在不使用*，/和pow()的情况下计算数字的平方

给定整数n，无需使用*，/和pow()即可计算数字的平方。

例子 ：

Input: n = 5
Output: 25

Input: 7
Output: 49

Input: n = 12
Output: 144

一个简单的解决方案是重复将n添加到结果中。

以下是此想法的实现。

C++

// Simple solution to calculate square without
// using * and pow()
#include 
using namespace std;
 
int square(int n)
{
    // handle negative input
    if (n < 0)
        n = -n;
 
    // Initialize result
    int res = n;
 
    // Add n to res n-1 times
    for (int i = 1; i < n; i++)
        res += n;
 
    return res;
}
 
// Driver code
int main()
{
    for (int n = 1; n <= 5; n++)
        cout << "n = " << n << ", n^2 = " << square(n)
             << endl;
    return 0;
}

Java

// Java Simple solution to calculate
// square without using * and pow()
import java.io.*;
 
class GFG {
 
    public static int square(int n)
    {
 
        // handle negative input
        if (n < 0)
            n = -n;
 
        // Initialize result
        int res = n;
 
        // Add n to res n-1 times
        for (int i = 1; i < n; i++)
            res += n;
 
        return res;
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        for (int n = 1; n <= 5; n++)
            System.out.println("n = " + n
                               + ", n^2 = " + square(n));
    }
}
 
// This code is contributed by sunnysingh

Python3

# Simple solution to
# calculate square without
# using * and pow()
 
def square(n):
 
    # handle negative input
    if (n < 0):
        n = -n
 
    # Initialize result
    res = n
 
    # Add n to res n-1 times
    for i in range(1, n):
        res += n
 
    return res
 
 
# Driver Code
for n in range(1, 6):
    print("n =", n, end=", ")
    print("n^2 =", square(n))
 
# This code is contributed by
# Smitha Dinesh Semwal

C#

// C# Simple solution to calculate
// square without using * and pow()
using System;
 
class GFG {
    public static int square(int n)
    {
 
        // handle negative input
        if (n < 0)
            n = -n;
 
        // Initialize result
        int res = n;
 
        // Add n to res n-1 times
        for (int i = 1; i < n; i++)
            res += n;
 
        return res;
    }
 
    // Driver code
    public static void Main()
    {
 
        for (int n = 1; n <= 5; n++)
            Console.WriteLine("n = " + n
                              + ", n^2 = " + square(n));
    }
}
 
// This code is contributed by Sam007

Javascript

 
// Simple solution to calculate square without
// using * and pow()
 
function square(n)
{
    // handle negative input
    if (n < 0)
        n = -n;
 
    // Initialize result
    let res = n;
 
    // Add n to res n-1 times
    for (let i = 1; i < n; i++)
        res += n;
 
    return res;
}
 
// Driver code
 
    for (let n = 1; n <= 5; n++)
        document.write("n= " + n +", n^2 = " + square(n)
            + "<br>");
     
 
//This code is contributed by Mayank Tyagi
 

输出

n = 1, n^2 = 1
n = 2, n^2 = 4
n = 3, n^2 = 9
n = 4, n^2 = 16
n = 5, n^2 = 25

上述解决方案的时间复杂度为O(n)。

方法二：

我们可以使用按位运算符在O(Logn)时间内完成此运算符。这个想法是基于以下事实。

square(n) = 0 if n == 0
  if n is even 
     square(n) = 4*square(n/2) 
  if n is odd
     square(n) = 4*square(floor(n/2)) + 4*floor(n/2) + 1 

Examples
  square(6) = 4*square(3)
  square(3) = 4*(square(1)) + 4*1 + 1 = 9
  square(7) = 4*square(3) + 4*3 + 1 = 4*9 + 4*3 + 1 = 49

这是如何运作的?

If n is even, it can be written as
  n = 2*x 
  n2 = (2*x)2 = 4*x2
If n is odd, it can be written as 
  n = 2*x + 1
  n2 = (2*x + 1)2 = 4*x2 + 4*x + 1

floor(n / 2)可以使用按位右移运算符来计算。可以计算2 * x和4 * x

下面是基于以上思想的实现。

C++

// Square of a number using bitwise operators
#include 
using namespace std;
 
int square(int n)
{
    // Base case
    if (n == 0)
        return 0;
 
    // Handle negative number
    if (n < 0)
        n = -n;
 
    // Get floor(n/2) using right shift
    int x = n >> 1;
 
    // If n is odd
    if (n & 1)
        return ((square(x) << 2) + (x << 2) + 1);
    else // If n is even
        return (square(x) << 2);
}
 
// Driver Code
int main()
{
    // Function calls
    for (int n = 1; n <= 5; n++)
        cout << "n = " << n << ", n^2 = " << square(n)
             << endl;
    return 0;
}

Java

// Square of a number using
// bitwise operators
class GFG {
    static int square(int n)
    {
 
        // Base case
        if (n == 0)
            return 0;
 
        // Handle negative number
        if (n < 0)
            n = -n;
 
        // Get floor(n/2) using
        // right shift
        int x = n >> 1;
 
        // If n is odd
        ;
        if (n % 2 != 0)
            return ((square(x) << 2) + (x << 2) + 1);
        else // If n is even
            return (square(x) << 2);
    }
 
    // Driver code
    public static void main(String args[])
    {
        // Function calls
        for (int n = 1; n <= 5; n++)
            System.out.println("n = " + n
                               + " n^2 = " + square(n));
    }
}
 
// This code is contributed by Sam007

Python3

# Square of a number using bitwise
# operators
 
 
def square(n):
 
    # Base case
    if (n == 0):
        return 0
 
    # Handle negative number
    if (n < 0):
        n = -n
 
    # Get floor(n/2) using
    # right shift
    x = n >> 1
 
    # If n is odd
    if (n & 1):
        return ((square(x) << 2)
                + (x << 2) + 1)
 
    # If n is even
    else:
        return (square(x) << 2)
 
 
# Driver Code
for n in range(1, 6):
    print("n = ", n, " n^2 = ",
          square(n))
# This code is contributed by Sam007

C＃

// Square of a number using bitwise
// operators
using System;
 
class GFG {
 
    static int square(int n)
    {
 
        // Base case
        if (n == 0)
            return 0;
 
        // Handle negative number
        if (n < 0)
            n = -n;
 
        // Get floor(n/2) using
        // right shift
        int x = n >> 1;
 
        // If n is odd
        ;
        if (n % 2 != 0)
            return ((square(x) << 2) + (x << 2) + 1);
        else // If n is even
            return (square(x) << 2);
    }
 
    // Driver code
    static void Main()
    {
        for (int n = 1; n <= 5; n++)
            Console.WriteLine("n = " + n
                              + " n^2 = " + square(n));
    }
}
 
// This code is contributed by Sam0007.

的PHP

> 1;
 
    // If n is odd
    if ($n & 1)
        return ((square($x) << 2) +
                    ($x << 2) + 1);
    else // If n is even
        return (square($x) << 2);
}
 
    // Driver Code
    for ($n = 1; $n <= 5; $n++)
        echo "n = ", $n, ", n^2 = ", square($n),"\n";
     
// This code is contributed by ajit
?>

Java脚本

 
// Square of a number using bitwise operators
 
function square(n)
{
    // Base case
    if (n == 0)
        return 0;
 
    // Handle negative number
    if (n < 0)
        n = -n;
 
    // Get floor(n/2) using right shift
    let x = n >> 1;
 
    // If n is odd
    if (n & 1)
        return ((square(x) << 2) + (x << 2) + 1);
    else // If n is even
        return (square(x) << 2);
}
 
// Driver Code
  // Function calls
    for (let n = 1; n <= 5; n++)
        document.write("n = " + n + ", n^2 = " + square(n)
             +"<br>");
    
 
//This code is contributed by Mayank Tyagi
 

输出

n = 1, n^2 = 1
n = 2, n^2 = 4
n = 3, n^2 = 9
n = 4, n^2 = 16
n = 5, n^2 = 25

上述解决方案的时间复杂度为O(Logn)。

方法3：

For a given number `num` we get square of it by multiplying number as `num * num`. 
Now write one of `num` in square `num * num` in terms of power of `2`. Check below examples.

Eg: num = 10, square(num) = 10 * 10 
                          = 10 * (8 + 2) = (10 * 8) + (10 * 2)
    num = 15, square(num) = 15 * 15 
                          = 15 * (8 + 4 + 2 + 1) = (15 * 8) + (15 * 4) + (15 * 2) + (15 * 1)

Multiplication with power of 2's can be done by left shift bitwise operator.

下面是基于以上思想的实现。

C++

// Simple solution to calculate square without
// using * and pow()
#include 
using namespace std;
 
int square(int num)
{
    // handle negative input
    if (num < 0) num = -num;
 
    // Initialize result
    int result = 0, times = num;
 
    while (times > 0)
    {
        int possibleShifts = 0, currTimes = 1;
 
        while ((currTimes << 1) <= times)
        {
            currTimes = currTimes << 1;
            ++possibleShifts;
        }
 
        result = result + (num << possibleShifts);
        times = times - currTimes;
    }
 
    return result;
}
 
// Driver code
int main()
{
    // Function calls
    for (int n = 10; n <= 15; ++n)
        cout << "n = " << n << ", n^2 = " << square(n) << endl;
    return 0;
}
 
// This code is contributed by sanjay235

Java

// Simple solution to calculate square
// without using * and pow()
import java.io.*;
 
class GFG{
     
public static int square(int num)
{
     
    // Handle negative input
    if (num < 0)
        num = -num;
 
    // Initialize result
    int result = 0, times = num;
 
    while (times > 0)
    {
        int possibleShifts = 0,
                 currTimes = 1;
 
        while ((currTimes << 1) <= times)
        {
            currTimes = currTimes << 1;
            ++possibleShifts;
        }
 
        result = result + (num << possibleShifts);
        times = times - currTimes;
    }
    return result;
}
 
// Driver code
public static void main(String[] args)
{
    for(int n = 10; n <= 15; ++n)
    {
        System.out.println("n = " + n +
                           ", n^2 = " +
                           square(n));
    }
}
}
 
// This code is contributed by RohitOberoi

Python3

# Simple solution to calculate square without
# using * and pow()
def square(num):
 
    # Handle negative input
    if (num < 0):
        num = -num
 
    # Initialize result
    result, times = 0, num
 
    while (times > 0):
        possibleShifts, currTimes = 0, 1
 
        while ((currTimes << 1) <= times):
            currTimes = currTimes << 1
            possibleShifts += 1
 
        result = result + (num << possibleShifts)
        times = times - currTimes
 
    return result
 
# Driver Code
 
# Function calls
for n in range(10, 16):
    print("n =", n, ", n^2 =", square(n))
 
# This code is contributed by divyesh072019

C＃

// Simple solution to calculate square
// without using * and pow()
using System;
class GFG {
     
    static int square(int num)
    {
          
        // Handle negative input
        if (num < 0)
            num = -num;
      
        // Initialize result
        int result = 0, times = num;
      
        while (times > 0)
        {
            int possibleShifts = 0,
                     currTimes = 1;
      
            while ((currTimes << 1) <= times)
            {
                currTimes = currTimes << 1;
                ++possibleShifts;
            }
      
            result = result + (num << possibleShifts);
            times = times - currTimes;
        }
        return result;
    }
     
  static void Main() {
        for(int n = 10; n <= 15; ++n)
        {
            Console.WriteLine("n = " + n +
                               ", n^2 = " +
                               square(n));
        }
  }
}
 
// This code is contributed by divyeshrabadiy07

Java脚本

 
// Simple solution to calculate square without
// using * and pow()
 
function square(num)
{
    // handle negative input
    if (num < 0) num = -num;
 
    // Initialize result
    let result = 0, times = num;
 
    while (times > 0)
    {
        let possibleShifts = 0, currTimes = 1;
 
        while ((currTimes << 1) <= times)
        {
            currTimes = currTimes << 1;
            ++possibleShifts;
        }
 
        result = result + (num << possibleShifts);
        times = times - currTimes;
    }
 
    return result;
}
 
// Driver code
 
    // Function calls
    for (let n = 10; n <= 15; ++n)
        document.write("n = " + n + ", n^2 = " + square(n) + "<br>");
 
//This code is contributed by Mayank Tyagi
 

输出

n = 10, n^2 = 100
n = 11, n^2 = 121
n = 12, n^2 = 144
n = 13, n^2 = 169
n = 14, n^2 = 196
n = 15, n^2 = 225

上述解决方案的时间复杂度为O(Log n)。感谢Sanjay提供的方法3解决方案。