查找给定的五位数加到5的最后五位数

给定一个五位数的数字N。任务是在将数字修改为以下数字后找到给定数字的后五位数提高到5的幂：

first digit, third digit, fifth digit, fourth digit, second digit.

例子：

Input : N = 12345
Output : 71232
Explanation : 
After modification the number becomes 13542. (13542)5 is 
455422043125550171232

Input : N = 10000
Output : 00000

方法：在此问题中，仅需要执行语句中描述的动作。但是，此问题有两个陷阱。
第一个陷阱是不能用64位整数表示五位数的第五次幂。但是我们实际上并不需要第五幂，我们需要第五幂以10 ^5为模。每次乘法后都可以应用mod运算。
第二个陷阱是您需要输出五个数字，而不是第五个幂模10 ⁵ 。区别在于从末尾开始的第五个数字为零。要输出，前导零的数字可以使用相应的格式(printf中的％05d)，也可以提取数字并一一输出。
下面是上述方法的实现：

C++

// CPP program to find last five digits
// of a five digit number raised to power five
 
#include 
using namespace std;
 
// Function to find the last five digits
// of a five digit number raised to power five
int lastFiveDigits(int n)
{
    n = (n / 10000) * 10000
        + ((n / 100) % 10)
              * 1000
        + (n % 10)
              * 100
        + ((n / 10) % 10)
              * 10
        + (n / 1000) % 10;
 
    int ans = 1;
    for (int i = 0; i < 5; i++) {
        ans *= n;
        ans %= 100000;
    }
 
    printf("%05d", ans);
}
 
// Driver code
int main()
{
    int n = 12345;
 
    lastFiveDigits(n);
 
    return 0;
}

Java

// Java program to find last five digits
// of a five digit number raised to power five
 
class GfG {
 
    // Function to find the last five digits
    // of a five digit number raised to power five
    static void lastFiveDigits(int n)
    {
        n = (n / 10000) * 10000
            + ((n / 100) % 10)
                  * 1000
            + (n % 10)
                  * 100
            + ((n / 10) % 10)
                  * 10
            + (n / 1000) % 10;
 
        int ans = 1;
        for (int i = 0; i < 5; i++) {
            ans *= n;
            ans %= 100000;
        }
 
        System.out.println(ans);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 12345;
 
        lastFiveDigits(n);
    }
}

Python3

# Python3 program to find last five digits
# of a five digit number raised to power five
 
# Function to find the last five digits
# of a five digit number raised to power five
def lastFiveDigits(n):
    n = ((int)(n / 10000) * 10000 +
        ((int)(n / 100) % 10) * 1000 + (n % 10) * 100 +
        ((int)(n / 10) % 10) * 10 + (int)(n / 1000) % 10)
    ans = 1
    for i in range(5):
        ans *= n
        ans %= 100000
    print(ans)
 
# Driver code
if __name__ == '__main__':
    n = 12345
 
    lastFiveDigits(n)
 
# This code contributed by PrinciRaj1992

C#

// C# program to find last five
// digits of a five digit number
// raised to power five
using System;
 
class GFG
{
 
    // Function to find the last
    // five digits of a five digit
    // number raised to power five
    public static void lastFiveDigits(int n)
    {
        n = (n / 10000) * 10000 +
           ((n / 100) % 10) * 1000 +
            (n % 10) * 100 +
           ((n / 10) % 10) * 10 +
            (n / 1000) % 10;
 
        int ans = 1;
        for (int i = 0; i < 5; i++)
        {
            ans *= n;
            ans %= 100000;
        }
 
        Console.WriteLine(ans);
    }
 
    // Driver code
    public static void Main(string[] args)
    {
        int n = 12345;
 
        lastFiveDigits(n);
    }
}
 
// This code is contributed
// by Shrikant13

Javascript

// JavaScript program to find last five digits
// of a five digit number raised to power five
 
// Function to find the last five digits
// of a five digit number raised to power five
function lastFiveDigits(n)
{
    n = (Math.floor(n / 10000)) * 10000
        + (Math.floor(n / 100) % 10)
            * 1000
        + (n % 10)
            * 100
        + (Math.floor(n / 10) % 10)
            * 10
        + Math.floor(n / 1000) % 10;
 
    let ans = 1;
    for (let i = 0; i < 5; i++) {
        ans *= n;
        ans %= 100000;
    }
 
    document.write(ans);
}
 
// Driver code
 
    let n = 12345;
 
    lastFiveDigits(n);
 
// This code is contributed by Manoj.
 

输出：

71232