HDU 4465 Candy

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Candy

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2520    Accepted Submission(s): 1100
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 \leq n \leq 2 \times 10^5) $and a real number p ($0\leq p \leq 1$, 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 Input10 0.400000 100 0.500000 124 0.432650 325 0.325100 532 0.487520 2276 0.720000 

 

Sample OutputCase 1: 3.528175Case 2: 10.326044Case 3: 28.861945Case 4: 167.965476Case 5: 32.601816Case 6: 1390.500000 

 

Source2012 Asia Chengdu Regional Contest

 解題：公式題

 首先可以算出公式

\[\sum_{i = 0}^{n-1}\binom{n}{n+i}\times (p^{n+1}\times (1-p)^i + (1-p)^{n+1}\times p^i)\times (n-i)\]

我們至少要拿n個，才會出現一個盒子裡面的糖果空，所以假設第一個為空白，枚舉第二個盒子剩餘的糖果數量，或者假設第二個盒子為空白，枚舉第一個盒子剩餘的糖果數量。

但是，更加狗血的事情發生了，會導致double溢出，你說用long double啊？呵呵，naive！還是要溢出，日了整個世界，好吧，使用log吧。換成對數，把*變成+

 算$\binom{n}{n+i}$ 是可以滾動著算的

可以發現在取對數後有：

$$C_{0} = 0,C_i = C_{i-1} + log(n + i) - log(i)$$

 1 #include <bits/stdc++.h> 2 using namespace std; 3 int main() { 4     int n,cs = 1; 5     double p; 6     while(~scanf("%d%lf",&n,&p)) { 7         double ret = 0,tmp,c = 0,logp = log(p),log_p = log(1 - p); 8         for(int i = 0; i < n; ++i) { 9             double logn_i = log(n-i);10             tmp = c + (n + 1)* logp + i*log_p + logn_i;11             ret += exp(tmp);12             tmp = c + (n + 1)*log_p + i*logp + logn_i;13             ret += exp(tmp);14             c += log(n + i + 1.0) - log(i + 1.0);15         }16         printf("Case %d: %.6f\n",cs++,ret);17     }18     return 0;19 }

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HDU 4465 Candy