EOJ(排序)——1852. Ordered Fractions

1852. Ordered Fractions

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

输入

One line with a single integer N.

输出

One fraction per line, sorted in order of magnitude.
样例

input

5

output

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

题目大意：

对分子，分母小于或等于N的分数集合进行排序，值相同的分数不计入最后的序列。

题目解析：

利用一个结构体存储分数的分子、分母，用二重循环读入各分数，最大公约数大于1的不保存。排序时可以将两分数通分，便于比较大小。

具体代码：

#include<iostream>
#include<algorithm>
using namespace std;

struct node{
	int a;//分子 
	int b;//分母 
}arr[100010];

bool cmp(node x,node y){
	return x.a*y.b<y.a*x.b;
}
int measure(int x, int y)
{	
	int z = y;
	while(x%y!=0)
	{
		z = x%y;
		x = y;
		y = z;	
	}
	return z;
}


int main() {
	freopen("data.in","r",stdin);
	int n,k=0;
	cin>>n;
	cout<<"0/1"<<endl;
	for(int i=1;i<=n;i++){
		for(int j=i+1;j<=n;j++){
			if(measure(i,j)<=1){
				arr[k].a=i;
				arr[k].b=j;
				k++;
			}
		}
	}
	sort(arr,arr+k,cmp);
	for(int i=0;i<k;i++)
		cout<<arr[i].a<<"/"<<arr[i].b<<endl;
	cout<<"1/1"<<endl;
	return 0;
}