7的幂的最后两位数

给定正N ，任务是找到7 ^N的后两位。
例子：

Input: N = 5 
Output: 07 
Explanation: 
The value of 7⁵ = 7 * 7 * 7 * 7 * 7 = 8507 
Therefore, the last two digits are 07.
Input: N = 12 
Output: 01 
Explanation: 
The value of 7¹² = 13841287201 
Therefore, the last two digits are 01. 
 

方法：找到X ^Y的最后K个数字的一般方法是讨论对数时间复杂度的本文。在本文中，我们将讨论固定时间解决方案。
下面是用于7 ^N对于N的一些值的值的观察：

7¹ = 7 last two digit = 07 
7² = 49 last two digit = 49 
7³ = 243 last two digit = 43 
7⁴ = 2401 lasr two digit = 01
7⁵ = 16807 last two digit = 07 
7⁶ = 117649 last two digit = 49 
7⁷ = 823543 last two digit = 43 
7⁸ = 5764801 last two digit = 01 
 

基于以上观察，我们有以下几种情况：

1. 如果7的最后两位^N = 07，则N = 4K + 3。
2. 如果N = 4K + 2，则7中的最后两位^N = 49。
3. 如果N = 4K + 1，则7中的最后两位^N = 43。
4. 如果7的最后两位^N = 01，则N = 4K。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find the last
// two digits of 7^N
string get_last_two_digit(int N)
{
    // Case 4
    if (N % 4 == 0)
        return "01";
 
    // Case 3
    else if (N % 4 == 1)
        return "07";
 
    // Case 2
    else if (N % 4 == 2)
        return "49";
 
    // Case 1
    return "43";
}
 
// Driver Code
int main()
{
    // Given Number
    int N = 12;
 
    // Function Call
    cout << get_last_two_digit(N);
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG{
 
// Function to find the last
// two digits of 7^N
public static String get_last_two_digit(int N)
{
    // Case 4
    if (N % 4 == 0)
        return "01";
 
    // Case 3
    else if (N % 4 == 1)
        return "07";
 
    // Case 2
    else if (N % 4 == 2)
        return "49";
 
    // Case 1
    return "43";
}
 
// Driver code
public static void main(String[] args)
{
    int N = 12;
 
    // Function Call
    System.out.println(get_last_two_digit(N));
}
}
 
// This code is contributed by grand_master

Python3

# Python3 program for the above approach
 
# Function to find the last
# two digits of 7 ^ N
def get_last_two_digit(N):
 
    # Case 4
    if (N % 4 == 0):
        return "01";
 
    # Case 3
    elif (N % 4 == 1):
        return "07";
 
    # Case 2
    elif (N % 4 == 2):
        return "49";
 
    # Case 1
    return "43";
     
# Driver Code
 
# Given number
N = 12;
 
# Function call
print( get_last_two_digit(N))
 
# This code is contributed by grand_master

C#

// C# program for the above approach
using System;
 
namespace GFG{
class GFG{
     
// Function to find the last
// two digits of 7^N
public static String get_last_two_digit(int N)
{
    // Case 4
    if (N % 4 == 0)
        return "01";
 
    // Case 3
    else if (N % 4 == 1)
        return "07";
 
    // Case 2
    else if (N % 4 == 2)
        return "49";
 
    // Case 1
    return "43";
}
 
// Driver code
public static void Main()
{
     
    // Given number
    int N = 12;
 
    // Function Call
    Console.Write(get_last_two_digit(N));
}
}
}
 
// This code is contributed by grand_master

Javascript

 
// Javascript program for the above approach
 
// Function to find the last
    // two digits of 7^N
    function get_last_two_digit(N)
    {
        // Case 4
        if (N % 4 == 0)
            return "01";
 
        // Case 3
        else if (N % 4 == 1)
            return "07";
 
        // Case 2
        else if (N % 4 == 2)
            return "49";
 
        // Case 1
        return "43";
    }
 
    // Driver code
     
        var N = 12;
 
        // Function Call
        document.write(get_last_two_digit(N));
 
// This code contributed by gauravrajput1
 

输出：

01

时间复杂度： O(1)
辅助空间： O(1)