2 次方乘法
给定两个数字 x 和 n，我们需要将 x 乘以 2 n 示例:
Input : x = 25, n = 3
Output : 200
25 multiplied by 2 raised to power 3
is 200.
Input : x = 70, n = 2
Output : 280
一个简单的解决方案是计算 2 的 n 次幂，然后乘以 x。
C++
// Simple C/C++ program
// to compute x * (2^n)
#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
// Returns 2 raised to power n
ll power2(ll n)
{
if (n == 0)
return 1;
if (n == 1)
return 2;
return power2(n / 2) *
power2(n / 2);
}
ll multiply(ll x, ll n)
{
return x * power2(n);
}
// Driven program
int main()
{
ll x = 70, n = 2;
cout<<multiply(x, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Simple Java program
// to compute x * (2^n)
import java.util.*;
class GFG {
// Returns 2 raised to power n
static long power2(long n)
{
if (n == 0)
return 1;
if (n == 1)
return 2;
return power2(n / 2)
* power2(n / 2);
}
static long multiply(long x, long n)
{
return x * power2(n);
}
/* Driver program */
public static void main(String[] args)
{
long x = 70, n = 2;
System.out.println(multiply(x, n));
}
}
// This code is contributed by Arnav Kr. Mandal.
Python 3
# Simple Python program
# to compute x * (2^n)
# Returns 2 raised to power n
def power2(n):
if (n == 0):
return 1
if (n == 1):
return 2
return power2(n / 2) *
power2(n / 2);
def multiply(x, n):
return x * power2(n);
# Driven program
x = 70
n = 2
print(multiply(x, n))
# This code is contributed by Smitha Dinesh Semwal
C
// Simple C# program
// to compute x * (2^n)
using System;
class GFG {
// Returns 2 raised to power n
static long power2(long n)
{
if (n == 0)
return 1;
if (n == 1)
return 2;
return power2(n / 2)
* power2(n / 2);
}
static long multiply(long x, long n)
{
return x * power2(n);
}
/* Driver program */
public static void Main()
{
long x = 70, n = 2;
Console.WriteLine(multiply(x, n));
}
}
// This code is contributed by Vt_m.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// Simple PHP program
// to compute x * (2^n)
// Returns 2 raised to power n
function power2($n)
{
if ($n == 0)
return 1;
if ($n == 1)
return 2;
return power2($n / 2) *
power2($n / 2);
}
function multiply( $x, $n)
{
return $x * power2($n);
}
// Driver Code
$x = 70; $n = 2;
echo multiply($x, $n);
// This code is contributed by ajit
?>
java 描述语言
<script>
// Simple JavaScript program
// to compute x * (2^n)
// Returns 2 raised to power n
function power2(n)
{
if (n == 0)
return 1;
if (n == 1)
return 2;
return power2(n / 2) *
power2(n / 2);
}
function multiply( x, n)
{
return x * power2(n);
}
// Driver Code
let x = 70
let n = 2;
document.write( multiply(x, n));
// This code is contributed by mohan
</script>
输出:
280
时间复杂度: O(Log n) 一个高效的解决方案是使用按位左移位运算符。我们知道 1<T9【n】表示 2 的幂 n
C++
// Efficient C/C++ program to compute x * (2^n)
#include <stdio.h>
typedef long long int ll;
ll multiply(ll x, ll n)
{
return x << n;
}
// Driven program to check above function
int main()
{
ll x = 70, n = 2;
printf("%lld", multiply(x, n));
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// JAVA Code for Multiplication with a
// power of 2
import java.util.*;
class GFG {
static long multiply(long x, long n)
{
return x << n;
}
/* Driver program to test above function */
public static void main(String[] args)
{
long x = 70, n = 2;
System.out.println(multiply(x, n));
}
}
//This code is contributed by Arnav Kr. Mandal.
Python 3
# Efficient Python3 code to compute x * (2^n)
def multiply( x , n ):
return x << n
# Driven code to check above function
x = 70
n = 2
print( multiply(x, n))
# This code is contributed by "Sharad_Bhardwaj".
C
// C# Code for Multiplication with a
// power of 2
using System;
class GFG {
static int multiply(int x, int n)
{
return x << n;
}
/* Driver program to test above function */
public static void Main()
{
int x = 70, n = 2;
Console.WriteLine(multiply(x, n));
}
}
//This code is contributed by vt_m.
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// Efficient PHP program to compute x * (2^n)
function multiply($x, $n)
{
return $x << $n;
}
// Driver Code
$x = 70;
$n = 2;
echo multiply($x, $n);
// This code is contributed by ajit
?>
java 描述语言
<script>
// Efficient JavaScript program to compute x * (2^n)
function multiply(x, n)
{
return x << n;
}
// Driver Code
let x = 70;
let n = 2;
document.write(multiply(x, n));
// This code is contributed by mohan
</script>
输出:
280
时间复杂度:O(1)
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