Given a:b and b:c. The task is to write a program to find ratio a:b:c
Examples:
Input: a:b = 2:3, b:c = 3:4 Output: 2:3:4 Input: a:b = 3:4, b:c = 8:9 Output: 6:8:9
Approach: The trick is to make the common term ‘b’ equal in both ratios. Therefore, multiply the first ratio by b2 (b term of second ratio) and the second ratio by b1.
Given: a:b1 and b2:c
Solution: a:b:c = (a*b2):(b1*b2):(c*b1)
For example:
If a : b = 5 : 9 and b : c = 7 : 4, then find a : b : c.
Solution:
Here, Make the common term ‘b’ equal in both ratios.
Therefore, multiply the first ratio by 7 and the second ratio by 9.
So, a : b = 35 : 63 and b : c = 63 : 36
Thus, a : b : c = 35 : 63 : 36
Below is the implementation of the above approach:
C++
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std;// Function to print a:b:cvoid solveProportion(int a, int b1, int b2, int c){ int A = a * b2; int B = b1 * b2; int C = b1 * c; // To print the given proportion // in simplest form. int gcd = __gcd(__gcd(A, B), C); cout << A / gcd << ":" << B / gcd << ":" << C / gcd;}// Driver codeint main(){ // Get the ratios int a, b1, b2, c; // Get ratio a:b1 a = 3; b1 = 4; // Get ratio b2:c b2 = 8; c = 9; // Find the ratio a:b:c solveProportion(a, b1, b2, c); return 0;} |
Java
// Java implementation of above approachimport java.util.*;import java.lang.*;import java.io.*;class GFG{static int __gcd(int a,int b){ return b==0 ? a : __gcd(b, a%b);} // Function to print a:b:cstatic void solveProportion(int a, int b1, int b2, int c){ int A = a * b2; int B = b1 * b2; int C = b1 * c; // To print the given proportion // in simplest form. int gcd = __gcd(__gcd(A, B), C); System.out.print( A / gcd + ":" + B / gcd + ":" + C / gcd);} // Driver codepublic static void main(String args[]){ // Get the ratios int a, b1, b2, c; // Get ratio a:b1 a = 3; b1 = 4; // Get ratio b2:c b2 = 8; c = 9; // Find the ratio a:b:c solveProportion(a, b1, b2, c);}} |
Python 3
# Python 3 implementation # of above approachimport math# Function to print a:b:cdef solveProportion(a, b1, b2, c): A = a * b2 B = b1 * b2 C = b1 * c # To print the given proportion # in simplest form. gcd1 = math.gcd(math.gcd(A, B), C) print( str(A // gcd1) + ":" + str(B // gcd1) + ":" + str(C // gcd1))# Driver codeif __name__ == "__main__": # Get ratio a:b1 a = 3 b1 = 4 # Get ratio b2:c b2 = 8 c = 9 # Find the ratio a:b:c solveProportion(a, b1, b2, c)# This code is contributed # by ChitraNayal |
C#
// C# implementation of above approachusing System;class GFG{static int __gcd(int a,int b){ return b == 0 ? a : __gcd(b, a % b);} // Function to print a:b:cstatic void solveProportion(int a, int b1, int b2, int c){ int A = a * b2; int B = b1 * b2; int C = b1 * c; // To print the given proportion // in simplest form. int gcd = __gcd(__gcd(A, B), C); Console.Write( A / gcd + ":" + B / gcd + ":" + C / gcd);}// Driver codepublic static void Main(){ // Get the ratios int a, b1, b2, c; // Get ratio a:b1 a = 3; b1 = 4; // Get ratio b2:c b2 = 8; c = 9; // Find the ratio a:b:c solveProportion(a, b1, b2, c);}}// This code is contributed // by Akanksha Rai(Abby_akku) |
PHP
<?php// PHP implementation of above approachfunction __gcd($a, $b){ return $b == 0 ? $a : __gcd($b, $a % $b);} // Function to print a:b:cfunction solveProportion($a, $b1, $b2, $c){ $A = $a * $b2; $B = $b1 * $b2; $C = $b1 * $c; // To print the given proportion // in simplest form. $gcd = __gcd(__gcd($A, $B), $C); echo ($A / $gcd) . ":" . ($B / $gcd) . ":" . ($C / $gcd);}// Driver code// Get the ratios// Get ratio a:b1$a = 3;$b1 = 4;// Get ratio b2:c$b2 = 8;$c = 9;// Find the ratio a:b:csolveProportion($a, $b1, $b2, $c);// This code is contributed by mits?> |
Javascript
<script> // Javascript implementation of above approach function __gcd(a, b) { return b == 0 ? a : __gcd(b, a % b); } // Function to print a:b:c function solveProportion(a, b1, b2, c) { let A = a * b2; let B = b1 * b2; let C = b1 * c; // To print the given proportion // in simplest form. let gcd = __gcd(__gcd(A, B), C); document.write( A / gcd + ":" + B / gcd + ":" + C / gcd); } // Get the ratios let a, b1, b2, c; // Get ratio a:b1 a = 3; b1 = 4; // Get ratio b2:c b2 = 8; c = 9; // Find the ratio a:b:c solveProportion(a, b1, b2, c); // This code is contributed by divyeshrabadiya07.</script> |
Output:
6:8:9
Time Complexity : O(log(A+B)) ,where A=a*b2 and B = b1*b2
Space Complexity : O(1), since no extra space has been taken.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
