[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 = ;
         //i代表分母,j代表分子
         for (i = ; i <= n; i++){
             for (j = ; j <= i; j++){
                 if (!j&&i != ) continue;
                 if (gcd(j, i) == ){
                     a[pos].x = j;
                     a[pos++].y = i;//  x/y
                 }
             }
         }
         sort(a, a + pos);
         for (i = ; i < pos; i++)
             printf("%d/%d\n", a[i].x, a[i].y);
     }
     return ;
 }