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.

One line with a single integer N.

One fraction per line, sorted in order of magnitude.

#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
struct A {int z;//分子 int m;//分母 double a;//分数的值，因为数字很小，可以直接double比较是否相等
};
int gcd(int x, int y)
{if (x % y)return gcd(y , x % y);return y;
}
bool cmp(A x,A y) //按照 分数的值 进行排序从小到大输出
{return x.a<y.a;
}
A symp[12882],temp;
int main(void)
{int n,i,j,j_m,i_z,g,x;scanf("%d",&n);x=1;symp[0].a=0;for(j=1; j<=n; j++){for(i=1; i<=j; i++){//分子大于分母就大于1，可以不用遍历枚举g=gcd(j,i); j_m=j/g;i_z=i/g;symp[x].a=i_z*1.0/j_m;symp[x].z=i_z;symp[x].m=j_m;x++;}} sort(symp,symp+x,cmp);cout<<"0/1"<<endl;for(i=1; i<x; i++) {if(symp[i].a==symp[i-1].a)continue;//和前一项相等的，表示重复了，可以不用输出 elseprintf("%d/%d\n",symp[i].z,symp[i].m);}return 0;
}

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
struct shu
{int fenzi;int fenmu;
}a[165*165];
int gcd(int x,int y)
{return y==0?x:gcd(y,x%y);
}
bool cmp(shu x,shu y)//分数通分后比较分子
{int z=gcd(x.fenmu,y.fenmu);z=x.fenmu*y.fenmu/z;return x.fenzi*(z/x.fenmu)<y.fenzi*(z/y.fenmu);
}
int main()
{int i,j,n,k;while(~scanf("%d",&n)){k=0;for(i=2;i<=n;i++){for(j=1;j<i;j++){if(gcd(i,j)!=1)continue;//判断是否是最简分数a[k].fenzi=j;a[k++].fenmu=i;}}sort(a,a+k,cmp);printf("0/1\n");for(i=0;i<k;i++){printf("%d/%d\n",a[i].fenzi,a[i].fenmu);}printf("1/1\n");// 1/1 的gcd等于1会出现多次，所以特判输出}
}