HDU 4254 A Famous Game (概率&组合数学公式)

A Famous Game

http://acm.hdu.edu.cn/showproblem.php?pid=4254

Time Limit: 2000/1000 MS (Java/Others) 

 Memory Limit: 32768/32768 K (Java/Others)

Problem Description

Mr. B and Mr. M like to play with balls. They have many balls colored in blue and red. Firstly, Mr. B randomly picks up N balls out of them and put them into a bag.
Mr. M knows that there are N+1 possible situations in which the number of red balls is ranged from 0 to N, and we assume the possibilities of the N+1 situations are
the same. But Mr. M does not know which situation occurs. Secondly, Mr. M picks up P balls out of the bag and examines them. There are Q red balls and P-Q blue
balls. The question is: if he picks up one more ball out of the bag, what is the possibility that this ball is red?

Input

Each test case contains only one line with three integers N, P and Q (2 <= N <= 100,000, 0 <= P <= N-1, 0 <= Q <= P).

Output

For each test case, display a single line containing the case number and the possibility of the next ball Mr. M picks out is red. The number should be rounded to four
decimal places.

Sample Input

   
   
   
   

    
    
    
    3 0 0
4 2 1
   
   
   
   

Sample Output

   
   
   
   

    
    
    
    Case 1: 0.5000
Case 2: 0.5000


    
    
    
    
 
     
 
      Hint 
     
 For example as the sample test one, there are three balls in the bag. The possibilities of the four possible situations are all 0.25. If there are no red balls in the bag, the possibility of the next ball are red is 0. If there is one red ball in the bag, the possibility is 1/3. If there are two red balls, the possibility is 2/3. Finally if all balls are red, the possibility is 1. So the answer is 0*(1/4)+(1/3)*(1/4)+(2/3)*(1/4)+1*(1/4)=0.5. 
    
    
    
    
     
   
   
   
   

Source

Fudan Local Programming Contest 2012 

解析见这，但我觉得里面有个公式有点问题。。

完整代码：

/*0ms,260KB*/

#include <cstdio>

int main(void)
{
	int T = 0, n, p, q;
	while (~scanf("%d%d%d", &n, &p, &q))
		printf("Case %d: %.4f\n", ++T, (q + 1.0) / (p + 2));
	return 0;
}