1062. 最简分数(20)

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

int GCD(int a , int b)
{
    int gcd;
    while( b!= 0){
        gcd = a % b;
        a = b ;
        b = gcd;
    }
    gcd = a;
    return gcd;
}

int GCM(int a , int b)
{
    int tmp1 = a;
    int tmp2 = b;
    int gcd = GCD(tmp1,tmp2);
    int gcm = a * b / gcd;
    return gcm; 
}


int main()
{
    int A,B,C,D;
    int num;
    scanf("%d/%d %d/%d %d",&A,&B,&C,&D,&num);

    int tmp1 = GCM(B,D);
    int gcm = GCM(tmp1,num);
    int mul = gcm / num;
    int start = A * gcm / B;
    int end = C * gcm / D;

    int cnt = 0;
    int Fenzi[num];
    if ( start > end){
        start = start + end;
        end = start - end;
        start = start - end;
    }

    for ( int i = 1 ; i < num ; i++){
        int tmp = i * mul;
        if ( GCD(i,num) == 1 && tmp > start && tmp < end){
            Fenzi[cnt++] = i;
        }
    }
    cout<<Fenzi[0]<<"/"<<num;
    for ( int i = 1 ; i < cnt ; i++){
        cout<<" "<<Fenzi[i]<<"/"<<num;
    }

    return 0;
}