Ordered Fractions

	Source : Unknown
	Time limit : 3 sec		Memory limit : 32 M

Submitted : 1227, Accepted: 383

Consider the set of all reduced fractions between 0 and 1 inclusive with denomin 

ators less than or equal to N.

Here is the set when N = 5:

0/1 1/5 1/4 1/3 2/5 1/2 3/5 2/3 3/4 4/5 1/1

Write a program that, given an integer N between 1 and 160 inclusive, prints the fractions in order of increasing magnitude.

INPUT FORMAT

Several lines with a single integer N in each line. Process to the end of file.

SAMPLE INPUT

5
5

OUTPUT FORMAT

One fraction per line, sorted in order of magnitude. print a blank line after each testcase.

SAMPLE OUTPUT

0/1
1/5
1/4
1/3
2/5
1/2
3/5
2/3
3/4
4/5
1/1

0/1
1/5
1/4
1/3
2/5
1/2
3/5
2/3
3/4
4/5
1/1

这个题开始提交时超时  后来才发现输入输出太多了  改成C语言的输入输出就没有超时了 0.12MS 相差几十倍 这个以后得注意点啊 嘿嘿 难怪曾哥那么喜欢用C语言 我却习惯用C++了 一时改不过来

#include <iostream>
#include <algorithm>
#include <fstream>
#include <stdio.h>
using namespace std;

#define max 170
struct pp
{
    int x,y;
    double b;
};

int  judge(int x,int y)
{
    int t;
    while (y)
    {
        t=y;
        y=x%y;
        x=t;
    }
    return x;
}
bool mycomp(const pp &x,const pp &y)
{
    if(x.b<y.b)  return true;
    return false;
}
int main()
{
//    ifstream cin("1.txt");
//    ofstream cout("2.txt");
//    freopen("output","w",stdout);
    int i,j,k;
    int p=0,q;
    int temp;
    //while (cin>>p)
        int m=0;
        pp t[max*max];
        for(i=1;i<max;i++)
        {
            for(j=0;j<max;j++)
            {
                t[m++].b=1;
            }
        }
        m=0;
        for(i=1;i<=170;i++)
        {
            for(j=i;j<=170;j++)
            {
                if(judge(i,j)>1) t[m].b=1;
                else
                {
                    t[m].b=i*1.0/j;
                    t[m].x=i;
                    t[m].y=j;
                    m++;
                }                
            }
        }
        sort(t,t+m,mycomp);
        while (scanf("%d",&p)!=EOF)
    //    while(cin>>p)
    {
                //    cout<<"0/1"<<endl;
            printf("0/1\n");
        for (i=0;i<m;i++)
        {
            if(t[i].b<1&&t[i].x<=p&&t[i].y<=p) //cout<<t[i].x<<"/"<<t[i].y<<endl;
                printf("%d/%d\n",t[i].x,t[i].y);
        }
        printf("1/1\n\n");
        //cout<<"1/1"<<endl;
    //    cout<<endl;
    }
    return 0;
}

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