检查给定的一对数字是否已订婚

给定两个正数N和M ，任务是检查给定的数字对(N，M)是否形成了订婚的数字。
例子：

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 berothered 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 berothered numbers. 
 

方法：

1. 求出给定数字N和M的适当除数的总和。
2. 如果N适当的除数的总和等于M + 1或M的适当的除数的总和是等于N + 1，则给定的对形成婚约数。
3. 否则它就不会形成一对订婚号码。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to check whether N is
// Perfect Square or not
bool 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 not
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 <= 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 Code
int main()
{
    int N = 9504;
    int M = 20734;
 
    // Function Call
    BetrothedNumbers(N, M);
 
    return 0;
}

Java

// Java program for the above approach
class GFG{
  
// Function to check whether N is
// Perfect Square or not
static 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 not
static 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 Code
public 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 approach
from math import sqrt,floor
 
# Function to check whether N is
# Perfect Square or not
def 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 not
def 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 Code
if __name__ == '__main__':
    N = 9504
    M = 20734
 
    # Function Call
    BetrothedNumbers(N, M)
 
# This code is contributed by Surendra_Gangwar

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to check whether N is
// perfect square or not
static 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 not
static 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 Code
public static void Main(String[] args)
{
    int N = 9504;
    int M = 20734;
 
    // Function Call
    BetrothedNumbers(N, M);
}
}
 
// This code is contributed by Rajput-Ji

Javascript

// Javascript program for the above approach
 
// Function to check whether N is
// Perfect Square or not
function 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 not
function 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.

输出：

NO

时间复杂度： O(√N+√M)