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📜  找到两个合成数字，使它们的差为N

📅  最后修改于: 2021-04-22 09:57:52             🧑  作者: Mango

给定数字N ，任务是找到两个复合数字X和Y ，以使它们之间的差等于N。请注意，此任务可以有多个答案。打印其中任何一个。
例子： 

Input: N = 4
Output: X = 36, Y = 32

Input: N = 1
Output: X = 9, Y = 8

方法：

  • 我们必须找到X – Y = N。
  • 我们知道， N的最小值可以是0或1 。如果它是0 ，那么我们可以将任何一个复合数字打印两次。
  • 如果N = 0 ，那么我们可以打印9 * N和8 * N ，因为这些复合数字之间的差异最小，即1 。
  • 我们还可以打印15 * N和16 * N ，但是我们必须打印任何两个复合数字，因此这些数字都是可能的。

下面是实现

C++
#include 
using namespace std;
// C++ code to Find two Composite Numbers
// such that there difference is N
 
// Function to find the two composite numbers
void find_composite_nos(int n)
{
    cout << 9 * n << " " << 8 * n;
}
 
// Driver code
int main()
{
    int n = 4;
 
    find_composite_nos(n);
 
    return 0;
}

Java
// Java code to Find two Composite Numbers
// such that there difference is N
class GFG
{
 
    // Function to find the two composite numbers
    static void find_composite_nos(int n)
    {
        System.out.println(9 * n + " " + 8 * n);
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int n = 4;
     
        find_composite_nos(n);
    }
}
 
// This code is contributed by AnkitRai01

Python3
# Python3 code to Find two Composite Numbers
# such that their difference is N
 
# Function to find the two composite numbers
def find_composite_nos(n) :
 
    print(9 * n, 8 * n);
 
# Driver code
if __name__ == "__main__" :
 
    n = 4;
 
    find_composite_nos(n);
 
# This code is contributed by AnkitRai01

C#
// C# code to Find two Composite Numbers
// such that there difference is N
using System;
 
class GFG
{
 
    // Function to find the two composite numbers
    static void find_composite_nos(int n)
    {
        Console.WriteLine(9 * n + " " + 8 * n);
    }
     
    // Driver code
    public static void Main()
    {
        int n = 4;
     
        find_composite_nos(n);
    }
}
 
// This code is contributed by AnkitRai01

Javascript

输出： 
36 32