Given two positive numbers N and M, the task is to check whether the given pairs of numbers (N, M) form a Betrothed Numbers or not.
Examples:
Input: N = 48, M = 75
Output: Yes
Explanation:
The proper divisors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24
Sum of proper divisors of 48 is 75(sum1)
The proper divisors of 75 are 1, 3, 5, 15, 25
Sum of proper divisors of 48 is 49(sum2)
Since sum2 = N + 1, therefore the given pairs form Betrothed numbers.
Input: N = 95, M = 55
Output: No
Explanation:
The proper divisors of 95 are 1, 5, 19
Sum of proper divisors of 48 is 25(sum1)
The proper divisors of 55 are 1, 5, 11
Sum of proper divisors of 48 is 17(sum2)
Since Neither sum2 is equals N + 1 nor sum1 is equals to M + 1, therefore the given pairs doesn’t form Betrothed numbers.
Approach:
- Find the sum of proper divisors of the given numbers N and M.
- If sum of proper divisors of N is equals to M + 1 or sum of proper divisors of M is equals to N + 1 then the given pairs form a Betrothed Numbers.
- Else it doesn’t forms a pair of Betrothed Numbers.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to check whether N is// Perfect Square or notbool isPerfectSquare(int N){ // Find sqrt double sr = sqrt(N); return (sr - floor(sr)) == 0;}// Function to check whether the given// pairs of numbers is Betrothed Numbers// or notvoid BetrothedNumbers(int n, int m){ int Sum1 = 1; int Sum2 = 1; // For finding the sum of all the // divisors of first number n for (int i = 2; i <= sqrt(n); i++) { if (n % i == 0) { Sum1 += i + (isPerfectSquare(n) ? 0 : n / i); } } // For finding the sum of all the // divisors of second number m for (int i = 2; i <= sqrt(m); i++) { if (m % i == 0) { Sum2 += i + (isPerfectSquare(m) ? 0 : m / i); } } if ((n + 1 == Sum2) && (m + 1 == Sum1)) { cout << "YES" << endl; } else { cout << "NO" << endl; }}// Driver Codeint main(){ int N = 9504; int M = 20734; // Function Call BetrothedNumbers(N, M); return 0;} |
Java
// Java program for the above approachclass GFG{ // Function to check whether N is// Perfect Square or notstatic boolean isPerfectSquare(int N){ // Find sqrt double sr = Math.sqrt(N); return (sr - Math.floor(sr)) == 0;} // Function to check whether the given// pairs of numbers is Betrothed Numbers// or notstatic void BetrothedNumbers(int n, int m){ int Sum1 = 1; int Sum2 = 1; // For finding the sum of all the // divisors of first number n for (int i = 2; i <= Math.sqrt(n); i++) { if (n % i == 0) { Sum1 += i + (isPerfectSquare(n) ? 0 : n / i); } } // For finding the sum of all the // divisors of second number m for (int i = 2; i <= Math.sqrt(m); i++) { if (m % i == 0) { Sum2 += i + (isPerfectSquare(m) ? 0 : m / i); } } if ((n + 1 == Sum2) && (m + 1 == Sum1)) { System.out.print("YES" +"\n"); } else { System.out.print("NO" +"\n"); }} // Driver Codepublic static void main(String[] args){ int N = 9504; int M = 20734; // Function Call BetrothedNumbers(N, M);}}// This code is contributed by 29AjayKumar |
Python3
# Python3 program for the above approachfrom math import sqrt,floor# Function to check whether N is# Perfect Square or notdef isPerfectSquare(N): # Find sqrt sr = sqrt(N) return (sr - floor(sr)) == 0# Function to check whether the given# pairs of numbers is Betrothed Numbers# or notdef BetrothedNumbers(n,m): Sum1 = 1 Sum2 = 1 # For finding the sum of all the # divisors of first number n for i in range(2,int(sqrt(n))+1,1): if (n % i == 0): if (isPerfectSquare(n)): Sum1 += i else: Sum1 += i + n/i # For finding the sum of all the # divisors of second number m for i in range(2,int(sqrt(m))+1,1): if (m % i == 0): if (isPerfectSquare(m)): Sum2 += i else: Sum2 += i + (m / i) if ((n + 1 == Sum2) and (m + 1 == Sum1)): print("YES") else: print("NO")# Driver Codeif __name__ == '__main__': N = 9504 M = 20734 # Function Call BetrothedNumbers(N, M)# This code is contributed by Surendra_Gangwar |
C#
// C# program for the above approachusing System;class GFG{// Function to check whether N is// perfect square or notstatic bool isPerfectSquare(int N){ // Find sqrt double sr = Math.Sqrt(N); return (sr - Math.Floor(sr)) == 0;}// Function to check whether the given// pairs of numbers is Betrothed numbers// or notstatic void BetrothedNumbers(int n, int m){ int Sum1 = 1; int Sum2 = 1; // For finding the sum of all the // divisors of first number n for(int i = 2; i <= Math.Sqrt(n); i++) { if (n % i == 0) { Sum1 += i + (isPerfectSquare(n) ? 0 : n / i); } } // For finding the sum of all the // divisors of second number m for(int i = 2; i <= Math.Sqrt(m); i++) { if (m % i == 0) { Sum2 += i + (isPerfectSquare(m) ? 0 : m / i); } } if ((n + 1 == Sum2) && (m + 1 == Sum1)) { Console.Write("YES" + "\n"); } else { Console.Write("NO" + "\n"); }}// Driver Codepublic static void Main(String[] args){ int N = 9504; int M = 20734; // Function Call BetrothedNumbers(N, M);}}// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript program for the above approach// Function to check whether N is// Perfect Square or notfunction isPerfectSquare(N){ // Find sqrt let sr = Math.sqrt(N); return (sr - Math.floor(sr)) == 0;}// Function to check whether the given// pairs of numbers is Betrothed Numbers// or notfunction BetrothedNumbers(n, m){ let Sum1 = 1; let Sum2 = 1; // For finding the sum of all the // divisors of first number n for (let i = 2; i <= Math.sqrt(n); i++) { if (n % i == 0) { Sum1 += i + (isPerfectSquare(n) ? 0 : parseInt(n / i)); } } // For finding the sum of all the // divisors of second number m for (let i = 2; i <= Math.sqrt(m); i++) { if (m % i == 0) { Sum2 += i + (isPerfectSquare(m) ? 0 : parseInt(m / i)); } } if ((n + 1 == Sum2) && (m + 1 == Sum1)) { document.write("YES"); } else { document.write("NO"); }}// Driver Code let N = 9504; let M = 20734; // Function Call BetrothedNumbers(N, M);// This code is contributed by rishavmahato348.</script> |
Output:
NO
Time Complexity: O(?N + ?M)
Auxiliary Space: O(1)
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