http://acm.hdu.edu.cn/showproblem.php?pid=1099
Lottery
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 19 Accepted Submission(s) : 12

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

我看不懂它的输出到底是什么规律呢？题意也看得很不明白！

帮你顶上去吧

呵呵！自己也顶顶！

啥东西啊

呵呵！是acm的一道题：

If the answer is an integer number, output the number. If the answer is not integer, then output the integer part of the answer(输出整数) followed by a space(空格) and then by the proper fraction in the format shown below.(输出小数部分,要是真分数) The fractional part should be irreducible. (分数不能约分)There should be no trailing spaces in any line of ouput. (每一行的最后不能有多余的空格)

我的理解是最少需要多少张的彩票才能将这n张彩票收集齐全，但是明显和答案不符，呵呵，想请问一下问什么会得到这个结果？

提示一下算法是怎样的？

据说算法是：sum=n/1+n/2+……+n/n  
其实我也没懂  网上搜了搜

给你一段c++的代码自己领悟吧：
#include<iostream>
using namespace std;
void tongfen(int &x,int &y)
{
    int m=x,n=y,r=m%n;
    while(r)
    {
        m=n;
        n=r;
        r=m%n;
    }
    x=x/n;
    y=y/n;
}
int main()
{
    int n,i,sumfz,sumfm,fz,fm,zheng,weif,weiz;
    while(scanf("%d",&n)!=EOF)
    {
        if(n==1){printf("1\n");continue;}
        sumfz=n;sumfm=1;zheng=0;
        for(i=2;i<=n;i++)
        {
            fz=n;fm=i;
            sumfz=sumfz*fm+fz*sumfm;
            sumfm=sumfm*fm;
            zheng=sumfz/sumfm+zheng;
            sumfz=sumfz%sumfm;
            if(sumfz!=0)tongfen(sumfm,sumfz);
        }
        if(sumfz==0)printf("%d\n",zheng);
        else
        {
            weif=0;
            weiz=0;
            i=sumfm;
            while(i)
            {
                i=i/10;
                weif++;
            }
            i=zheng;
            while(i)
            {
                i=i/10;
                weiz++;
            }
            for(i=0;i<=weiz;i++)
                printf(" ");
            printf("%d\n",sumfz);
            printf("%d ",zheng);
            for(i=0;i<weif;i++)
                printf("-");
            printf("\n");
            for(i=0;i<=weiz;i++)
                printf(" ");
            printf("%d\n",sumfm);
        }
    }
    return 0;
}