HDU 1099.Lottery【数学题，求期望】【12月31】

Lottery

Problem Description

Eddy's company publishes a kind of lottery.This set of lottery which are numbered 1 to n, and a set of one of each is required for a prize .With one number per lottery, how many lottery on average are required to make a complete set of n coupons?

 

Input

Input consists of a sequence of lines each containing a single positive integer n, 1<=n<=22, giving the size of the set of coupons.

 

Output

For each input line, output the average number of lottery required to collect the complete set of n coupons. If the answer is an integer number, output the number. If the answer is not integer, then output the integer part of the answer followed by a space and then by the proper fraction in the format shown below. The fractional part should be irreducible. There should be no trailing spaces in any line of ouput.

 

Sample Input

   
   
   
   

    
    
    
    2
5
17
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    3 
   5
11 --
   12
   340463
58 ------
   720720
   
   
   
   
   
   
   
   


   
   
   
   

题意读了好久····就是求n*(1  + 1/2 +  1/3 + ...... + 1/n)，分析如下：（http://www.cnblogs.com/arno-my-boke/archive/2015/01/21/4238137.html）

题目的大概意思是说一套彩票有编号1到n共n种，张数不限，问你平均买多少张能把编号为1到n的n中彩票全买下来，也就是求期望。

举个例子，当n=5时，第一张有用的概率为1，买一张就行了，第二张有用的概率为4/5，所以买5/4张彩票能买上对你有用的，一次类推，sum=1+5/4+5/3+5/2+5/1=11…5/12

AC代码如下：

#include<iostream>
#include<cstdio>
using namespace std;
long long gcd(long long a, long long b)
{
    long long  r;
    while(b > 0)
    {
         r = a % b;
         a = b;
         b = r;
    }
    return a;
}
int main()
{

    int n;
    while(cin >> n)
    {
        unsigned long long y = 1;
        unsigned long long x = 0;
        for(int i = 1;i <= n; ++i)
        {
            x = y+i*x;
            y *= i;
            long long GCD = gcd(x, y);
            x /= GCD;
            y /= GCD;
        }
        long long GCD = gcd(n, y);
        y /= GCD;
        n /= GCD;
        x *= n;
        //格式控制太操蛋！
        if(x%y == 0) cout << x/y << endl;
        else
        {
            if(x/y > 9) cout <<" ";
            cout <<"  "<<  x%y <<endl;
            cout << x/y <<" ";
            long long zz = y;
            while(zz)
            {
                cout <<"-";
                zz /= 10;
            }
            cout << endl;
            if(x/y > 9) cout <<" ";
            cout <<"  "<< y <<endl;
        }
    }
    return 0;
}