Dados los enteros positivos k, ayb, necesitamos imprimir los últimos k dígitos de a^b, es decir, pow(a, b).
Input Constraint: k <= 9, a <= 10^6, b<= 10^6
Ejemplos:
Input : a = 11, b = 3, k = 2 Output : 31 Explanation : a^b = 11^3 = 1331, hence last two digits are 31 Input : a = 10, b = 10000, k = 5 Output : 00000 Explanation : a^b = 1000..........0 (total zeros = 10000), hence last 5 digits are 00000
Solución ingenua
Primero calcule a^b, luego tome los últimos k dígitos tomando módulo con 10^k. La solución anterior falla cuando a^b es demasiado grande, ya que podemos contener como máximo 2^64 -1 en C/C++.
Solución eficiente
La forma eficiente es mantener solo k dígitos después de cada multiplicación. Esta idea es muy similar a la discutida en Modular Exponentiation donde discutimos una forma general de encontrar (a^b)%c, aquí en este caso c es 10^k.
Aquí está la implementación.
C++
// C++ code to find last k digits of a^b
#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 k digits of a^b
void printLastKDigits(int a, int b, int k)
{
cout << "Last " << k;
cout << " digits of " << a;
cout << "^" << b << " = ";
// Generating 10^k
int temp = 1;
for (int i = 1; i <= k; i++)
temp *= 10;
// Calling modular exponentiation
temp = power(a, b, temp);
// Printing leftmost zeros. Since (a^b)%k
// can have digits less than k. In that
// case we need to print zeros
for (int i = 0; i < k - 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 a = 11;
int b = 3;
int k = 2;
printLastKDigits(a, b, k);
return 0;
}
Java
// Java code to find last k digits of a^b
public class GFG
{
/* Iterative Function to calculate (x^y)%p in O(log y) */
static int power(long x, long y, long p)
{
long 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) != 0)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return (int) res;
}
// Method to print last k digits of a^b
static void printLastKDigits(int a, int b, int k)
{
System.out.print("Last " + k + " digits of " + a +
"^" + b + " = ");
// Generating 10^k
int temp = 1;
for (int i = 1; i <= k; i++)
temp *= 10;
// Calling modular exponentiation
temp = power(a, b, temp);
// Printing leftmost zeros. Since (a^b)%k
// can have digits less than k. In that
// case we need to print zeros
for (int i = 0; i < k - Integer.toString(temp).length() ; 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.print(temp);
}
// Driver Method
public static void main(String[] args)
{
int a = 11;
int b = 3;
int k = 2;
printLastKDigits(a, b, k);
}
}
Python3
# Python 3 code to find last
# k digits of a^b
# 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 k digits of a^b
def printLastKDigits( a, b, k):
print("Last " + str( k)+" digits of " +
str(a) + "^" + str(b), end = " = ")
# Generating 10^k
temp = 1
for i in range(1, k + 1):
temp *= 10
# Calling modular exponentiation
temp = power(a, b, temp)
# Printing leftmost zeros. Since (a^b)%k
# can have digits less than k. In that
# case we need to print zeros
for i in range( k - numberOfDigits(temp)):
print("0")
# If temp is not zero then print temp
# If temp is zero then already printed
if (temp):
print(temp)
# Driver Code
if __name__ == "__main__":
a = 11
b = 3
k = 2
printLastKDigits(a, b, k)
# This code is contributed
# by ChitraNayal
C#
// C# code to find last k digits of a^b
using System;
class GFG
{
// Iterative Function to calculate
// (x^y)%p in O(log y)
static int power(long x, long y, long p)
{
long 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) != 0)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return (int) res;
}
// Method to print last k digits of a^b
static void printLastKDigits(int a, int b, int k)
{
Console.Write("Last " + k + " digits of " +
a + "^" + b + " = ");
// Generating 10^k
int temp = 1;
for (int i = 1; i <= k; i++)
temp *= 10;
// Calling modular exponentiation
temp = power(a, b, temp);
// Printing leftmost zeros. Since (a^b)%k
// can have digits less than k. In that
// case we need to print zeros
for (int i = 0;
i < k - temp.ToString().Length; i++)
Console.WriteLine(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 a = 11;
int b = 3;
int k = 2;
printLastKDigits(a, b, k);
}
}
// This code is contributed
// by 29AjayKumar
PHP
<?php
// PHP code to find last k digits of a^b
/* 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 $res;
}
// function to calculate
// number of digits in x
function numberOfDigits($x)
{
$i = 0;
while ($x)
{
$x = (int)$x / 10;
$i++;
}
return $i;
}
// function to print last k digits of a^b
function printLastKDigits( $a, $b, $k)
{
echo "Last ",$k;
echo " digits of " ,$a;
echo "^" , $b , " = ";
// Generating 10^k
$temp = 1;
for ($i = 1; $i <= $k; $i++)
$temp *= 10;
// Calling modular exponentiation
$temp = power($a, $b, $temp);
// Printing leftmost zeros. Since
// (a^b)%k can have digits less
// then k. In that case we need
// to print zeros
for ($i = 0;
$i < $k - numberOfDigits($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
$a = 11;
$b = 3;
$k = 2;
printLastKDigits($a, $b, $k);
// This code is contributed by ajit
?>
Javascript
<script>
// javascript code to find last k digits of a^b
/* Iterative Function to calculate (x^y)%p in O(log y) */
function power(x , y , p) {
var 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) != 0)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return parseInt( res);
}
// Method to print last k digits of a^b
function printLastKDigits(a , b , k) {
document.write("Last " + k + " digits of " + a + "^" + b + " = ");
// Generating 10^k
var temp = 1;
for (i = 1; i <= k; i++)
temp *= 10;
// Calling modular exponentiation
temp = power(a, b, temp);
// Printing leftmost zeros. Since (a^b)%k
// can have digits less than k. In that
// case we need to print zeros
for (i = 0; i < k - temp.toString().length; 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 Method
var a = 11;
var b = 3;
var k = 2;
printLastKDigits(a, b, k);
// This code contributed by gauravrajput1
</script>
Producción:
Last 2 digits of 11^3 = 31
Complejidad de tiempo: O(log b)
Complejidad de espacio: O(1)
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA