Dado un número n, necesitamos encontrar los dos últimos dígitos de 2 n .
Ejemplos :
Input : n = 7 Output : 28 Input : n = 72 Output : 96 2^72 = 4722366482869645213696
Un enfoque ingenuo es encontrar el valor de 2^n iterativamente o usando la función pow . Una vez que se calcule el valor de 2^n, busque los dos últimos dígitos e imprímalos.
Nota: este enfoque solo funcionará durante 2 n dentro de un cierto rango, ya que se produce un desbordamiento .
A continuación se muestra la implementación del enfoque anterior.
C++
// C++ code to find last 2 digits of 2^n
#include <bits/stdc++.h>
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
Python3
# 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
PHP
<?php
// PHP code to find last
// 2 digits of 2^n
// Find the first digit
function 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
$n = 10;
$num = 1;
// pow function used
$num = pow(2, $n);
echo ("Last " . 2);
echo (" digits of " . 2);
echo("^" . $n . " = ");
echo( LastTwoDigit($num)) ;
// This code is contributed
// by Shivi_Aggarwal
?>
Javascript
<script>
// 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
</script>
Producción:
Last 2 digits of 2^10 = 24
Enfoque eficiente: la forma eficiente es mantener solo 2 dígitos después de cada multiplicación. Esta idea es muy similar a la discutida en Exponenciación modular donde se discute una forma general de encontrar (a^b)%c, aquí en este caso c es 10^2 ya que solo se necesitan los dos últimos dígitos.
A continuación se muestra la implementación del enfoque anterior.
C++
// C++ code to find last 2 digits of 2^n
#include <iostream>
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 than 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 than 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
Python3
# 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 than 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
PHP
<?php
// PHP code to find last
// 2 digits of 2^n
/* Iterative Function to
calculate (x^y)%p in O(log y) */
function 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 $res;
}
// PHP function to calculate
// number of digits in x
function numberOfDigits($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, $temp);
// Printing leftmost zeros.
// Since (2^n)%2 can have
// digits less than 2. In
// that case we need to
// print zeros
for ($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
?>
Javascript
<script>
// Javascript code to find last
// 2 digits of 2^n
/* Iterative Function to
calculate (x^y)%p in O(log y) */
function power(x, y, p)
{
let 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
let r = y & 1;
if (r == 1)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
// JavaScript function to calculate
// number of digits in x
function numberOfDigits(x)
{
let i = 0;
while (x != 0)
{
x /= 10;
i++;
}
return i;
}
// JavaScript function to print
// last 2 digits of 2^n
function LastTwoDigit(n)
{
document.write("Last " + 2 +
" digits of " + 2 + "^");
document.write(n +" = ");
// Generating 10^2
let temp = 1;
for (let 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 than 2. In that case
// we need to print zeros
for (let i = 0;
i < ( 2 - numberOfDigits(temp)); i++)
document.write(0 + " ");
// If temp is not zero then
// print temp. If temp is zero
// then already printed
if (temp != 0)
document.write(temp);
}
// driver program
let n = 72;
LastTwoDigit(n);
</script>
Producción:
Last 2 digits of 2^72 = 96
Complejidad de tiempo: O (log n)
Publicación traducida automáticamente
Artículo escrito por vishal9619 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA