可能的对形成具有给定值的毕达哥拉斯三重奏

给定一个整数C ，任务是找到范围[1，C)中的所有可能的对(A，B) ，使得：

1. A ² + B ² = C ²
2. A < B

例子：

Input: C = 5 
Output:(3, 4) 
Explanation: 
(3)² + (4)² = 9 + 16 = 25 = 5²

Input: C = 25 
Output:(15, 20), (7, 24) 
Explanation: Both the pairs satisfy the necessary conditions. 

方法：

1. 检查[1，C)范围内A和B的所有可能值。
2. 存储满足给定条件的所有对。

下面是上述方法的实现：

C++

// C++ program to compute
// all the possible
// pairs that forms a
// pythagorean triple
// with a given value
#include 
using namespace std;
 
// Function to generate all
// possible pairs
vector > Pairs(int C)
{
    // Vector to store all the
    // possible pairs
    vector > ans;
 
    // Checking all the possible
    // pair in the range of [1, c)
    for (int i = 1; i < C; i++) {
        for (int j = i + 1; j < C;
             j++) {
 
            // If the pair satisfies
            // the condition push it
            // in the vector
            if ((i * i) + (j * j) == (C * C)) {
                ans.push_back(make_pair(i, j));
            }
        }
    }
    return ans;
}
 
// Driver Program
int main()
{
    int C = 13;
    vector > ans
        = Pairs(C);
 
    // If no valid pair exist
    if (ans.size() == 0) {
        cout << "No valid pair exist"
             << endl;
        return 0;
    }
 
    // Print all valid pairs
    for (auto i = ans.begin();
         i != ans.end(); i++) {
        cout << "(" << i->first << ", "
             << i->second << ")" << endl;
    }
 
    return 0;
}

Java

// Java program to compute all
// the possible pairs that forms
// a pythagorean triple with a
// given value
import java.util.*;
 
class GFG{
static class pair
{
    int first, second;
     
    public pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}
 
// Function to generate all
// possible pairs
static Vector Pairs(int C)
{
     
    // Vector to store all the
    // possible pairs
    Vector ans = new Vector();
 
    // Checking all the possible
    // pair in the range of [1, c)
    for(int i = 1; i < C; i++)
    {
       for(int j = i + 1; j < C; j++)
       {
            
          // If the pair satisfies
          // the condition push it
          // in the vector
          if ((i * i) + (j * j) == (C * C))
          {
              ans.add(new pair(i, j));
          }
       }
    }
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    int C = 13;
    Vector ans = Pairs(C);
 
    // If no valid pair exist
    if (ans.size() == 0)
    {
        System.out.print("No valid pair " +
                         "exist" + "\n");
        return;
    }
 
    // Print all valid pairs
    for(pair i:ans)
    {
       System.out.print("(" + i.first +
                       ", " + i.second +
                        ")" + "\n");
    }
}
}
 
// This code is contributed by gauravrajput1

Python3

# Python3 program to compute all
# the possible pairs that forms a
# pythagorean triple with a given value
 
# Function to generate all
# possible pairs
def Pairs(C):
 
    # Vector to store all the
    # possible pairs
    ans = []
   
    # Checking all the possible
    # pair in the range of [1, c)
    for i in range(C):
        for j in range(i + 1, C):
   
            # If the pair satisfies
            # the condition push it
            # in the vector
            if ((i * i) + (j * j) == (C * C)):
                ans.append([i, j])
             
    return ans;
     
# Driver code
if __name__=="__main__":
     
    C = 13;
    ans = Pairs(C);
   
    # If no valid pair exist
    if (len(ans) == 0):
        print("No valid pair exist")
        exit()
   
    # Print all valid pairs
    for i in range(len(ans)):
        print("(" + str(ans[i][0]) +
             ", " + str(ans[i][1]) + ")")
 
# This code is contributed by rutvik_56

C#

// C# program to compute all
// the possible pairs that forms
// a pythagorean triple with a
// given value
using System;
using System.Collections.Generic;
 
class GFG{
class pair
{
    public int first, second;
     
    public pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}
 
// Function to generate all
// possible pairs
static List Pairs(int C)
{
     
    // List to store all the
    // possible pairs
    List ans = new List();
 
    // Checking all the possible
    // pair in the range of [1, c)
    for(int i = 1; i < C; i++)
    {
        for(int j = i + 1; j < C; j++)
        {
                 
            // If the pair satisfies
            // the condition push it
            // in the vector
            if ((i * i) + (j * j) == (C * C))
            {
                ans.Add(new pair(i, j));
            }
        }
    }
    return ans;
}
 
// Driver code
public static void Main(String[] args)
{
    int C = 13;
    List ans = Pairs(C);
 
    // If no valid pair exist
    if (ans.Count == 0)
    {
        Console.Write("No valid pair " +
                      "exist" + "\n");
        return;
    }
 
    // Print all valid pairs
    foreach(pair i in ans)
    {
    Console.Write("(" + i.first +
                 ", " + i.second +
                  ")" + "\n");
    }
}
}
 
// This code is contributed by PrinciRaj1992

输出：

(5, 12)

时间复杂度： O(C ² )