HDU4465(数学期望+大数处理)

Candy

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 507    Accepted Submission(s): 232
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 × 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
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
    //求数学期望，运用公式，然后结合对数指数处理超界问题
   
   
   
   
   
   
   
   

    
    
    
    
#include<iostream> #include<math.h> using namespace std; double f[400000+10]; int n;
    
    
    
    
void func() {  int i;  f[0]=0;  for(i=1;i<=400000+10;i++)   f[i]=f[i-1]+log(1.0*i);
    
    
    
    
}
    
    
    
    
double Log(int i) {  return f[n+i]-f[n]-f[i]; }
    
    
    
    
int main() {  int i,tag=0;;  double p,count;  func();  while(~scanf("%d%lf",&n,&p))  {   count=0;   for(i=0;i<=n;i++)   {    count+=(n-i)*(exp(Log(i)+i*log(1.0*(1-p))+(n+1)*log(1.0*p))+exp(Log(i)+i*log(1.0*p)+(n+1)*log(1.0*(1-p))));   }   printf("Case %d: %.6lf\n",++tag,count);  }  return 0; }