查找满足给定条件的数字a和b

给定一个整数n ，任务是找到两个满足以下条件的整数a和b ：

1. a％b = 0
2. a * b> n
3. a / b

如果没有一对满足上述条件，则打印-1 。
注意：对于n ，可以有多个(a，b)对满足上述条件。

例子：

Input: n = 10
Output: a = 90, b = 10
90 % 10 = 0
90 * 10 = 900 > 10
90 / 10 = 9 < 10
All three conditions are satisfied.

Input: n = 1
Output: -1

方法：假设b = n ，通过采用此假设，可以根据给定条件找到a ：

• (a％b = 0) => a应该是b的倍数。
• (a / b => a / b = n – 1 ，即 。
• (a * b> n) => a = n

下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
  
// Function to print the required numbers
void find(int n)
{
    // Suppose b = n and we want a % b = 0 and also
    // (a / b) < n so a = b * (n - 1)
    int b = n;
    int a = b * (n - 1);
  
    // Special case if n = 1
    // we get a = 0 so (a * b) < n
    if (a * b > n && a / b < n) {
        cout << "a = " << a << ", b = " << b;
    }
  
    // If no pair satisfies the conditions
    else
        cout << -1 << endl;
}
  
// Driver code
int main()
{
    int n = 10;
    find(n);
    return 0;
}

Java

// Java implementation of the above approach 
  
public class GFG{
  
    // Function to print the required numbers 
    static void find(int n) 
    { 
        // Suppose b = n and we want a % b = 0 and also 
        // (a / b) < n so a = b * (n - 1) 
        int b = n; 
        int a = b * (n - 1); 
      
        // Special case if n = 1 
        // we get a = 0 so (a * b) < n 
        if (a * b > n && a / b < n) { 
            System.out.print("a = " + a + ", b = " + b); 
        } 
      
        // If no pair satisfies the conditions 
        else
            System.out.println(-1); 
    } 
      
    // Driver code 
    public static void main(String []args)
    { 
        int n = 10; 
        find(n); 
    } 
    // This code is contributed by Ryuga
}

Python3

# Python3 implementation of the above approach 
  
# Function to print the required numbers 
def find(n): 
  
    # Suppose b = n and we want a % b = 0 
    # and also (a / b) < n so a = b * (n - 1) 
    b = n 
    a = b * (n - 1) 
  
    # Special case if n = 1 
    # we get a = 0 so (a * b) < n 
    if a * b > n and a // b < n: 
        print("a = {}, b = {}" . format(a, b)) 
      
    # If no pair satisfies the conditions 
    else:
        print(-1) 
  
# Driver Code
if __name__ == "__main__": 
  
    n = 10
    find(n) 
  
# This code is contributed by Rituraj Jain

C#

// C# implementation of the above approach 
using System;
  
class GFG
{
  
// Function to print the required numbers 
static void find(int n) 
{ 
    // Suppose b = n and we want a % b = 0 
    // and also (a / b) < n so a = b * (n - 1) 
    int b = n; 
    int a = b * (n - 1); 
  
    // Special case if n = 1 
    // we get a = 0 so (a * b) < n 
    if (a * b > n && a / b < n) 
    { 
        Console.Write("a = " + a + ", b = " + b); 
    } 
  
    // If no pair satisfies the conditions 
    else
        Console.WriteLine(-1); 
} 
  
// Driver code 
public static void Main()
{ 
    int n = 10; 
    find(n); 
} 
}
  
// This code is contributed 
// by Akanksha Rai

PHP

 $n && $a / $b <$n) {
        echo "a = " , $a , ", b = " , $b;
    }
  
    // If no pair satisfies the conditions
    else
        echo -1 ;
}
  
// Driver code
  
    $n = 10;
    find($n);
// This code is contributed 
// by inder_verma..
?>

输出：

a = 90, b = 10