hdu4465 Candy(快速排列组合+概率)

链接：http://acm.hdu.edu.cn/showproblem.php?pid=4465

Candy

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2524    Accepted Submission(s): 1103
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


题意：有两个盒子各有n个糖，每天随机选一个（概率分别为p,1-p)，然后吃一颗糖。直到有一天，打开一个盒子看没糖l！输入n,p，求此时另一个盒子里糖的个数数学期望。

分析：根据期望的定义，不妨设最后打开第一个盒子，此时第二个盒子有i颗，则这之前打开过n+(n-i)次盒子，其中有n次取的是盒子1，其余n-i次取的盒子2，概率为C(2n-i,n)P^(n+1)*(1-P)^(n-i)。注意P的指数是n+1，因为除了前面打开过n次盒子1之外，最后又打开了一次。
          当然光计算出这个概率，这题还是很难实现的，我们需要优化一下。我们可以利用对数，设v1(i)=ln(C(2n-i,n))+(n+1)ln(P)+(n-i)ln(1-P)，则“最后打开第一个盒子”对应的数学期望为　i*e^v1(i)。
         同理，当最后打开的是第二个盒子，对数为 v2(i)=ln(C(2n-i,n))+(n+1)ln(1-p)+(n-i)ln(p)，对应的数学期望为 i*e^v2(i)。
        根据数学期望的定义，最终答案为 sum{i*(e^v1(i)+e^v2(i))。                                    （参考算法竞赛入门经典333页）

 

 #include<iostream>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 using namespace std;
 const int maxn=400040;
 double nlog[maxn],p,q; //nlog[n]记录log[1]+log[2]+...+log[n]
 int n;
 double cal(int n,int m)
 {
     return nlog[n]-nlog[m]-nlog[n-m];
 }
 int main()
 {
     int cas=0;
     memset(nlog,0,sizeof(nlog));
     for(int i=1;i<maxn;i++) // 对nlog[n]做预处理
        nlog[i] +=nlog[i-1] + log(i);
     while(scanf("%d %lf",&n,&p) != EOF)
     {
         q=log(1-p); // 直接对（1-p)取对数，后面运用到
         p=log(p);
         double ans=0;
         for(int i=0;i<=n;i++)
         {
             ans+=i*(exp(cal(2*n-i,n)+(n+1)*p+(n-i)*q) + exp(cal(2*n-i,n)+(n+1)*q+(n-i)*p));
         }
         printf("Case %d: %lf\n",++cas,ans);
     }
     return 0;
 }