Ordered Fractions

Problem:

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.

PROGRAM NAME: frac1

INPUT FORMAT

One line with a single integer N.

SAMPLE INPUT (file frac1.in)

5

OUTPUT FORMAT

One fraction per line, sorted in order of magnitude.

SAMPLE OUTPUT (file frac1.out)

0/1
1/51/4
1/3
2/51/2
3/52/3
3/4
4/51/1

———————————————————————————————————————————————————————————————————————

My Answer：

/*
ID:iby071
LANG:C++
TASK:frac1
*/

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

struct frac
{	friend bool operator<(frac,frac);
int a;
int b;
frac(int n1=0,int n2=0):a(n1),b(n2){}
};

bool operator<(frac f1,frac f2)			//定义分数大小的比较
{	if(f1.a*f2.b<f1.b*f2.a) return true;
else return false;
}

int divide(int low,int high,frac f[])
{	frac k=f[low];
do
{	while(low<high && k<f[high]) --high;
if(low<high) {f[low]=f[high];++low;}
while(low<high && f[low]<k) ++low;
if(low<high) {f[high]=f[low];--high;}
}while(low!=high);
f[low]=k;
return low;
}

void quickSort(int low,int high,frac f[])
{	if(low>=high) return;
int mid=divide(low,high,f);
quickSort(low,mid-1,f);
quickSort(mid+1,high,f);
}

int gcd(int a,int b)		//背下来：辗转相除法（by欧几里得）求最大公约数！
{	if(b==0)
return a;
else
return gcd(b,a%b);
}

bool check(int a,int b)			//判断是否最简分数
{	if(gcd(a,b)==1) return true;
else return false;
}

int main()
{	ifstream fin("frac1.in");
ofstream fout("frac1.out");
int n,size=0;
frac *f;
fin>>n;
f=new frac[n*(n-1)/2+2];
f[0]=frac(0,1);					//特别的：0/0
++size;

for(int b=n;b>1;--b)
for(int a=1;a<b;++a)
if(check(a,b)) {f[size]=frac(a,b);++size;}

f[size]=frac(1,1);				//特别的：1/1
++size;

quickSort(0,size-1,f);

for(int i=0;i<size;++i)
fout<<f[i].a<<'/'<<f[i].b<<endl;

return 0;
}