检查数字是否可以由两个平方之和表示

我们有一个数字n。我们需要确定数字n是否可以由两个平方之和表示。

例子 ：

Input  : n = 17
Output : Yes
4^2 + 1^2 = 17

Input  : n = 169
Output : Yes
5^2 + 12^2 = 169

Input : n = 24
Output : No

蛮力法– O(n)
我们使用两个for循环运行直到n的平方根，然后每次检查两个循环数的平方和是否等于N.如果找到该组合，则将输出Yes，否则输出No.

for i=1 to sqrt(n)
   for j=i to sqrt(n)
       if (i*i+j*j == n)
           return true;
return false;

C++

// A brute force approach based implementation
// to find if a number can be written as sum
// of two squares.
#include 
using namespace std;
 
// function to check if there exist two
// numbers sum of whose squares is n.
bool sumSquare(int n)
{
    for (long i = 1; i * i <= n; i++)
        for (long j = 1; j * j <= n; j++)
            if (i * i + j * j == n) {
                cout << i << "^2 + "
                     << j << "^2" << endl;
                return true;
            }
    return false;
}
 
// driver Program
int main()
{
    int n = 25;
    if (sumSquare(n))
        cout << "Yes";
    else
        cout << "No";
}

Java

// A brute force approach based implementation
// to find if a number can be written as sum
// of two squares.
 
class GFG {
     
    // function to check if there exist two
    // numbers sum of whose squares is n.
    static boolean sumSquare(int n)
    {
        for (long i = 1; i * i <= n; i++)
            for (long j = 1; j * j <= n; j++)
                if (i * i + j * j == n) {
                    System.out.println(i + "^2 + "
                                       + j + "^2");
                    return true;
     }
 
        return false;
    }
      
    // driver Program
    public static void main(String args[])
    {
        int n = 25;
        if (sumSquare(n))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
 
 
/*This code is contributed by Nikita Tiwari.*/

Python3

# A brute force approach based
# implementation to find if a number
# can be written as sum of two squares.
 
# function to check if there exist two
# numbers sum of whose squares is n.
def sumSquare( n) :
    i = 1
    while i * i <= n :
        j = 1
        while(j * j <= n) :
            if (i * i + j * j == n) :
                print(i, "^2 + ", j , "^2" )
                return True
            j = j + 1
        i = i + 1
         
    return False
  
# driver Program
n = 25
if (sumSquare(n)) :
    print("Yes")
else :
    print( "No")
     
 
# This code is contributed by Nikita Tiwari.

C#

// A brute force approach based
// implementation to find if a
// number can be written as sum
// of two squares
using System;
 
class GFG {
     
    // function to check if there exist two
    // numbers sum of whose squares is n
    static bool sumSquare(int n)
    {
        for (long i = 1; i * i <= n; i++)
            for (long j = 1; j * j <= n; j++)
                if (i * i + j * j == n)
                {
                    Console.Write(i + "^2 + "
                                    + j + "^2");
                    return true;
                  }
        return false;
    }
     
    // Driver Code
    public static void Main(String []args)
    {
        int n = 25;
        if (sumSquare(n))
            Console.Write("\nYes");
        else
            Console.Write("\nNo");
    }
}
 
// This code is contributed by Smitha Dinesh Semwal.

Javascript

// A brute force approach based implementation
// to find if a number can be written as sum
// of two squares.
    // function to check if there exist two
    // numbers sum of whose squares is n.
    function sumSquare(n) {
        for (i = 1; i * i <= n; i++)
            for (j = 1; j * j <= n; j++)
                if (i * i + j * j == n) {
                    document.write(i + "^2 + " + j + "^2"+"<br/>");
                    return true;
                }
 
        return false;
    }
 
    // driver Program
        var n = 25;
        if (sumSquare(n))
            document.write("Yes");
        else
            document.write("No");
 
// This code is contributed by gauravrajput1

输出：

3^2 + 4^2
Yes

我们也可以在O(sqrt(n))中解决这个问题
为了解决sqrt(n)复杂性的问题，我们使用了一个哈希表，其中将存储直到sqrt(n)的数字平方，并且每次我们在哈希表中搜索(n – sqrt(i))(如果存在)，然后返回是的，其他返回否。

unordered_map hashmap;
for i = 1 to sqrt(n)
      hashmap[i*i]=1;
      if (hashmap.find(N-i*i) != hashmap.end())
         return true;
return false;
 

C++

// An efficient approach based implementation
// to find if a number can be written as sum
// of two squares.
#include 
using namespace std;
 
// function to check if there exist two
// numbers sum of whose squares is n.
bool sumSquare(int n)
{
    unordered_map s;
    for (int i = 0; i * i <= n; ++i) {
 
        // store square value in hashmap
        s[i * i] = 1;
        if (s.find(n - i * i) != s.end()) {
            cout << sqrt(n - i * i) << "^2 + "
                 << i << "^2" << endl;
            return true;
        }
    }
    return false;
}
 
// driver Program
int main()
{
    int n = 169;
    if (sumSquare(n))
        cout << "Yes";
    else
        cout << "No";
}

Java

// An efficient approach based implementation
// to find if a number can be written as sum
// of two squares.
import java.util.*;
class GFG
{
 
// function to check if there exist two
// numbers sum of whose squares is n.
static boolean sumSquare(int n)
{
    HashMap s = new HashMap();
    for (int i = 0; i * i <= n; ++i)
    {
 
        // store square value in hashmap
        s.put(i * i, 1);
        if (s.containsKey(n - i * i))
        {
            System.out.println((int)Math.sqrt(n - i * i) +
                                      "^2 + " + i + "^2");
            return true;
        }
    }
    return false;
}
 
// Driver Code
public static void main(String[] args)
{
    int n = 169;
    System.out.print(sumSquare(n) ?
                          "YES\n" : "NO\n");
}
}
 
// This code is contributed by Princi Singh

Python3

# An efficient approach based implementation
# to find if a number can be written as sum
# of two squares.
 
# function to check if there exist two
# numbers sum of whose squares is n.
def sumSquare(n):
 
    s = dict()
    for i in range(n):
 
        if i * i > n:
            break
 
        # store square value in hashmap
        s[i * i] = 1
 
        if (n - i * i) in s.keys():
            print((n - i * i)**(1 / 2),
                       "^2 +", i, "^2")
            return True
         
    return False
 
# Driver Code
n = 169
if (sumSquare(n)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by Mohit Kumar

C＃

// An efficient approach based implementation
// to find if a number can be written as sum
// of two squares.
using System;
using System.Collections.Generic;
class GFG
{
 
// function to check if there exist two
// numbers sum of whose squares is n.
static bool sumSquare(int n)
{
    Dictionary s = new Dictionary();
    for (int i = 0; i * i <= n; ++i)
    {
 
        // store square value in hashmap
        s.Add(i * i, 1);
        if (s.ContainsKey(n - i * i))
        {
            Console.WriteLine((int)Math.Sqrt(n - i * i) +
                                     "^2 + " + i + "^2");
            return true;
        }
    }
    return false;
}
 
// Driver Code
public static void Main(String[] args)
{
    int n = 169;
    Console.WriteLine(sumSquare(n) ?
                             "YES" : "NO");
}
}
 
// This code is contributed by Princi Singh

输出 ：

5^2 + 12^2
Yes

我们也可以在O(sqrt(n)log n)中解决这个问题
该方法由Sagar Shukla贡献。

二进制搜索方法：
另一种检查方法$c - a^2$是一个完美的正方形，是通过使用Binary Search实现的。该方法与典型的二进制搜索来查找数字的方法相同。唯一的区别在于，我们需要找到一个介于该范围中间的整数$[0, c - a^2]$这样这个数字就是…的平方根$c - a^2$换句话说，我们需要在范围内找到一个整数mid $[0, c - a^2]$ ，这样midxmid = $c - a^2$

下面是上述方法的实现：

C++

// C++ program for Check whether a number can be
// represented by sum of two squares using binary search.
#include
#include
using namespace std;
 
// Function for binary search.
bool binary_search(int num2, int se, int num)
{
    int mid;
    int ss=0;
 
    while(ss<=se)
    {
        // Calculating mid.
        mid=(ss+se)/2;
        if ((pow(mid,2)+pow(num2,2))==num)
        {
            return true;
        }
        else if ((pow(mid,2)+pow(num2,2))>num)
        {
            se=mid-1;
        }
        else
        {
            ss=mid+1;
        }
    }
    return false;
}
 
// Driver code
int main()
{
    int rt;
    int num=169;
    rt=sqrt(num);
    bool flag=false;
    for (int i = rt; i >=0; --i)
    {
        if (binary_search(i,i,num))
        {
            flag=true;
            break;
        }
    }
    if (flag)
    {
        cout<<"true";
    }
    else cout<<"false";
    return 0;
}
// This code is contributed by Dhruv Gupta

Java

// Java program for Check whether a number can be
// represented by sum of two squares using binary search.
import java.util.*;
import java.lang.*;
 
public class GfG {
    public static boolean judgeSquareSum(int c)
    {
 
        // Iterating loop from 0 to c - a * a.
        for (long a = 0; a * a <= c; a++) {
            int b = c - (int)(a * a);
 
            // If b is a square root of c - a * a
            // then return true.
            if (binary_search(0, b, b))
                return true;
        }
        return false;
    }
 
    // Function for binary search.
    public static boolean binary_search(long s, long e, int n)
    {
 
        // If lower limit exceeds upper limit.
        if (s > e)
            return false;
 
        // Calculating mid.
        long mid = s + (e - s) / 2;
        if (mid * mid == n)
            return true;
        if (mid * mid > n)
            return binary_search(s, mid - 1, n);
        return binary_search(mid + 1, e, n);
    }
 
    // Driver function
    public static void main(String argc[])
    {
        int c = 17;
        System.out.println(judgeSquareSum(c));
    }
}

Python3

# Python3 program for Check whether 
# a number can be represented by
# sum of two squares using binary search.
   
def judgeSquareSum(c):
   
    # Iterating loop from 0 to c - a * a.
    a = 0;
    while(a * a <= c):
        b = c - int(a * a);
   
        # If b is a square root of 
        # c - a * a then return true.
        if (binary_search(0, b, b)):
            return 1;
        a+=1;
    return 0;
   
# Function for binary search.
def binary_search(s, e, n):
    # If lower limit exceeds 
    # upper limit.
    if (s > e):
        return 0;
     
    # Calculating mid.
    mid = s + int((e - s) / 2);
    if (int(mid * mid) == n):
        return 1;
     
    if (int(mid * mid) > n):
        return binary_search(s, mid - 1, n);
     
    return binary_search(mid + 1, e, n);
   
# Driver Code
c = 17;
if(judgeSquareSum(c)):
    print("true");
else:
    print("false");
   
# This code is contributed by mits

C＃

// C# program for Check whether a
// number can be represented by
// sum of two squares using
// binary search.
using System;
class GFG {
 
public static bool judgeSquareSum(int c)
    {
        // Iterating loop from 0 to c - a * a.
        for (long a = 0; a * a <= c; a++)
        {
            int b = c - (int)(a * a);
 
            // If b is a square root of c - a * a
            // then return true.
            if (binary_search(0, b, b))
                return true;
        }
        return false;
    }
 
    // Function for binary search.
    public static bool binary_search(long s,
                              long e, int n)
    {
 
        // If lower limit exceeds upper limit.
        if (s > e)
            return false;
 
        // Calculating mid.
        long mid = s + (e - s) / 2;
         
        if (mid * mid == n)
            return true;
             
        if (mid * mid > n)
            return binary_search(s, mid - 1, n);
             
        return binary_search(mid + 1, e, n);
    }
     
    // Driver Code
    public static void Main()
    {
        int c = 17;
        Console.WriteLine(judgeSquareSum(c));
    }
}
 
// This code is contributed by Sam007

的PHP

 $e)
            return 0;
 
        // Calculating mid.
        $mid = $s + intval(($e - $s) / 2);
        if (intval($mid * $mid) == $n)
            return 1;
             
        if (intval($mid * $mid) > $n)
            return binary_search($s, $mid - 1, $n);
             
        return binary_search($mid + 1, $e, $n);
    }
 
// Driver Code
$c = 17;
if(judgeSquareSum($c))
    echo "true";
else
    echo "false";
 
// This code is contributed by Sam007
?>

Java脚本

 
// javascript program for Check whether a number can be
// represented by sum of two squares using binary search.
 
    function judgeSquareSum(c) {
 
        // Iterating loop from 0 to c - a * a.
        for (a = 0; a * a <= c; a++) {
            var b = c - parseInt( (a * a));
 
            // If b is a square root of c - a * a
            // then return true.
            if (binary_search(0, b, b))
                return true;
        }
        return false;
    }
 
    // Function for binary search.
    function binary_search(s , e , n) {
 
        // If lower limit exceeds upper limit.
        if (s > e)
            return false;
 
        // Calculating mid.
        var mid = s + parseInt((e - s) / 2);
        if (mid * mid == n)
            return true;
        if (mid * mid > n)
            return binary_search(s, mid - 1, n);
        return binary_search(mid + 1, e, n);
    }
 
    // Driver function
        var c = 17;
        document.write(judgeSquareSum(c));
 
// This code is contributed by Rajput-Ji

输出 ：

true

时间复杂度：O(sqrt(c)log(c))
该方法由Sagar Shukla贡献。

费马定理方法：

该方法基于以下陈述，该陈述基于费马定理：
“当且仅当n的素数分解时，形式(4k + 3)的每个素数出现偶数次，任何正数n才能表示为两个平方的和。”
利用上述定理，我们可以直接找出给定数n是否可以表示为两个平方的和。
为此，我们只需找到给定数n的所有素数，范围可以从$[2,\sqrt{n}]$连同这些因素的计数，通过反复划分。如果在任何步骤中我们发现形式(4k + 3)的任何素数出现的次数都是奇数次，则可以返回False值。
如果n本身是质数，则不会被n中的任何质数整除。 $[2,\sqrt{n}]$ 。因此，我们需要检查n是否可以以(4k + 3)的形式表示。如果是这样，我们需要返回一个False值，指示此素数出现奇数(1)次。
否则，我们可以返回True值。

C++

// Check whether a number can be represented
// by sum of two squares using Fermat Theorem.
#include
using namespace std;
 
bool judgeSquareSum(int n)
{
for (int i = 2;
        i * i <= n; i++)
{
    int count = 0;
    if (n % i == 0)
    {
        // Count all the prime factors.
        while (n % i == 0)
        {
            count++;
            n /= i;
        }
 
        // Ifany prime factor of the form
        // (4k+3)(4k+3) occurs an odd
        // number of times.
        if (i % 4 == 3 && count % 2 != 0)
            return false;
    }
}
 
// If n itself is a x prime number and
// can be expressed in the form of
// 4k + 3 we return false.
return n % 4 != 3;
}
 
// Driver Code
int main()
{
    int n = 17;
    if(judgeSquareSum(n))
        cout << "Yes";
    else
        cout << "No";
}
         
// This code is contributed by
// prabhat kumar singh

Java

// Java program to Check whether a number 
// can be represented by sum of two
// squares using Fermat Theorem.
import java.util.*;
import java.lang.*;
 
class GFG
{
public static boolean judgeSquareSum(int n)
{
    for (int i = 2; i * i <= n; i++)
    {
        int count = 0;
        if (n % i == 0)
        {
            // Count all the prime factors.
            while (n % i == 0)
            {
                count++;
                n /= i;
            }
             
            // Ifany prime factor of the form
            // (4k+3)(4k+3) occurs an odd
            // number of times.
            if (i % 4 == 3 && count % 2 != 0)
                return false;
        }
    }
     
    // If n itself is a prime number and can
    // be expressed in the form of 4k + 3
    // we return false.
    return n % 4 != 3;
}
 
// Driver Code
public static void main(String argc[])
{
    int n = 17;
    if(judgeSquareSum(n))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}

Python3

# Check whether a number can be represented
# by sum of two squares using Fermat Theorem.
def judgeSquareSum(n):
 
    i = 2;
    while (i * i <= n):
        count = 0;
        if (n % i == 0):
             
            # Count all the prime factors.
            while (n % i == 0):
                count += 1;
                n = int(n / i);
     
            # Ifany prime factor of the
            # form (4k+3)(4k+3) occurs
            # an odd number of times.
            if (i % 4 == 3 and count % 2 != 0):
                return False;
        i += 1;
     
    # If n itself is a x prime number and
    # can be expressed in the form of 4k + 3
    # we return false.
    return n % 4 != 3;
 
# Driver Code
n = 17;
if(judgeSquareSum(n)):
    print("Yes");
else:
    print("No");
     
# This code is contributed by mits

C＃

// C# program to Check whether a number
// can be represented by sum of two
// squares using Fermat Theorem.
using System;
 
class GFG
{
public static bool judgeSquareSum(int n)
{
    for (int i = 2; i * i <= n; i++)
    {
        int count = 0;
        if (n % i == 0)
        {
            // Count all the prime factors.
            while (n % i == 0)
            {
                count++;
                n /= i;
            }
             
            // If any prime factor of the
            // form (4k+3)(4k+3) occurs an
            // odd number of times.
            if (i % 4 == 3 && count % 2 != 0)
                return false;
        }
    }
     
    // If n itself is a prime number and
    // can be expressed in the form of
    // 4k + 3 we return false.
    return n % 4 != 3;
}
 
// Driver Code
static public void Main ()
{
    int n = 17;
    if(judgeSquareSum(n))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code is contributed
// by akt_mit

Java脚本

 
// Javascript program to Check whether a number 
// can be represented by sum of two
// squares using Fermat Theorem.
function judgeSquareSum(n)
{
    for(i = 2; i * i <= n; i++)
    {
        var count = 0;
         
        if (n % i == 0)
        {
             
            // Count all the prime factors.
            while (n % i == 0)
            {
                count++;
                n = parseInt(n / i);
            }
             
            // If any prime factor of the form
            // (4k+3)(4k+3) occurs an odd
            // number of times.
            if (i % 4 == 3 && count % 2 != 0)
                return false;
        }
    }
     
    // If n itself is a prime number and can
    // be expressed in the form of 4k + 3
    // we return false.
    return n % 4 != 3;
}
 
// Driver Code
var n = 17;
if (judgeSquareSum(n))
    document.write("Yes");
else
    document.write("No");
 
// This code is contributed by Rajput-Ji
 

输出 ：

Yes