两个n位数字的最小和最大和

给定整数N≥1 ，任务是找到两个N位数字的最小和最大和。
例子：

Input: N = 1 
Output: 
Largest = 18 
Smallest = 0 
Largest 1-digit number is 9 and 9 + 9 = 18 
Smallest 1-digit number is 0 and 0 + 0 = 0
Input: N = 2 
Output: 
Largest = 198 
Smallest = 20 
 

方法：

• 对于最大的：答案将是2 *(10 ^N – 1)，因为两个n位数字之和的序列将继续进行，例如2 * 9、2 * 99、2 * 999，…
• 对于最小的：
  • 如果N = 1，则答案将为0 。
  • 如果N> 1，则答案将是2 *(10 ^{N – 1} )，因为两个n位数字之和的序列将继续为0、20、200、2000等。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the smallest sum
// of 2 n-digit numbers
int smallestSum(int n)
{
    if (n == 1)
        return 0;
    return (2 * pow(10, n - 1));
}
 
// Function to return the largest sum
// of 2 n-digit numbers
int largestSum(int n)
{
    return (2 * (pow(10, n) - 1));
}
 
// Driver code
int main()
{
    int n = 4;
    cout << "Largest = " << largestSum(n) << endl;
    cout << "Smallest = " << smallestSum(n);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG {
 
    // Function to return the smallest sum
    // of 2 n-digit numbers
    static int smallestSum(int n)
    {
        if (n == 1)
            return 0;
        return (2 * (int)Math.pow(10, n - 1));
    }
 
    // Function to return the largest sum
    // of 2 n-digit numbers
    static int largestSum(int n)
    {
        return (2 * ((int)Math.pow(10, n) - 1));
    }
 
    // Driver code
    public static void main(String args[])
    {
        int n = 4;
        System.out.println("Largest = " + largestSum(n));
        System.out.print("Smallest = " + smallestSum(n));
    }
}

Python3

# Python3 implementation of the approach
 
# Function to return the smallest sum
# of 2 n-digit numbers
def smallestSum(n):
  
    if (n == 1):
        return 0
    return (2 * pow(10, n - 1))
 
# Function to return the largest sum
# of 2 n-digit numbers
def largestSum(n):
    return (2 * (pow(10, n) - 1))
 
# Driver code
n = 4
print("Largest = ", largestSum(n))
print("Smallest = ", smallestSum(n))

C#

// C# implementation of the approach
using System;
class GFG {
 
    // Function to return the smallest sum
    // of 2 n-digit numbers
    static int smallestSum(int n)
    {
        if (n == 1)
            return 0;
        return (2 * (int)Math.Pow(10, n - 1));
    }
 
    // Function to return the largest sum
    // of 2 n-digit numbers
    static int largestSum(int n)
    {
        return (2 * ((int)Math.Pow(10, n) - 1));
    }
 
    // Driver code
    public static void Main()
    {
        int n = 4;
        Console.WriteLine("Largest = " + largestSum(n));
        Console.Write("Smallest = " + smallestSum(n));
    }
}

Javascript

 
// Javascript implementation of the approach
 
// Function to return the smallest sum
// of 2 n-digit numbers
function smallestSum(n)
{
    if (n == 1)
        return 0;
    return (2 * Math.pow(10, n - 1));
}
 
// Function to return the largest sum
// of 2 n-digit numbers
function largestSum(n)
{
    return (2 * (Math.pow(10, n) - 1));
}
 
// Driver code
var n = 4;
document.write("Largest = " + largestSum(n) + "<br>");
document.write("Smallest = " + smallestSum(n));
 
// This code is contributed by noob2000.

输出：

Largest = 19998
Smallest = 2000