HDU-博彩问题-带分数

问题及代码：

/* 
*Copyright (c)2014,烟台大学计算机与控制工程学院 
*All rights reserved. 
*文件名称：number.cpp 
*作    者：单昕昕 
*完成日期：2015年2月3日 
*版 本 号：v1.0 
* 
*问题描述：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?
*程序输入：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.
*程序输出：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

*/ 
//<span style="font-family: Arial, Helvetica, sans-serif;">AC代码。</span>
#include <iostream>
using namespace std;
int n,s,a1,b1,a2,b2,s1,s2;
int gcd(int a,int b)
{
    return (a%b!=0?(gcd(b,a%b)):b);//求最大公倍数
}
int f(int x,int y)
{
    int t1=a1,t2=b1;
    a1=t1*y+b1*x;
    b1=t2*y;
    int t=a1/b1;
    s+=t;//整数部分
    a1-=t*b1;
    t=gcd(a1,b1);
    a1=a1/t;
    b1=b1/t;
}
int main()
{
    while (cin>>n)
    {
        s=0;
        a1=0;
        b1=1;
        for (int i=1; i<=n; i++)
        {
            f(n,i);
        }
        if (a1==0)
            cout <<s<<endl;
        else
        {
            int t1=0,t2=0,temp1=s,temp2=b1;
            while (temp1!=0)
            {
                t1++;
                temp1/=10;
            }
            t1++;
            while (temp2!=0)
            {
                t2++;
                temp2/=10;
            }
            for (int i=1; i<=t1; i++)
                cout <<" ";
            cout <<a1<<endl;
            cout <<s<<" ";
            for (int i=t2; i>=1; i--)
                cout <<"-";
            cout <<endl;
            for (int i=1; i<=t1; i++)
                cout <<" ";
            cout <<b1<<endl;
        }
    }
    return 0;
}
//我写的这个代码<span style="font-family: Arial, Helvetica, sans-serif;">明明答案什么的都对，可就是WA了。。真心难过。。</span>
#include <iostream>
using namespace std;
int gcd(int a,int b)
{
    return (a%b!=0?(gcd(b,a%b)):b);
}
int lcm(int u,int v)  //求最大公倍数
{
    int h;
    h=gcd(u,v);
    return(u*v/h);
}
int main()
{
    int n;
    while(cin>>n)
    {
        int a[n],i;
        long long temp,s=0,m,fz,mm,ff,gys,count,flag;
        a[0]=0;
        for(i=1; i<=n; ++i)
            a[i]=i;
        temp=a[1];
        for(i=2; i<=n; i++)
            temp=lcm(temp,a[i]);
        for(i=1; i<=n; ++i)
            s+=(temp/i);
        s*=n;
        m=s/temp;
        mm=m;
        fz=s-(temp*m);
        gys=gcd(fz,temp);
        if(fz==0)cout<<m<<" "<<endl;
        else
        {
            for(count = 0; ; count++)
            {
                mm=mm/10;
                if(mm <= 0)
                {
                    break;
                }
            }
            for(i=0; i<=count+1; ++i)
                cout<<" ";
            flag=count;
            cout<<fz/gys<<endl;
            cout<<m<<" ";
            ff=temp/gys;
            for(count = 0; ; count++)
            {
                ff=ff/10;
                if(ff <= 0)
                {
                    break;
                }
            }
            for(i=0; i<=count; ++i)
                cout<<"-";
            cout<<endl;
            for(i=0; i<=flag+1; ++i)
                cout<<" ";
            cout<<temp/gys<<endl;
        }
    }
    return 0;
}

运行结果：

知识点总结：
就是求n*(1/1+1/2+1/3+...+1/n)这个式子的值，然后化成带分数。

学习心得：

格式输出很变态，一定要注意：

1、分数前面有个空格；

2、分数线的长度和分母一样；

3、分数线是由减号构成的。