当数字的第一位除以最后一位时，求出余数

给定数字N，请找到N的第一个数字除以最后一个数字时的余数。
例子：

Input: N = 1234
Output: 1
First digit = 1
Last digit = 4
Remainder = 1 % 4 = 1

Input: N = 5223
Output: 2
First digit = 5
Last digit = 3
Remainder = 5 % 3 = 2

方法：找到数字的第一位和最后一位。当第一个数字除以最后一个数字时，找到余数。
下面是上述方法的实现：

C++

// C++ program to find the remainder
// when the First digit of a number
// is divided by its Last digit
 
#include 
using namespace std;
 
// Function to find the remainder
void findRemainder(int n)
{
    // Get the last digit
    int l = n % 10;
 
    // Get the first digit
    while (n >= 10)
        n /= 10;
    int f = n;
 
    // Compute the remainder
    int remainder = f % l;
 
    cout << remainder << endl;
}
 
// Driver code
int main()
{
 
    int n = 5223;
 
    findRemainder(n);
 
    return 0;
}

Java

// Java program to find the remainder
// when the First digit of a number
// is divided by its Last digit
class GFG
{
     
// Function to find the remainder
static void findRemainder(int n)
{
    // Get the last digit
    int l = n % 10;
 
    // Get the first digit
    while (n >= 10)
        n /= 10;
    int f = n;
 
    // Compute the remainder
    int remainder = f % l;
 
    System.out.println(remainder);
}
 
// Driver code
public static void main(String[] args)
{
    int n = 5223;
    findRemainder(n);
}
}
 
// This code is contributed by Code_Mech

Python3

# Python3 program to find the remainder
# when the First digit of a number
# is divided by its Last digit
 
# Function to find the remainder
def findRemainder(n):
     
    # Get the last digit
    l = n % 10
  
    # Get the first digit
    while (n >= 10):
        n //= 10
    f = n
 
    # Compute the remainder
    remainder = f % l
 
    print(remainder)
 
# Driver code
n = 5223
 
findRemainder(n)
 
# This code is contributed by Mohit Kumar

C#

// C# program to find the remainder
// when the First digit of a number
// is divided by its Last digit
using System;
 
class GFG
{
     
// Function to find the remainder
static void findRemainder(int n)
{
    // Get the last digit
    int l = n % 10;
 
    // Get the first digit
    while (n >= 10)
        n /= 10;
    int f = n;
 
    // Compute the remainder
    int remainder = f % l;
 
    Console.WriteLine(remainder);
}
 
// Driver code
public static void Main()
{
    int n = 5223;
    findRemainder(n);
}
}
 
// This code is contributed by Code_Mech

输出：

2

时间复杂度： O(L)，其中L是十进制表示形式的数字长度

辅助空间： O(1)