Método 1 (usando bucles anidados)
Podemos calcular la potencia usando sumas repetidas.
Por ejemplo para calcular 5^6.
1) Primero 5 veces sumamos 5, obtenemos 25. (5^2)
2) Luego 5 veces sumamos 25, obtenemos 125. (5^3)
3) Luego 5 veces sumamos 125, obtenemos 625 (5^4)
4) Luego 5 veces sumamos 625, obtenemos 3125 (5^5)
5) Luego 5 veces sumamos 3125, obtenemos 15625 (5^6)
C++
// C++ code for power function
#include <bits/stdc++.h>
using namespace std;
/* Works only if a >= 0 and b >= 0 */
int pow(int a, int b)
{
if (b == 0)
return 1;
int answer = a;
int increment = a;
int i, j;
for(i = 1; i < b; i++)
{
for(j = 1; j < a; j++)
{
answer += increment;
}
increment = answer;
}
return answer;
}
// Driver Code
int main()
{
cout << pow(5, 3);
return 0;
}
// This code is contributed
// by rathbhupendra
C
#include<stdio.h>
/* Works only if a >= 0 and b >= 0 */
int pow(int a, int b)
{
//base case : anything raised to the power 0 is 1
if (b == 0)
return 1;
int answer = a;
int increment = a;
int i, j;
for(i = 1; i < b; i++)
{
for(j = 1; j < a; j++)
{
answer += increment;
}
increment = answer;
}
return answer;
}
/* driver program to test above function */
int main()
{
printf("\n %d", pow(5, 3));
getchar();
return 0;
}
Java
import java.io.*;
class GFG {
/* Works only if a >= 0 and b >= 0 */
static int pow(int a, int b)
{
if (b == 0)
return 1;
int answer = a;
int increment = a;
int i, j;
for (i = 1; i < b; i++) {
for (j = 1; j < a; j++) {
answer += increment;
}
increment = answer;
}
return answer;
}
// driver program to test above function
public static void main(String[] args)
{
System.out.println(pow(5, 3));
}
}
// This code is contributed by vt_m.
Python
# Python 3 code for power # function # Works only if a >= 0 and b >= 0 def pow(a,b): if(b==0): return 1 answer=a increment=a for i in range(1,b): for j in range (1,a): answer+=increment increment=answer return answer # driver code print(pow(5,3)) # this code is contributed # by Sam007
C#
using System;
class GFG
{
/* Works only if a >= 0 and b >= 0 */
static int pow(int a, int b)
{
if (b == 0)
return 1;
int answer = a;
int increment = a;
int i, j;
for (i = 1; i < b; i++) {
for (j = 1; j < a; j++) {
answer += increment;
}
increment = answer;
}
return answer;
}
// driver program to test
// above function
public static void Main()
{
Console.Write(pow(5, 3));
}
}
// This code is contributed by Sam007
PHP
<?php
// Works only if a >= 0
// and b >= 0
function poww($a, $b)
{
if ($b == 0)
return 1;
$answer = $a;
$increment = $a;
$i;
$j;
for($i = 1; $i < $b; $i++)
{
for($j = 1; $j < $a; $j++)
{
$answer += $increment;
}
$increment = $answer;
}
return $answer;
}
// Driver Code
echo( poww(5, 3));
// This code is contributed by nitin mittal.
?>
Javascript
<script>
/* Works only if a >= 0 and b >= 0 */
function pow(a , b)
{
if (b == 0)
return 1;
var answer = a;
var increment = a;
var i, j;
for (i = 1; i < b; i++)
{
for (j = 1; j < a; j++)
{
answer += increment;
}
increment = answer;
}
return answer;
}
// driver program to test above function
document.write(pow(5, 3));
// This code is contributed by shikhasingrajput
</script>
Producción :
C++
#include<bits/stdc++.h>
using namespace std;
/* A recursive function to get x*y */
int multiply(int x, int y)
{
if(y)
return (x + multiply(x, y - 1));
else
return 0;
}
/* A recursive function to get a^b
Works only if a >= 0 and b >= 0 */
int pow(int a, int b)
{
if(b)
return multiply(a, pow(a, b - 1));
else
return 1;
}
// Driver Code
int main()
{
cout << pow(5, 3);
getchar();
return 0;
}
// This code is contributed
// by Akanksha Rai
C
#include<stdio.h>
/* A recursive function to get a^b
Works only if a >= 0 and b >= 0 */
int pow(int a, int b)
{
if(b)
return multiply(a, pow(a, b-1));
else
return 1;
}
/* A recursive function to get x*y */
int multiply(int x, int y)
{
if(y)
return (x + multiply(x, y-1));
else
return 0;
}
/* driver program to test above functions */
int main()
{
printf("\n %d", pow(5, 3));
getchar();
return 0;
}
Java
import java.io.*;
class GFG {
/* A recursive function to get a^b
Works only if a >= 0 and b >= 0 */
static int pow(int a, int b)
{
if (b > 0)
return multiply(a, pow(a, b - 1));
else
return 1;
}
/* A recursive function to get x*y */
static int multiply(int x, int y)
{
if (y > 0)
return (x + multiply(x, y - 1));
else
return 0;
}
/* driver program to test above functions */
public static void main(String[] args)
{
System.out.println(pow(5, 3));
}
}
// This code is contributed by vt_m.
Python3
def pow(a,b): if(b): return multiply(a, pow(a, b-1)); else: return 1; # A recursive function to get x*y * def multiply(x, y): if (y): return (x + multiply(x, y-1)); else: return 0; # driver program to test above functions * print(pow(5, 3)); # This code is contributed # by Sam007
C#
using System;
class GFG
{
/* A recursive function to get a^b
Works only if a >= 0 and b >= 0 */
static int pow(int a, int b)
{
if (b > 0)
return multiply(a, pow(a, b - 1));
else
return 1;
}
/* A recursive function to get x*y */
static int multiply(int x, int y)
{
if (y > 0)
return (x + multiply(x, y - 1));
else
return 0;
}
/* driver program to test above functions */
public static void Main()
{
Console.Write(pow(5, 3));
}
}
// This code is contributed by Sam007
PHP
<?php
/* A recursive function to get a^b
Works only if a >= 0 and b >= 0 */
function p_ow( $a, $b)
{
if($b)
return multiply($a,
p_ow($a, $b - 1));
else
return 1;
}
/* A recursive function
to get x*y */
function multiply($x, $y)
{
if($y)
return ($x + multiply($x, $y - 1));
else
return 0;
}
// Driver Code
echo pow(5, 3);
// This code is contributed by anuj_67.
?>
Javascript
<script>
// A recursive function to get a^b
// Works only if a >= 0 and b >= 0
function pow(a, b)
{
if (b > 0)
return multiply(a, pow(a, b - 1));
else
return 1;
}
// A recursive function to get x*y
function multiply(x, y)
{
if (y > 0)
return (x + multiply(x, y - 1));
else
return 0;
}
// Driver code
document.write(pow(5, 3));
// This code is contributed by gauravrajput1
</script>
C++
#include <iostream>
using namespace std;
//function calculating power
long long pow(int a, int n){
int ans=1;
while(n>0){
// calculate last bit(right most) bit of n
int last_bit = n&1;
//if last bit is 1 then multiply ans and a
if(last_bit){
ans = ans*a;
}
//make a equal to square of a as on every succeeding bit it got squared like a^0, a^1, a^2, a^4, a^8
a = a*a;
n = n >> 1;
}
return ans;
}
//driver code
int main() {
cout<<pow(3,5);
return 0;
}
C
#include <stdio.h>
// function calculating power
long long pow_(int a, int n){
int ans = 1;
while(n > 0)
{
// calculate last bit(right most) bit of n
int last_bit = n&1;
// if last bit is 1 then multiply ans and a
if(last_bit){
ans = ans*a;
}
//make a equal to square of a as on every succeeding bit it got squared like a^0, a^1, a^2, a^4, a^8
a = a*a;
n = n >> 1;
}
return ans;
}
// driver code
int main()
{
// pow is an inbuilt function so I have used pow_ as a function name
printf("%lld",pow_(3,5));
return 0;
}
// This code is contributed by akashish_.
Java
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG
{
// function calculating power
static int pow(int a, int n){
int ans = 1;
while(n > 0)
{
// calculate last bit(right most) bit of n
int last_bit = n&1;
//if last bit is 1 then multiply ans and a
if(last_bit != 0){
ans = ans*a;
}
//make a equal to square of a as on every succeeding bit it got squared like a^0, a^1, a^2, a^4, a^8
a = a*a;
n = n >> 1;
}
return ans;
}
// Driver Code
public static void main(String[] args)
{
System.out.print(pow(3,5));
}
}
// This code is contributed by code_hunt.
Python3
# function calculating power def pow(a, n): ans = 1 while(n > 0): # calculate last bit(right most) bit of n last_bit = n&1 # if last bit is 1 then multiply ans and a if(last_bit): ans = ans*a # make a equal to square of a as on # every succeeding bit it got squared # like a^0, a^1, a^2, a^4, a^8 a = a*a n = n >> 1 return ans # driver code print(pow(3, 5)) # This code is contributed by shinjanpatra
C#
// C# code to implement the approach
using System;
using System.Numerics;
using System.Collections.Generic;
public class GFG {
// function calculating power
static int pow(int a, int n){
int ans = 1;
while(n > 0)
{
// calculate last bit(right most) bit of n
int last_bit = n&1;
//if last bit is 1 then multiply ans and a
if(last_bit != 0){
ans = ans*a;
}
//make a equal to square of a as on every succeeding bit it got squared like a^0, a^1, a^2, a^4, a^8
a = a*a;
n = n >> 1;
}
return ans;
}
// Driver Code
public static void Main(string[] args)
{
Console.Write(pow(3,5));
}
}
// This code is contributed by sanjoy_62.
Javascript
<script>
// function calculating power
function pow(a, n){
let ans = 1;
while(n > 0)
{
// calculate last bit(right most) bit of n
let last_bit = n&1;
// if last bit is 1 then multiply ans and a
if(last_bit)
{
ans = ans*a;
}
// make a equal to square of a as on
// every succeeding bit it got squared
// like a^0, a^1, a^2, a^4, a^8
a = a*a;
n = n >> 1;
}
return ans;
}
// driver code
document.write(pow(3,5),"</br>");
// This code is contributed by shinjanpatra
</script>
PHP
<?php
// function calculating power
function p_ow($a, $n){
$ans = 1;
while($n > 0)
{
// calculate last bit(right most) bit of n
$last_bit = $n&1;
// if last bit is 1 then multiply ans and a
if($last_bit)
{
$ans = $ans*$a;
}
// make a equal to square of a as on
// every succeeding bit it got squared
// like a^0, a^1, a^2, a^4, a^8
$a = $a*$a;
$n = $n >> 1;
}
return $ans;
}
// driver code
echo(p_ow(5,3));
?>
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA