HDU 4465 Candy 纯数学

Candy

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2900    Accepted Submission(s): 1310
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

 

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 总共有两种情况，先假设第一个盒子空了，第二个盒子还剩i个，那么这种情况下总共拿了n+(n-i)+1(+1是最后打开发现没了)次，这个情况下对应的概率为C(2n-i,)*p^(n+1)*(1-p)^(n-i),期望就是这个概率乘以i，另一种情况和这个类似，这样这个式子在数学上就是没问题i。但在计算机中有问题，p^(n+1)和(1-p)^(n-i) 都无限接近于0，直接这么和c相乘误差太大了，而且c（2n-i,i）也超级大，直接暴力求，300位整数都不一定够，这就很伤。想了半天也没想起来，看了看书，才知道通过log函数可以把乘法变加法，把除法变减法

然后打一个ln的表，把原来的式子拆开，借助x=e^( ln(x) ) 完美变形，就可以ac了

my ugly code

#include <cstdio>#include <cmath>#include <algorithm>#include <cstring>#include <string>#include <map>#include <iostream>using namespace std;const int maxn=2e5+10;int n;double p;int cas=1;double sum[maxn]={0.0};void INIT(){    sum[0]=0.0;    for(int i=1;i<=2e5;i++){        sum[i]=sum[i-1]+log(i);    }    /*for(int i=1;i<=2e5;i++)        printf("**%f**\n",sum[i]);*/}double CC(int k1,int k2){    double ans1,ans2;    ans1=sum[k1]-sum[k1-k2];    ans2=sum[k2];    return ans1-ans2;}int main(){    INIT();    while(~scanf("%d%lf",&n,&p)){        double ans=0.0;        for(int i=1;i<=n;i++){            double tmp=CC(2*n-i,n);            ans+=i*( exp(tmp+(n+1)*log(p)+(n-i)*log(1-p))+exp(tmp+(n+1)*log(1-p)+(n-i)*log(p)) );        }        printf("Case %d: %.6f\n",cas++,ans);    }    return 0;}