[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：

 1 #include <iostream>

 2 #include <algorithm>

 3 #include <cstdio>

 4 #define maxn 1000010

 5 using namespace std;

 6 struct node{

 7     int x, y;

 8     bool operator<(const node &tmp)const{

 9         if (x != tmp.x)

10             return x*tmp.y < y*tmp.x;

11         return y > tmp.y;

12     }

13 }a[maxn];

14 int gcd(int a, int b){

15     return !b ? a : gcd(b, a%b);

16 }

17 int main(){

18     int n, i, j;

19     while (~scanf("%d", &n)){

20         int pos = 0;

21         //i代表分母,j代表分子

22         for (i = 1; i <= n; i++){

23             for (j = 0; j <= i; j++){

24                 if (!j&&i != 1) continue;

25                 if (gcd(j, i) == 1){

26                     a[pos].x = j;

27                     a[pos++].y = i;//  x/y

28                 }

29             }

30         }

31         sort(a, a + pos);

32         for (i = 0; i < pos; i++)

33             printf("%d/%d\n", a[i].x, a[i].y);

34     }

35     return 0;

36 }

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