HIToj--1076--Ordered Fractions(水题)-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_29963431/article/details/50253551

此博客提供了一个程序，用于将小于等于N的整数分母的最简分数按大小顺序排列。输入为整数N（1到160），输出为排序后的分数。

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Ordered Fractions

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

Submitted : 1510, Accepted : 496

Consider the set of all reduced fractions between 0 and 1 inclusivewith denominators 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

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct node
{
	int a,b;
	double s;
}num[100010];
int cmp(node s1,node s2)
{
	return s1.s<s2.s;
}
int gcd(int a,int b)
{
	
	return b==0?a:gcd(b,a%b);
}
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		int cnt=0;
		for(int i=1;i<=n;i++)
		{
			for(int j=0;j<=i;j++)
			{
				if(gcd(i,j)==1)
				{
					num[cnt].a=j;
					num[cnt].b=i;
					num[cnt].s=j*1.0/i;
					cnt++;
				}
			}
			
		}
		sort(num,num+cnt,cmp);
		for(int i=0;i<cnt;i++)
		printf("%d/%d\n",num[i].a,num[i].b);
		printf("\n");
	}
	return 0;
}