HDU 4465 Candy_lazychild is a lazy child who likes candy very muc-优快云博客

Candy

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1908 Accepted Submission(s): 833
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

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

题目大意 :两个盒子开始都有n块糖，每次从一个中拿糖的概率为p，求当一个盒子空时，另一个盒子中糖的数目期望

题目分析 :不难得到公式: ans = 求和(i = 0~n)(n-i) * C(i, n + i) * ( (p ^ n* (1 - p) ^ i) + (p ^ i + (1 - p) ^ n))，本题n上限为20000，组合数存不下故中间过程要用递推公式即an = an * n / (n + 1) *(n + i) / i * p这样还是会卡精度有点坑爹，可以采用一个很简单的技巧。见程序

#include <cstdio>
#include <cmath>
int main()
{
    int n, ca = 0;
    double p;
    while(scanf("%d %lf", &n, &p) != EOF)
    {
        double tmp1 = n, tmp2 = n ;
        // printf("tmp1 = %f  tmp2 = %f\n", tmp1, tmp2);
        double ans1 = tmp1 * pow(1 - p, n + 1);
        double ans2 = tmp2 * pow(p, n + 1);
        int now = n;
        int re = n + 1;
        for(int i = 1; i < n; i++)
        {
            now = n - i; //系数
            //递推公式
            tmp1 *= 1.0  / (double)(now + 1) * p / i * (n + i) * now;
            tmp2 *= 1.0  / (double)(now + 1) * (1 - p) / i * (n + i) * now;
            //当tmp大于某个设定的值时让它乘p(或(1-p))来保证其数字不会暴精度
            //关键技巧
            while(tmp1 > n || tmp2 > n)
            {
                tmp1 *= (1 - p);
                tmp2 *= p;
                re--;
            }
            //因为每一项都要乘pow(1-p, n+1)或pow(p,n+1)
            //我们在此补上剩余未乘的p(或(1-p))的幂数即可
            ans1 += tmp1 * pow(1 - p, re);
            ans2 += tmp2 * pow(p, re);
           // printf("tmp1 = %lf   tmp2 = %lf  \n", tmp1, tmp2);
        }
        printf("Case %d: %lf\n", ++ca, ans1 + ans2);
    }
}