Given an integer N, the task is to print all proper fractions such that the denominator is less than or equal to N.
Proper Fractions: A fraction is said to be a proper fraction if the numerator is less than the denominator.
Examples:
Input: N = 3
Output: 1/2, 1/3, 2/3Input: N = 4
Output: 1/2, 1/3, 1/4, 2/3, 3/4
Approach:
Traverse all numerators over [1, N-1] and, for each of them, traverse over all denominators in the range [numerator+1, N] and check if the numerator and denominator are coprime or not. If found to be coprime, then print the fraction.
Below is the implementation of the above approach:
C++14
// C++ program to implement the // above approach #include <bits/stdc++.h> using namespace std; // Function to print all // proper fractions void printFractions( int n) { for ( int i = 1; i < n; i++) { for ( int j = i + 1; j <= n; j++) { // If the numerator and the // denominator are coprime if (__gcd(i, j) == 1) { string a = to_string(i); string b = to_string(j); cout << a + "/" + b << ", " ; } } } } // Driver Code int main() { int n = 3; printFractions(n); return 0; } |
Java
// Java program to implement the // above approach class GFG{ // Function to print all // proper fractions static void printFractions( int n) { for ( int i = 1 ; i < n; i++) { for ( int j = i + 1 ; j <= n; j++) { // If the numerator and the // denominator are coprime if (__gcd(i, j) == 1 ) { String a = String.valueOf(i); String b = String.valueOf(j); System.out.print(a + "/" + b + ", " ); } } } } static int __gcd( int a, int b) { return b == 0 ? a : __gcd(b, a % b); } // Driver code public static void main(String[] args) { int n = 3 ; printFractions(n); } } // This code is contributed by 29AjayKumar |
Python3
# Python3 program for the # above approach # Function to print # all proper functions def printfractions(n): for i in range ( 1 , n): for j in range (i + 1 , n + 1 ): # If the numerator and # denominator are coprime if __gcd(i, j) = = 1 : a = str (i) b = str (j) print (a + '/' + b, end = ", " ) def __gcd(a, b): if b = = 0 : return a else : return __gcd(b, a % b) # Driver code if __name__ = = '__main__' : n = 3 printfractions(n) # This code is contributed by virusbuddah_ |
C#
// C# program to implement the // above approach using System; class GFG{ // Function to print all // proper fractions static void printFractions( int n) { for ( int i = 1; i < n; i++) { for ( int j = i + 1; j <= n; j++) { // If the numerator and the // denominator are coprime if (__gcd(i, j) == 1) { string a = i.ToString(); string b = j.ToString(); Console.Write(a + "/" + b + ", " ); } } } } static int __gcd( int a, int b) { return b == 0 ? a : __gcd(b, a % b); } // Driver code public static void Main( string [] args) { int n = 3; printFractions(n); } } // This code is contributed by rutvik_56 |
Javascript
<script> // Javascript program for the above approach // Function to print all proper functions const printFractions = (n) => { for ( var i = 1; i < n; i++) { for ( var j = i + 1; j <= n; j++) { // If the numerator and denominator are coprime if (__gcd(i, j) == 1){ let a = `${i}`; let b = `${j}`; document.write(`${a}/${b}, `) } } } } const __gcd = (a, b) => { if (b == 0){ return a; } else { return __gcd(b, a % b); } } // Driver code let n = 3; printFractions(n); // This article is contributed by _saurabh_jaiswal </script> |
1/2, 1/3, 2/3,
Time Complexity: O(N2 log N)
Auxiliary Space: O(1)
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