计算平方和为N(a ^ 2 + b ^ 2 = N)的对(a，b)

给定数字N，任务是对满足条件a ^ 2 + b ^ 2 = N的所有’a’和’b’进行计数。
注意：-( a，b)和(b，a)被认为是两个不同的对，并且(a，a)也有效并且仅被考虑一次。
例子：

Input: N = 10
Output:  2
1^2 + 3^2 = 9
3^2 + 1^2 = 9

Input: N = 8
Output: 1
2^2 + 2^2 = 8

方法：

1. 从1遍历到N的平方根。
   • 从N中减去当前数字的平方，并检查它们的差是否是一个完美的平方。
   • 如果它是完美的平方，则增加计数。
2. 返回计数。

下面是上述方法的实现：

C++

// C++ program to count pairs whose sum
// of squares is N
#include 
using namespace std;
 
// Function to count the pairs satisfying
// a ^ 2 + b ^ 2 = N
int countPairs(int N)
{
    int count = 0;
 
    // Check for each number 1 to sqrt(N)
    for (int i = 1; i <= sqrt(N); i++) {
 
        // Store square of a number
        int sq = i * i;
 
        // Subtract the square from given N
        int diff = N - sq;
 
        // Check if the difference is also
        // a perfect square
        int sqrtDiff = sqrt(diff);
 
        // If yes, then increment count
        if (sqrtDiff * sqrtDiff == diff)
            count++;
    }
 
    return count;
}
 
// Driver code
int main()
{
    // Loop to Count no. of pairs satisfying
    // a ^ 2 + b ^ 2 = i for N = 1 to 10
    for (int i = 1; i <= 10; i++)
        cout << "For n = " << i << ", "
             << countPairs(i) << " pair exists\n";
 
    return 0;
}

Java

// Java program to count pairs whose sum
// of squares is N
 
import java.io.*;
 
class GFG {
 
 
 
// Function to count the pairs satisfying
// a ^ 2 + b ^ 2 = N
static int countPairs(int N)
{
    int count = 0;
 
    // Check for each number 1 to sqrt(N)
    for (int i = 1; i <= (int)Math.sqrt(N); i++)
    {
 
        // Store square of a number
        int sq = i * i;
 
        // Subtract the square from given N
        int diff = N - sq;
 
        // Check if the difference is also
        // a perfect square
        int sqrtDiff = (int)Math.sqrt(diff);
 
        // If yes, then increment count
        if (sqrtDiff * sqrtDiff == diff)
            count++;
    }
 
    return count;
}
 
    // Driver code
    public static void main (String[] args)
    {
    // Loop to Count no. of pairs satisfying
    // a ^ 2 + b ^ 2 = i for N = 1 to 10
    for (int i = 1; i <= 10; i++)
        System.out.println( "For n = " + i + ", "
            + countPairs(i) + " pair exists\n");
    }
}
// This code is contributed by inder_verma.

Python 3

# Python 3 program to count pairs whose sum
# of squares is N
 
# From math import everything
from math import *
 
# Function to count the pairs satisfying
# a ^ 2 + b ^ 2 = N
def countPairs(N) :
    count = 0
 
    # Check for each number 1 to sqrt(N)
    for i in range(1, int(sqrt(N)) + 1) :
 
        # Store square of a number
        sq = i * i
 
        # Subtract the square from given N
        diff = N - sq
 
        #  Check if the difference is also
        # a perfect square
        sqrtDiff = int(sqrt(diff))
 
        # If yes, then increment count
        if sqrtDiff * sqrtDiff == diff :
            count += 1
 
    return count
 
# Driver code    
if __name__ == "__main__" :
 
    # Loop to Count no. of pairs satisfying
    # a ^ 2 + b ^ 2 = i for N = 1 to 10
    for i in range(1,11) :
        print("For n =",i,", ",countPairs(i),"pair exists")
 
  
# This code is contributed by ANKITRAI1

C#

// C# program to count pairs whose sum
// of squares is N
  
 
using System;
class GFG {
  
  
  
// Function to count the pairs satisfying
// a ^ 2 + b ^ 2 = N
static int countPairs(int N)
{
    int count = 0;
  
    // Check for each number 1 to Sqrt(N)
    for (int i = 1; i <= (int)Math.Sqrt(N); i++)
    {
  
        // Store square of a number
        int sq = i * i;
  
        // Subtract the square from given N
        int diff = N - sq;
  
        // Check if the difference is also
        // a perfect square
        int sqrtDiff = (int)Math.Sqrt(diff);
  
        // If yes, then increment count
        if (sqrtDiff * sqrtDiff == diff)
            count++;
    }
  
    return count;
}
  
    // Driver code
    public static void Main ()
    {
    // Loop to Count no. of pairs satisfying
    // a ^ 2 + b ^ 2 = i for N = 1 to 10
    for (int i = 1; i <= 10; i++)
        Console.Write( "For n = " + i + ", "
            + countPairs(i) + " pair exists\n");
    }
}

Javascript

// Javascript program to count pairs whose sum
// of squares is N
 
// Function to count the pairs satisfying
// a ^ 2 + b ^ 2 = N
function countPairs(N)
{
    let count = 0;
 
    // Check for each number 1 to sqrt(N)
    for (let i = 1; i <= Math.sqrt(N); i++) {
 
        // Store square of a number
        let sq = i * i;
 
        // Subtract the square from given N
        let diff = N - sq;
 
        // Check if the difference is also
        // a perfect square
        let sqrtDiff = Math.sqrt(diff);
 
        // If yes, then increment count
        if (sqrtDiff * sqrtDiff == diff)
            count++;
    }
 
    return count;
}
 
// Driver code
    // Loop to Count no. of pairs satisfying
    // a ^ 2 + b ^ 2 = i for N = 1 to 10
    for (let i = 1; i <= 10; i++)
        document.write("For n = " + i + ", "
             + countPairs(i) + " pair exists<br>");
 
// This code is contributed by rishavmahato348.

输出：

For n = 1, 1 pair exists
For n = 2, 1 pair exists
For n = 3, 0 pair exists
For n = 4, 1 pair exists
For n = 5, 2 pair exists
For n = 6, 0 pair exists
For n = 7, 0 pair exists
For n = 8, 1 pair exists
For n = 9, 1 pair exists
For n = 10, 2 pair exists

时间复杂度： O(sqrt(N))