑n+i的情况即可。

当然包括了两种情况第一个盒子吃完第二个盒子吃完，所以应该ln后将两种情况加起来。

ln的式子是： e ln(C(n+i, i)) + (n+1)*ln(p) + (i)*ln(1-p) 。

这是其中一个的概率，另外一个P和（1-P)换一下位置。

#include<stdio.h>#include<iostream> #include <algorithm>#include<string.h>#include<vector>#include<math.h>#include<queue>#include<set>#define LL long long#define INf 0x3f3f3f3f#define mod 1000000007using namespace std;double t[400010];//存放Log的概率 void in()//预处理 Log {t[0]=0;for(int i=1;i<=400000;i++)t[i]=t[i-1]+log(i);} double logc(int n,int m)//数学公式 {return t[n]-t[m]-t[n-m];}int main(){int n,count=1;in();double p,p1;while(scanf("%d%lf",&n,&p)!=EOF){p1=1-p;p=log(p);p1=log(p1);double ans=0;double ans1=0;double ans2=0;for(int i=0;i<n;i++)//从另外一个箱子取走的糖果数目 {ans=logc(n+i,i)+(n+1)*p+i*p1;ans1=logc(n+i,i)+(n+1)*p1+i*p; ans2=ans2+(exp(ans)+exp(ans1))*(n-i);//exp是e为底的函数 }printf("Case %d: %.6lf\n",count++,ans2);} return 0;} 

特别说明下：（n-i)代表的是剩下的糖果，期望等于概率乘以变量。