B - Candy
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit Status Practice HDU 4465
Appoint description:
System Crawler (2014-10-17)
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 × 10 5) 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
http://mikewell.blog.163.com/blog/static/192914172201210180306917/
公式快速排列组合函数：logC(m,n)，zuhe[i]其实就是i的阶乘，然后给取log值这样zuhe[m]-zuhe[n]-zuhe[m-n]就是C(m,n)的log值然后把其余也取对数，然后再求exp就好了！
快速排列组合函数：logC(m,n)，zuhe[i]其实就是i的阶乘，然后给取log值这样zuhe[m]-zuhe[n]-zuhe[m-n]就是C(m,n)的log值然后把其余也取对数，然后再求exp就好了！ 1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<cmath>
5 #define M(a,b) memset(a,b,sizeof(a))
6 #define INF 0x3f3f3f3f
7
8 using namespace std;
9
10 int n;
11 double p;
12 double zuhe[400005];
13
14 double logC(int m,int n)
15 {
16 return zuhe[m]-zuhe[n]-zuhe[m-n];
17 }
18
19 int main()
20 {
21 int cas = 1;
22 zuhe[0] = 0;
23 for(int i = 1;i<400005;i++) zuhe[i] = zuhe[i-1]+log(i*1.0);
24 while(scanf("%d%lf",&n,&p)==2)
25 {
26 double res = 0;
27 for(int i = 0;i<=n;i++)
28 {
29 res+=(n-i)*exp(logC(n+i,i)+(n+1)*log(p)+(i)*log(1-p));
30 res+=(n-i)*exp(logC(n+i,i)+(n+1)*log(1-p)+(i)*log(p));
31 }
32 printf("Case %d: %.16f\n",cas++,res);
33 }
34 return 0;
35 }