程序来查找2 ^ n的后两位数字

给定数字n，我们需要找到2 ⁿ的最后两位数字。
例子：

Input : n = 7
Output : 28

Input : n = 72
Output : 96
2^72 = 4722366482869645213696

天真的方法是迭代地找到2 ^ n的值或使用pow函数。计算完2 ^ n的值后，找到最后两位数字并打印出来。
注意：由于发生溢出，该方法仅在特定范围内工作2 ⁿ 。
下面是上述方法的实现。

C++

// C++ code to find last 2 digits of 2^n
#include 
using namespace std;
 
// Find the first digit
int LastTwoDigit(long long int num)
{
    // Get the last digit from the number
    int one = num % 10;
 
    // Remove last digit from number
    num /= 10;
 
    // Get the last digit from
    // the number(last second of num)
    int tens = num % 10;
 
    // Take last digit to ten's position
    // i.e. last second digit
    tens *= 10;
 
    // Add the value of ones and tens to
    // make it complete 2 digit number
    num = tens + one;
 
    // return the first digit
    return num;
}
 
// Driver program
int main()
{
    int n = 10;
    long long int num = 1;
 
    // pow function used
    num = pow(2, n);
 
    cout << "Last " << 2;
 
    cout << " digits of " << 2;
 
    cout << "^" << n << " = ";
 
    cout << LastTwoDigit(num) << endl;
    return 0;
}

Java

// Java code to find last 2 digits of 2^n
 
class Geeks {
     
// Find the first digit
static long LastTwoDigit(long num)
{
     
    // Get the last digit from the number
    long one = num % 10;
 
    // Remove last digit from number
    num /= 10;
 
    // Get the last digit from
    // the number(last second of num)
    long tens = num % 10;
 
    // Take last digit to ten's position
    // i.e. last second digit
    tens *= 10;
 
    // Add the value of ones and tens to
    // make it complete 2 digit number
    num = tens + one;
 
    // return the first digit
    return num;
}
 
    // Driver code
    public static void main(String args[])
    {
        int n = 10;
        long num = 1;
     
        // pow function used
        num = (long)Math.pow(2, n);
     
        System.out.println("Last 2 digits of 2^10 = "
                                 +LastTwoDigit(num));
     
    }
}
 
// This code is contributed by ankita_saini

Python 3

# Python 3 code to find
# last 2 digits of 2^n
 
# Find the first digit
def LastTwoDigit(num):
     
    # Get the last digit from the number
    one = num % 10
 
    # Remove last digit from number
    num //= 10
 
    # Get the last digit from
    # the number(last second of num)
    tens = num % 10
 
    # Take last digit to ten's position
    # i.e. last second digit
    tens *= 10
 
    # Add the value of ones and tens to
    # make it complete 2 digit number
    num = tens + one
 
    # return the first digit
    return num
 
# Driver Code
if __name__ == "__main__":
    n = 10
    num = 1
 
    # pow function used
    num = pow(2, n);
 
    print("Last " + str(2) + " digits of " +
                    str(2) + "^" + str(n) +
                           " = ", end = "")
 
    print(LastTwoDigit(num))
 
# This code is contributed
# by ChitraNayal

C#

// C# code to find last
// 2 digits of 2^n
using System;
 
class GFG
{
     
// Find the first digit
static long LastTwoDigit(long num)
{
     
    // Get the last digit
    // from the number
    long one = num % 10;
 
    // Remove last digit
    // from number
    num /= 10;
 
    // Get the last digit
    // from the number(last
    // second of num)
    long tens = num % 10;
 
    // Take last digit to
    // ten's position i.e.
    // last second digit
    tens *= 10;
 
    // Add the value of ones
    // and tens to make it
    // complete 2 digit number
    num = tens + one;
 
    // return the first digit
    return num;
}
 
    // Driver code
    public static void Main(String []args)
    {
        int n = 10;
        long num = 1;
     
        // pow function used
        num = (long)Math.Pow(2, n);
     
        Console.WriteLine("Last 2 digits of 2^10 = " +
                                   LastTwoDigit(num));
    }
}
 
// This code is contributed
// by Ankita_Saini

Javascript

 
// Javascript code to find last 2 digits of 2^n
 
// Find the first digit
function LastTwoDigit(num)
{
    // Get the last digit from the number
    let one = num % 10;
 
    // Remove last digit from number
    num = Math.floor(num/10);
 
    // Get the last digit from
    // the number(last second of num)
    let tens = num % 10;
 
    // Take last digit to ten's position
    // i.e. last second digit
    tens *= 10;
 
    // Add the value of ones and tens to
    // make it complete 2 digit number
    num = tens + one;
 
    // return the first digit
    return num;
}
 
// Driver program
 
    let n = 10;
    let num = 1;
 
    // pow function used
    num = Math.pow(2, n);
 
    document.write("Last " + 2);
 
    document.write(" digits of " + 2);
 
    document.write("^" + n + " = ");
 
    document.write(LastTwoDigit(num) + "<br>");
 
// This code is contributed by Mayank Tyagi
 

输出：

Last 2 digits of 2^10 = 24

高效的方法：高效的方法是每次乘法后仅保留2位数字。这个想法与“模幂”中讨论的想法非常相似，在这里讨论了一种一般的方法来找到(a ^ b)％c，在这种情况下，由于仅需要最后两位，所以c为10 ^ 2。
下面是上述方法的实现。

C++

// C++ code to find last 2 digits of 2^n
#include 
using namespace std;
 
/* Iterative Function to calculate (x^y)%p in O(log y) */
int power(long long int x, long long int y, long long int p)
{
    long long int res = 1; // Initialize result
 
    x = x % p; // Update x if it is more than or
    // equal to p
 
    while (y > 0) {
 
        // If y is odd, multiply x with result
        if (y & 1)
            res = (res * x) % p;
 
        // y must be even now
        y = y >> 1; // y = y/2
        x = (x * x) % p;
    }
    return res;
}
 
// C++ function to calculate
// number of digits in x
int numberOfDigits(int x)
{
    int i = 0;
    while (x) {
        x /= 10;
        i++;
    }
    return i;
}
 
// C++ function to print last 2 digits of 2^n
void LastTwoDigit(int n)
{
    cout << "Last " << 2;
    cout << " digits of " << 2;
    cout << "^" << n << " = ";
 
    // Generating 10^2
    int temp = 1;
    for (int i = 1; i <= 2; i++)
        temp *= 10;
 
    // Calling modular exponentiation
    temp = power(2, n, temp);
 
    // Printing leftmost zeros. Since (2^n)%2
    // can have digits less then 2. In that
    // case we need to print zeros
    for (int i = 0; i < 2 - numberOfDigits(temp); i++)
        cout << 0;
 
    // If temp is not zero then print temp
    // If temp is zero then already printed
    if (temp)
        cout << temp;
}
 
// Driver program to test above functions
int main()
{
    int n = 72;
    LastTwoDigit(n);
    return 0;
}

Java

// Java code to find last
// 2 digits of 2^n
class GFG
{
 
/* Iterative Function to
calculate (x^y)%p in O(log y) */
static int power(long x, long y,
                         long p)
{
int res = 1; // Initialize result
 
x = x % p; // Update x if it is more
           // than or equal to p
 
while (y > 0)
{
 
    // If y is odd, multiply
    // x with result
    long r = y & 1;
 
    if (r == 1)
        res = (res * (int)x) % (int)p;
 
    // y must be even now
    y = y >> 1; // y = y/2
    x = (x * x) % p;
}
return res;
}
 
// Java function to calculate
// number of digits in x
static int numberOfDigits(int x)
{
int i = 0;
while (x != 0)
{
    x /= 10;
    i++;
}
return i;
}
 
// Java function to print
// last 2 digits of 2^n
static void LastTwoDigit(int n)
{
System.out.print("Last " + 2 +
                 " digits of " + 2 + "^");
System.out.print(n +" = ");
 
// Generating 10^2
int temp = 1;
for (int i = 1; i <= 2; i++)
    temp *= 10;
 
// Calling modular exponentiation
temp = power(2, n, temp);
 
// Printing leftmost zeros.
// Since (2^n)%2 can have digits
// less then 2. In that case
// we need to print zeros
for (int i = 0;
         i < ( 2 - numberOfDigits(temp)); i++)
    System.out.print(0 + " ");
 
// If temp is not zero then
// print temp. If temp is zero
// then already printed
if (temp != 0)
    System.out.println(temp);
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 72;
    LastTwoDigit(n);
}
}
 
// This code is contributed
// by ChitraNayal

的Python 3

# Python 3 code to find
# last 2 digits of 2^n
 
# Iterative Function to
# calculate (x^y)%p in O(log y)
def power(x, y, p):
 
    res = 1 # Initialize result
 
    x = x % p # Update x if it is more
              # than or equal to p
 
    while (y > 0):
 
        # If y is odd, multiply
        # x with result
        if (y & 1):
            res = (res * x) % p
 
        # y must be even now
        y = y >> 1 # y = y/2
        x = (x * x) % p
         
    return res
 
# function to calculate
# number of digits in x
def numberOfDigits(x):
 
    i = 0
    while (x):
        x //= 10
        i += 1
     
    return i
 
# function to print
# last 2 digits of 2^n
def LastTwoDigit(n):
 
    print("Last " + str(2) +
          " digits of " + str(2), end = "")
    print("^" + str(n) + " = ", end = "")
 
    # Generating 10^2
    temp = 1
    for i in range(1, 3):
        temp *= 10
 
    # Calling modular exponentiation
    temp = power(2, n, temp)
 
    # Printing leftmost zeros.
    # Since (2^n)%2 can have digits
    # less then 2. In that case we
    # need to print zeros
    for i in range(2 - numberOfDigits(temp)):
        print(0, end = "")
 
    # If temp is not zero then print temp
    # If temp is zero then already printed
    if temp:
        print(temp)
 
# Driver Code
if __name__ == "__main__":
    n = 72
    LastTwoDigit(n)
 
# This code is contributed
# by ChitraNayal

C＃

// C# code to find last
// 2 digits of 2^n
using System;
 
class GFG
{
 
/* Iterative Function to calculate
   (x^y)%p in O(log y) */
static int power(long x, long y,
                         long p)
{
int res = 1; // Initialize result
 
x = x % p; // Update x if it is more
           // than or equal to p
 
while (y > 0)
{
 
    // If y is odd, multiply
    // x with result
    long r = y & 1;
 
    if (r == 1)
        res = (res * (int)x) % (int)p;
 
    // y must be even now
    y = y >> 1; // y = y/2
    x = (x * x) % p;
}
return res;
}
 
// C# function to calculate
// number of digits in x
static int numberOfDigits(int x)
{
    int i = 0;
    while (x != 0)
    {
        x /= 10;
        i++;
    }
    return i;
}
 
// C# function to print
// last 2 digits of 2^n
static void LastTwoDigit(int n)
{
Console.Write("Last " + 2 +
              " digits of " + 2 + "^");
Console.Write(n + " = ");
 
// Generating 10^2
int temp = 1;
for (int i = 1; i <= 2; i++)
    temp *= 10;
 
// Calling modular exponentiation
temp = power(2, n, temp);
 
// Printing leftmost zeros. Since
// (2^n)%2 can have digits less
// then 2. In that case we need
// to print zeros
for (int i = 0;
         i < ( 2 - numberOfDigits(temp)); i++)
    Console.Write(0 + " ");
 
// If temp is not zero then print temp
// If temp is zero then already printed
if (temp != 0)
    Console.Write(temp);
}
 
// Driver Code
public static void Main()
{
    int n = 72;
    LastTwoDigit(n);
}
}
 
// This code is contributed
// by ChitraNayal

的PHP

 0)
    {
 
        // If y is odd, multiply
        // x with result
        if ($y & 1)
            $res = ($res * $x) % $p;
 
        // y must be even now
        $y = $y >> 1; // y = y/2
        $x = ($x * $x) % $p;
    }
    return $res;
}
 
// PHP function to calculate
// number of digits in x
function numberOfDigits($x)
{
    $i = 0;
    while ($x)
    {
        $x /= 10;
        $i++;
    }
    return $i;
}
 
// PHP function to print
// last 2 digits of 2^n
function LastTwoDigit($n)
{
    echo("Last " . 2);
    echo(" digits of " . 2);
    echo("^" . $n ." = ");
 
    // Generating 10^2
    $temp = 1;
    for ($i = 1; $i <= 2; $i++)
        $temp *= 10;
 
    // Calling modular
    // exponentiation
    $temp = power(2, $n, $temp);
 
    // Printing leftmost zeros.
    // Since (2^n)%2 can have
    // digits less then 2. In
    // that case we need to
    // print zeros
    for ($i = 0;
         $i < 2 - numberOfDigits($temp); $i++)
        echo (0);
 
    // If temp is not zero then
    // print temp. If temp is zero
    // then already printed
    if ($temp)
        echo ($temp);
}
 
// Driver Code
$n = 72;
LastTwoDigit($n);
 
// This code is contributed
// by Shivi_Aggarwal
?>

输出：

Last 2 digits of 2^72 = 96

时间复杂度： O(log n)