Candy HDU 4665 概率期望问题

Candy

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1159    Accepted Submission(s): 508
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

 

Recommend

liuyiding

 公式很好推。。。

主要是组合数的计算

/*
 * Author:  
 * Created Time:  2013/10/16 18:29:51
 * File Name: A.cpp
 * solve: A.cpp
 */
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<string>
#include<map>
#include<stack>
#include<set>
#include<iostream>
#include<vector>
#include<queue>
//ios_base::sync_with_stdio(false);
//#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;
#define sz(v) ((int)(v).size())
#define rep(i, a, b) for (int i = (a); i < (b); ++i)
#define repf(i, a, b) for (int i = (a); i <= (b); ++i)
#define repd(i, a, b) for (int i = (a); i >= (b); --i)
#define clr(x) memset(x,0,sizeof(x))
#define clrs( x , y ) memset(x,y,sizeof(x))
#define out(x) printf(#x" %d
", x)
#define sqr(x) ((x) * (x))
typedef long long LL;

const int INF = 1000000000;
const double eps = 1e-8;
const int maxn = 210000;

int sgn(const double &x) {  return (x > eps) - (x < -eps); }

double lnx[maxn*2];
void init()
{
    lnx[0] = log(1.0);
    lnx[1] = log(1.0);
    repf(i,2,400010)
    {
        lnx[i] = lnx[i - 1] + log((double)i);
    }
}
double C(int n,int m)
{
    return lnx[n] - lnx[m] - lnx[n - m];
}
int main() 
{
    //freopen("in.txt","r",stdin);
    int n;
    double p;
    int k = 1;
    init();
    while(scanf("%d%lf",&n,&p) == 2)
    {
        double ans = 0;
        double P = log(p);
        double Q = log(1 - p);
        rep(i,0,n)
        {
            double tmp = C(n+i,i);
            ans += (n-i)*(exp(tmp + (n+1)*P + i*Q) + exp(tmp + (n+1)*Q + i*P));
        }
        printf("Case %d: %.6lf
",k++,ans);
        
    }
    return 0;
}