计算三元组(a，b，c)的数量，使得a ^ 2 + b ^ 2 = c ^ 2且1 <= a <= b <= c <= n

给定整数N ，任务是计算三元组(a，b，c)的数量，以使a ² + b ² = c ²且1≤a≤b≤c≤N 。

例子：

Input: N = 5 
Output: 1 
The only possible triplet pair is (3, 4, 5) 
3^2 + 4^2 = 5^2 i.e. 9 + 16 = 25

Input: N = 10 
Output: 2 
(3, 4, 5) and (6, 8, 10) are the required triplet pairs. 
 

方法1：
运行两个循环，第一个循环从i = 1到N，第二个循环从j = i + 1到N。考虑每一对，找到i * i + j * j并检查这是否是一个完美的平方且其平方根小于N.如果是，则增加计数。

下面是上述方法的实现：

C++

// C++ program to Find number of
// Triplets 1 <= a <= b<= c <= n,
// Such that a^2 + b^2 = c^2
#include 
using namespace std;
 
// function to ind number of
// Triplets 1 <= a <= b<= c <= n,
// Such that a^2 + b^2 = c^2
int Triplets(int n)
{
    // to store required answer
    int ans = 0;
 
    // run nested loops for first two numbers.
    for (int i = 1; i <= n; ++i) {
        for (int j = i; j <= n; ++j) {
            int x = i * i + j * j;
 
            // third number
            int y = sqrt(x);
 
            // check if third number is perfect
            // square and less than n
            if (y * y == x && y <= n)
                ++ans;
        }
    }
 
    return ans;
}
 
// Driver code
int main()
{
 
    int n = 10;
 
    // function call
    cout << Triplets(n);
 
    return 0;
}

Java

// Java program to Find number of
// Triplets 1 <= a <= b<= c <= n,
// Such that a^2 + b^2 = c^2
class Solution
{
// function to ind number of
// Triplets 1 <= a <= b<= c <= n,
// Such that a^2 + b^2 = c^2
static int Triplets(int n)
{
    // to store required answer
    int ans = 0;
 
    // run nested loops for first two numbers.
    for (int i = 1; i <= n; ++i) {
        for (int j = i; j <= n; ++j) {
            int x = i * i + j * j;
 
            // third number
            int y =(int) Math.sqrt(x);
 
            // check if third number is perfect
            // square and less than n
            if (y * y == x && y <= n)
                ++ans;
        }
    }
 
    return ans;
}
 
// Driver code
public static void main(String args[])
{
 
    int n = 10;
 
    // function call
    System.out.println(Triplets(n));
 
}
}
//contributed by Arnab Kundu

Python3

# Python3 program to Find number of
# Triplets 1 <= a <= b<= c <= n,
# Such that a^2 + b^2 = c^2
import math
 
# function to ind number of
# Triplets 1 <= a <= b<= c <= n,
# Such that a^2 + b^2 = c^2
def Triplets(n):
 
    # to store required answer
    ans = 0
 
    # run nested loops for first two numbers.
    for i in range(1, n + 1):
        for j in range(i, n + 1):
            x = i * i + j * j
 
            # third number
            y = int(math.sqrt(x))
 
            # check if third number is perfect
            # square and less than n
            if (y * y == x and y <= n):
                ans += 1
    return ans
 
# Driver code
if __name__ == "__main__":
    n = 10
 
    # function call
    print(Triplets(n))
 
# This code is contributed
# by ChitraNayal

C#

// C# program to Find number of
// Triplets 1 <= a <= b<= c <= n,
// Such that a^2 + b^2 = c^2
using System;
 
class GFG
{
// function to ind number of
// Triplets 1 <= a <= b<= c <= n,
// Such that a^2 + b^2 = c^2
static int Triplets(int n)
{
    // to store required answer
    int ans = 0;
 
    // run nested loops for first two numbers.
    for (int i = 1; i <= n; ++i)
    {
        for (int j = i; j <= n; ++j)
        {
            int x = i * i + j * j;
 
            // third number
            int y = (int)Math.Sqrt(x);
 
            // check if third number is perfect
            // square and less than n
            if (y * y == x && y <= n)
                ++ans;
        }
    }
 
    return ans;
}
 
// Driver code
static void Main()
{
    int n = 10;
    Console.WriteLine(Triplets(n));
}
}
 
// This code is contributed by ANKITRAI1

Javascript

// javascript program to Find number of
// Triplets 1 <= a <= b<= c <= n,
// Such that a^2 + b^2 = c^2
 
    // function to ind number of
    // Triplets 1 <= a <= b<= c <= n,
    // Such that a^2 + b^2 = c^2
    function Triplets(n)
    {
     
        // to store required answer
        var ans = 0;
 
        // run nested loops for first two numbers.
        for (let i = 1; i <= n; ++i) {
            for (let j = i; j <= n; ++j) {
                var x = i * i + j * j;
 
                // third number
                var y = parseInt( Math.sqrt(x));
 
                // check if third number is perfect
                // square and less than n
                if (y * y == x && y <= n)
                    ++ans;
            }
        }
        return ans;
    }
 
    // Driver code
    var n = 10;
 
    // function call
    document.write(Triplets(n));
 
// This code is contributed by shikhasingrajput

输出：

2

方法2：

• 查找直到n ^2的所有理想平方并将其保存到ArrayList中。
• 现在，对于从1到n的每个a ，执行以下操作：
  • 从先前计算的理想平方的列表中选择c ² 。
  • 然后，可以将b ²计算为b ² = c ² – a ² 。
  • 现在检查上一步中计算的a <= b <= c和b ^2是否必须是理想平方。
  • 如果满足以上条件，则增加计数。
• 最后打印计数。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include
using namespace std;
 
// Function to return an Array containing
// all the perfect squares upto n
vector getPerfectSquares(int n)
{
    vector perfectSquares;
    int current = 1, i = 1;
 
    // While current perfect square
    // is less than or equal to n
    while (current <= n)
    {
        perfectSquares.push_back(current);
        current = pow(++i, 2);
    }
    return perfectSquares;
}
 
// Function to return the count of
// triplet (a, b, c) pairs such that
// a^2 + b^2 = c^2 and
// 1 <= a <= b <= c <= n
int countTriplets(int n)
{
     
    // Vector of perfect squares upto n^2
    vector perfectSquares = getPerfectSquares(
                                 pow(n, 2));
 
    int count = 0;
    for(int a = 1; a <= n; a++)
    {
        int aSquare = pow(a, 2);
 
        for(int i = 0; i < perfectSquares.size(); i++)
        {
            int cSquare = perfectSquares[i];
 
            // Since, a^2 + b^2 = c^2
            int bSquare = abs(cSquare - aSquare);
            int b = sqrt(bSquare);
            int c = sqrt(cSquare);
 
            // If c < a or bSquare is not a
            // perfect square
            if (c < a || (find(perfectSquares.begin(),
                               perfectSquares.end(),
                               bSquare) ==
                               perfectSquares.end()))
                continue;
 
            // If triplet pair (a, b, c) satisfy
            // the given condition
            if ((b >= a) && (b <= c) &&
                (aSquare + bSquare == cSquare))
                count++;
        }
    }
    return count;
}
 
// Driver code
int main()
{
    int n = 10;
     
    cout << countTriplets(n);
 
    return 0;
}
 
// This code is contributed by himanshu77

Java

// Java implementation of the approach
import java.util.*;
 
public class GFG {
 
    // Function to return an ArrayList containing
    // all the perfect squares upto n
    public static ArrayList getPerfectSquares(int n)
    {
        ArrayList perfectSquares = new ArrayList<>();
        int current = 1, i = 1;
 
        // while current perfect square is less than or equal to n
        while (current <= n) {
            perfectSquares.add(current);
            current = (int)Math.pow(++i, 2);
        }
        return perfectSquares;
    }
 
    // Function to return the count of triplet (a, b, c) pairs
    // such that a^2 + b^2 = c^2 and 1 <= a <= b <= c <= n
    public static int countTriplets(int n)
    {
        // List of perfect squares upto n^2
        ArrayList perfectSquares
            = getPerfectSquares((int)Math.pow(n, 2));
 
        int count = 0;
        for (int a = 1; a <= n; a++) {
            int aSquare = (int)Math.pow(a, 2);
            for (int i = 0; i < perfectSquares.size(); i++) {
                int cSquare = perfectSquares.get(i);
 
                // Since, a^2 + b^2 = c^2
                int bSquare = cSquare - aSquare;
                int b = (int)Math.sqrt(bSquare);
                int c = (int)Math.sqrt(cSquare);
 
                // If c < a or bSquare is not a perfect square
                if (c < a || !perfectSquares.contains(bSquare))
                    continue;
 
                // If triplet pair (a, b, c) satisfy the given condition
                if ((b >= a) && (b <= c) && (aSquare + bSquare == cSquare))
                    count++;
            }
        }
        return count;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 10;
 
        System.out.println(countTriplets(n));
    }
}

Python3

# Python3  implementation of the approach
import math
 
# Function to return an ArrayList containing
# all the perfect squares upto n
def getPerfectSquares(n):
 
    perfectSquares = []
    current = 1
    i = 1
 
    # while current perfect square is less than or equal to n
    while (current <= n) :
        perfectSquares.append(current)
        i += 1
        current = i** 2
     
    return perfectSquares
 
# Function to return the count of triplet (a, b, c) pairs
# such that a^2 + b^2 = c^2 and 1 <= a <= b <= c <= n
def countTriplets(n):
 
    # List of perfect squares upto n^2
    perfectSquares= getPerfectSquares(n**2)
 
    count = 0
    for a in range(1, n +1 ):
        aSquare = a**2
        for i in range(len(perfectSquares)):
            cSquare = perfectSquares[i]
 
            # Since, a^2 + b^2 = c^2
            bSquare = abs(cSquare - aSquare)
            b = math.sqrt(bSquare)
            b = int(b)
            c = math.sqrt(cSquare)
            c = int(c)
             
            # If c < a or bSquare is not a perfect square
            if (c < a or (bSquare not in perfectSquares)):
                continue
 
            # If triplet pair (a, b, c) satisfy the given condition
            if ((b >= a) and (b <= c) and (aSquare + bSquare == cSquare)):
                count += 1
                 
    return count
 
# Driver code
if __name__ == "__main__":
 
    n = 10
 
    print(countTriplets(n))
 
# This code is contributed by chitranayal

C＃

// C# implementation of the approach
using System.Collections;
using System;
 
class GFG
{
 
// Function to return an ArrayList containing
// all the perfect squares upto n
public static ArrayList getPerfectSquares(int n)
{
    ArrayList perfectSquares = new ArrayList();
    int current = 1, i = 1;
 
    // while current perfect square is less
    // than or equal to n
    while (current <= n)
    {
        perfectSquares.Add(current);
        current = (int)Math.Pow(++i, 2);
    }
    return perfectSquares;
}
 
// Function to return the count of triplet
// (a, b, c) pairs such that a^2 + b^2 = c^2
// and 1 <= a <= b <= c <= n
public static int countTriplets(int n)
{
    // List of perfect squares upto n^2
    ArrayList perfectSquares = getPerfectSquares((int)Math.Pow(n, 2));
 
    int count = 0;
    for (int a = 1; a <= n; a++)
    {
        int aSquare = (int)Math.Pow(a, 2);
        for (int i = 0; i < perfectSquares.Count; i++)
        {
            int cSquare = (int)perfectSquares[i];
 
            // Since, a^2 + b^2 = c^2
            int bSquare = cSquare - aSquare;
            int b = (int)Math.Sqrt(bSquare);
            int c = (int)Math.Sqrt(cSquare);
 
            // If c < a or bSquare is not a perfect square
            if (c < a || !perfectSquares.Contains(bSquare))
                continue;
 
            // If triplet pair (a, b, c) satisfy
            // the given condition
            if ((b >= a) && (b <= c) &&
                (aSquare + bSquare == cSquare))
                count++;
        }
    }
    return count;
}
 
// Driver code
public static void Main()
{
    int n = 10;
 
    Console.WriteLine(countTriplets(n));
}
}
 
// This code is contributed by mits.

Java脚本

 
// Javascript implementation of the approach
 
// Function to return an Array containing
// all the perfect squares upto n
function getPerfectSquares(n)
{
    var perfectSquares = [];
    var current = 1, i = 1;
 
    // While current perfect square
    // is less than or equal to n
    while (current <= n)
    {
        perfectSquares.push(current);
        current = Math.pow(++i, 2);
    }
    return perfectSquares;
}
 
// Function to return the count of
// triplet (a, b, c) pairs such that
// a^2 + b^2 = c^2 and
// 1 <= a <= b <= c <= n
function countTriplets(n)
{
     
    // Vector of perfect squares upto n^2
    var perfectSquares = getPerfectSquares(
                                 Math.pow(n, 2));
    var count = 0;
    for(var a = 1; a <= n; a++)
    {
        var aSquare = Math.pow(a, 2);
 
        for(var i = 0;
                i < perfectSquares.length;
                i++)
        {
            var cSquare = perfectSquares[i];
 
            // Since, a^2 + b^2 = c^2
            var bSquare = Math.abs(cSquare -
                                   aSquare);
            var b = Math.sqrt(bSquare);
            var c = Math.sqrt(cSquare);
 
            // If c < a or bSquare is not a
            // perfect square
            if (c < a ||
                !perfectSquares.includes(bSquare))
                continue;
 
            // If triplet pair (a, b, c) satisfy
            // the given condition
            if ((b >= a) && (b <= c) &&
                (aSquare + bSquare == cSquare))
                count++;
        }
    }
    return count;
}
 
// Driver code
var n = 10;
 
document.write(countTriplets(n));
 
// This code is contributed by noob2000
 

输出：

2

如果您希望与行业专家一起参加现场课程，请参阅《 Geeks现场课程》和《 Geeks现场课程美国》。