[Swust OJ 801]--Ordered Fractions

题目链接：http://acm.swust.edu.cn/problem/801/

Time limit(ms): 1000　　　　　　Memory limit(kb): 10000

 

Description

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.

Input

One line with a single integer N.

 

Output

One fraction per line, sorted in order of magnitude.

 

Sample Input

5

 

Sample 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：给定一个数，代表分母的最大值，从小到大，输出所有分母小于N,值小于等于1的所有分数

直接上代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#define maxn 1000010
using namespace std;
struct node{
    int x, y;
    bool operator<(const node &tmp)const{
        if (x != tmp.x)
            return x*tmp.y < y*tmp.x;
        return y > tmp.y;
    }
}a[maxn];
int gcd(int a, int b){
    return !b ? a : gcd(b, a%b);
}
int main(){
    int n, i, j;
    while (~scanf("%d", &n)){
        int pos = 0;
        //i代表分母,j代表分子
        for (i = 1; i <= n; i++){
            for (j = 0; j <= i; j++){
                if (!j&&i != 1) continue;
                if (gcd(j, i) == 1){
                    a[pos].x = j;
                    a[pos++].y = i;//  x/y
                }
            }
        }
        sort(a, a + pos);
        for (i = 0; i < pos; i++)
            printf("%d/%d
", a[i].x, a[i].y);
    }
    return 0;
}