在一个范围内找到一个非传递性互质三元组

给定L和R，找到一个可能的非传递三元组(a，b，c)，使得对(a，b)是互质的，而对(b，c)是互质的，而(a，c)不是互质的。
例如：(2，5，6)是不可传递的三元组，因为对(2，5)是互质的，而对(5，6)是互质的，而对(2，6)不是互质的

例子：

Input : L = 2, R = 10
Output : a = 4, b = 7, c = 8 is one such triplet
Explanation (4, 7, 8) is a possible triplet (while there are also other such triplets present in this range), Here, pair (4, 7) is coprime and pair (7, 8) is coprime but the pair (4, 8) is not coprime

Input : L = 21, R = 47
Output : a = 23, b = 25, c = 46 is one such triplet
Explanation (23, 25, 46) is a possible triplet (while there are also other such triplets present in this range), Here, pair (23, 25) is coprime and pair (25, 46) is coprime but the pair (23, 46) is not coprime

方法1(蛮力)：

我们在L和R之间生成所有可能的三元组，并检查该属性是否成立，对(a，b)是互素的，而对(b，c)是互素的，而对(a，c)则不是。

C++

// C++ program to find possible non transitive triplets btw L and R
#include 
using namespace std;
  
// Function to return gcd of a and b
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
  
// function to check for gcd
bool coprime(int a, int b)
{
    // a and b are coprime if their gcd is 1.
    return (gcd(a, b) == 1);
}
  
/* Checks if any possible triplet (a, b, c) satisfying the condition
   that (a, b) is coprime, (b, c) is coprime but (a, c) isnt */
void possibleTripletInRange(int L, int R)
{
  
    bool flag = false;
    int possibleA, possibleB, possibleC;
  
    // Generate and check for all possible triplets
    // between L and R
    for (int a = L; a <= R; a++) {
        for (int b = a + 1; b <= R; b++) {
            for (int c = b + 1; c <= R; c++) {
  
                // if we find any such triplets set flag to true
                if (coprime(a, b) && coprime(b, c) && !coprime(a, c)) {
                    flag = true;
                    possibleA = a;
                    possibleB = b;
                    possibleC = c;
                    break;
                }
            }
        }
    }
  
    // flag = True indicates that a pair exists
    // between L and R
    if (flag == true) {
        cout << "(" << possibleA << ", " << possibleB
             << ", " << possibleC << ")"
             << " is one such possible triplet between "
             << L << " and " << R << "\n";
    }
    else {
        cout << "No Such Triplet exists between "
             << L << " and " << R << "\n";
    }
}
  
// Driver code
int main()
{
    int L, R;
  
    // finding possible Triplet between 2 and 10
    L = 2;
    R = 10;
    possibleTripletInRange(L, R);
  
    // finding possible Triplet between 23 and 46
    L = 23;
    R = 46;
    possibleTripletInRange(L, R);
  
    return 0;
}

Java

// Java program to find possible non 
// transitive triplets btw L and R
class GFG {
      
    // Function to return gcd of a and b
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
              
        return gcd(b % a, a);
    }
  
    // function to check for gcd
    static boolean coprime(int a, int b)
    {
          
        // a and b are coprime if their 
        // gcd is 1.
        return (gcd(a, b) == 1);
    }
  
    // Checks if any possible triplet 
    // (a, b, c) satifying the condition
    // that (a, b) is coprime, (b, c) is
    // coprime but (a, c) isnt */
    static void possibleTripletInRange(int L, int R)
    {
  
        boolean flag = false;
        int possibleA = 0, possibleB = 0, 
                           possibleC = 0;
  
        // Generate and check for all possible
        // triplets between L and R
        for (int a = L; a <= R; a++) {
            for (int b = a + 1; b <= R; b++) {
                for (int c = b + 1; c <= R; c++)
                {
  
                    // if we find any such triplets
                    // set flag to true
                    if (coprime(a, b) && coprime(b, c)
                                    && !coprime(a, c))
                    {
                        flag = true;
                        possibleA = a;
                        possibleB = b;
                        possibleC = c;
                        break;
                    }
                }
            }
        }
  
        // flag = True indicates that a pair exists
        // between L and R
        if (flag == true) {
            System.out.println("(" + possibleA + ", " 
                  + possibleB + ", " + possibleC + ")"
                    + " is one such possible triplet "
                      + "between " + L + " and " + R);
        }
        else {
            System.out.println("No Such Triplet exists"
                      + "between " + L + " and " + R);
        }
    }
  
    // Driver code
    public static void main(String[] args)
    {
          
        int L, R;
  
        // finding possible Triplet between
        // 2 and 10
        L = 2;
        R = 10;
        possibleTripletInRange(L, R);
  
        // finding possible Triplet between 
        // 23 and 46
        L = 23;
        R = 46;
        possibleTripletInRange(L, R);
    }
}
  
// This code is contributed by
// Smitha DInesh Semwal

Python3

# Python3 program to find possible non 
# transitive triplets btw L and R
  
# Function to return gcd of a and b
def gcd(a, b):
  
    if (a == 0):
        return b;
    return gcd(b % a, a);
  
# function to check for gcd
def coprime(a, b):
  
    # a and b are coprime if
    # their gcd is 1.
    return (gcd(a, b) == 1);
  
# Checks if any possible triplet 
# (a, b, c) satifying the condition
# that (a, b) is coprime, (b, c) 
# is coprime but (a, c) isnt 
def possibleTripletInRange(L, R):
  
    flag = False;
    possibleA = 0;
    possibleB = 0;
    possibleC = 0;
  
    # Generate and check for all 
    # possible triplets between L and R
    for a in range(L, R + 1): 
        for b in range(a + 1, R + 1): 
            for c in range(b + 1, R + 1):
                  
                # if we find any such triplets 
                # set flag to true
                if (coprime(a, b) and coprime(b, c) and     
                                      coprime(a, c) == False):
                    flag = True;
                    possibleA = a;
                    possibleB = b;
                    possibleC = c;
                    break;
  
    # flag = True indicates that a 
    # pair exists between L and R
    if (flag == True): 
        print("(", possibleA, ",", possibleB, 
              ",", possibleC, ") is one such", 
              "possible triplet between", L, "and", R);
    else:
        print("No Such Triplet exists between",
                                  L, "and", R);
  
# Driver Code
  
# finding possible Triplet
# between 2 and 10
L = 2;
R = 10;
possibleTripletInRange(L, R);
  
# finding possible Triplet 
# between 23 and 46
L = 23;
R = 46;
possibleTripletInRange(L, R);
  
# This code is contributed by mits

C#

// C# program to find possible 
// non transitive triplets 
// btw L and R
using System;
class GFG 
{
    // Function to return
    // gcd of a and b
    static int gcd(int a,
                   int b)
    {
        if (a == 0)
            return b;
              
        return gcd(b % a, a);
    }
  
    // function to
    // check for gcd
    static bool coprime(int a, 
                        int b)
    {
          
        // a and b are coprime 
        // if their gcd is 1.
        return (gcd(a, b) == 1);
    }
  
    // Checks if any possible 
    // triplet (a, b, c) satifying 
    // the condition that (a, b) 
    // is coprime, (b, c) is
    // coprime but (a, c) isnt */
    static void possibleTripletInRange(int L, 
                                       int R)
    {
  
        bool flag = false;
        int possibleA = 0, 
            possibleB = 0, 
            possibleC = 0;
  
        // Generate and check for
        // all possible triplets 
        // between L and R
        for (int a = L; a <= R; a++) 
        {
            for (int b = a + 1; 
                     b <= R; b++)
            {
                for (int c = b + 1; 
                         c <= R; c++)
                {
  
                    // if we find any 
                    // such triplets
                    // set flag to true
                    if (coprime(a, b) && 
                        coprime(b, c) && 
                       !coprime(a, c))
                    {
                        flag = true;
                        possibleA = a;
                        possibleB = b;
                        possibleC = c;
                        break;
                    }
                }
            }
        }
  
        // flag = True indicates 
        // that a pair exists
        // between L and R
        if (flag == true)
        {
            Console.WriteLine("(" + possibleA + ", " + 
                                    possibleB + ", " + 
                                    possibleC + ")" + 
                    " is one such possible triplet " + 
                        "between " + L + " and " + R);
        }
        else 
        {
            Console.WriteLine("No Such Triplet exists" +
                          "between " + L + " and " + R);
        }
    }
  
    // Driver code
    public static void Main()
    {
        int L, R;
  
        // finding possible 
        // Triplet between
        // 2 and 10
        L = 2;
        R = 10;
        possibleTripletInRange(L, R);
  
        // finding possible 
        // Triplet between 
        // 23 and 46
        L = 23;
        R = 46;
        possibleTripletInRange(L, R);
    }
}
  
// This code is contributed
// by anuj_67.

---

输出：

(8, 9, 10) is one such possible triplet between 2 and 10
(44, 45, 46) is one such possible triplet between 23 and 46

蛮力解的时间复杂度为O(n ³ log(A))，其中A是三元组的最小数目。
注意：复杂度的对数因子是针对一对数字计算GCD的对数因子。

方法2(有效) ：

由于我们仅需要一对这样的可能对，因此我们可以使用它来进一步分解我们的复杂性。

我们只需要确定一些情况并寻求解决这些问题即可解决此问题。

情况1：L和R之间的数字少于3。
这种情况很简单，我们不能形成任何三元组，所以答案是这种情况永远是“不可能”

情况2：L和R之间有三个以上的数字。
现在，
这是众所周知的证明，连续数字始终是互质的。我们甚至可以轻松证明这一点。

Proof:
Given that N and N + 1 are two consecutive integers. 
Now suppose gcd(n, n + 1) = X,
? X divides n and X also divides (n + 1). 
Which implies that X divides ((n + 1) - n) or X divides 1.
But, There is no number which divides 1 except 1.
? X = 1, or we can also say that gcd(n, n + 1) = 1 

Thus, n and n + 1 are coprime.

因此，如果我们采用三个连续的2k，2k + 1、2k + 2形式的数字，那么我们最终总是会得到可能的三元组，因为如上所述，对(2k，2k + 1)和(2k + 1，2k + 2)连续数对是互质数，而对(2k，2k + 2)的gcd为2(因为它们是偶数)。

情况3：当L和R之间恰好有3个数字时
这是案例3的扩展，现在这个案例可以有2个案例，

情况3.1当三个数字的格式为2k，2k + 1，2k + 2
我们已经在案例2中研究了这种情况。因此，这是唯一的三元组，也是L和R之间的有效三元组。

情况3.2当三个数字的形式为2k – 1，2k，2k + 1
我们已经看到，作为连续数字对的(2k – 1，2k)和(2k，2k + 1)是互质对，因此我们需要检查(2k – 1，2k + 1)对是否互质。
可以证明，对(2k – 1，2k + 1)始终是互质的，如下所示

Proof:
Given that 2k - 1 and 2k + 1 are two numbers 
Now suppose gcd((2k - 1), (2k + 1)) = X,
? X divides (2k - 1) and X also divides (2k + 1). 
Which implies that X divides ((2k + 1) - (2k - 1)) or X divides 2.
2 being a prime is only divisible by 1 and 2 itself. 
But, 2k - 1 and 2k + 1 are odd numbers so X can never be equal to 2.
? X = 1, or we can also say that gcd((2k -1), (2k + 1)) = 1 

Thus, 2k - 1 and 2k + 1 are coprime.

因此，在这种情况下，我们将无法找到任何可能的有效三元组。

下面是上述方法的实现：

C++

/* C++ program to find a non transitive co-prime
   triplets between L and R */
#include 
using namespace std;
  
/* Checks if any possible triplet (a, b, c) satisfying the condition
   that (a, b) is coprime, (b, c) is coprime but (a, c) isnt */
  
void possibleTripletInRange(int L, int R)
{
  
    bool flag = false;
    int possibleA, possibleB, possibleC;
  
    int numbersInRange = (R - L + 1);
  
    /* Case 1 : Less than 3 numbers between L and R */
    if (numbersInRange < 3) {
        flag = false;
    }
  
    /* Case 2: More than 3 numbers between L and R */
    else if (numbersInRange > 3) {
        flag = true;
  
        // triplets should always be of form (2k, 2k + 1, 2k + 2)
        if (L % 2) {
            L++;
        }
  
        possibleA = L;
        possibleB = L + 1;
        possibleC = L + 2;
    }
  
    else {
        /* Case 3.1: Exactly 3 numbers in range of form
                     (2k, 2k + 1, 2k + 2) */
        if (!(L % 2)) {
            flag = true;
            possibleA = L;
            possibleB = L + 1;
            possibleC = L + 2;
        }
        else {
            /* Case 3.2: Exactly 3 numbers in range of form
                         (2k - 1, 2k, 2k + 1) */
            flag = false;
        }
    }
  
    // flag = True indicates that a pair exists between L and R
    if (flag == true) {
        cout << "(" << possibleA << ", " << possibleB
             << ", " << possibleC << ")"
             << " is one such possible triplet between "
             << L << " and " << R << "\n";
    }
    else {
        cout << "No Such Triplet exists between "
             << L << " and " << R << "\n";
    }
}
  
// Driver code
int main()
{
    int L, R;
  
    // finding possible Triplet between 2 and 10
    L = 2;
    R = 10;
    possibleTripletInRange(L, R);
  
    // finding possible Triplet between 23 and 46
    L = 23;
    R = 46;
    possibleTripletInRange(L, R);
  
    return 0;
}

Java

// Java program to find a 
// non transitive co-prime
// triplets between L and R 
import java.io.*;
  
class GFG 
{
  
// Checks if any possible triplet 
// (a, b, c) satifying the condition
// that (a, b) is coprime, (b, c) 
// is coprime but (a, c) isnt
static void possibleTripletInRange(int L, 
                                   int R)
{
    boolean flag = false;
    int possibleA = 0, 
        possibleB = 0, 
        possibleC = 0;
   
    int numbersInRange = (R - L + 1);
  
    // Case 1 : Less than 3 
    // numbers between L and R
    if (numbersInRange < 3) 
    {
        flag = false;
    }
  
    // Case 2: More than 3 
    // numbers between L and R 
    else if (numbersInRange > 3)
    {
        flag = true;
  
        // triplets should always 
        // be of form (2k, 2k + 1,
        // 2k + 2)
        if (L % 2 > 0) 
        {
            L++;
        }
  
        possibleA = L;
        possibleB = L + 1;
        possibleC = L + 2;
    }
  
    else 
    {
        /* Case 3.1: Exactly 3 numbers 
                      in range of form
                     (2k, 2k + 1, 2k + 2) */
        if (!(L % 2 > 0)) 
        {
            flag = true;
            possibleA = L;
            possibleB = L + 1;
            possibleC = L + 2;
        }
        else 
        {
            /* Case 3.2: Exactly 3 numbers
                         in range of form
                         (2k - 1, 2k, 2k + 1) */
            flag = false;
        }
    }
  
    // flag = True indicates 
    // that a pair exists 
    // between L and R
    if (flag == true) 
    {
        System.out.println("(" + possibleA +
                          ", " + possibleB + 
                          ", " + possibleC + 
             ")" + " is one such possible" + 
                       " triplet between " + 
                          L + " and " + R );
    }
    else {
        System.out.println("No Such Triplet" + 
                          " exists between " + 
                             L + " and " + R);
    }
}
  
// Driver code
public static void main (String[] args) 
{
int L, R;
  
// finding possible Triplet 
// between 2 and 10
L = 2;
R = 10;
possibleTripletInRange(L, R);
  
// finding possible Triplet
// between 23 and 46
L = 23;
R = 46;
possibleTripletInRange(L, R);
}
}
  
// This code is contributed
// by anuj_67.

Python3

# Python3 program to find a non transitive 
# co-prime triplets between L and R 
  
# Checks if any possible triplet (a, b, c) 
# satifying the condition that (a, b) is
# coprime, (b, c) is coprime but (a, c) isnt 
def possibleTripletInRange(L, R):
  
    flag = False;
    possibleA = 0;
    possibleB = 0;
    possibleC = 0;
  
    numbersInRange = (R - L + 1);
  
    # Case 1 : Less than 3 numbers
    # between L and R 
    if (numbersInRange < 3):
        flag = False;
  
    # Case 2: More than 3 numbers 
    # between L and R 
    elif (numbersInRange > 3):
        flag = True;
  
        # triplets should always be of 
        # form (2k, 2k + 1, 2k + 2)
        if ((L % 2) > 0):
            L += 1;
  
        possibleA = L;
        possibleB = L + 1;
        possibleC = L + 2;
  
    else:
          
        # Case 3.1: Exactly 3 numbers in range
        #            of form (2k, 2k + 1, 2k + 2)
        if ((L % 2) == 0):
            flag = True;
            possibleA = L;
            possibleB = L + 1;
            possibleC = L + 2;
        else:
              
            # Case 3.2: Exactly 3 numbers in range 
            #            of form (2k - 1, 2k, 2k + 1) 
            flag = False;
  
    # flag = True indicates that a pair 
    # exists between L and R
    if (flag == True):
        print("(", possibleA, ",", possibleB, 
              ",", possibleC, ") is one such", 
              "possible triplet between", L, "and", R);
    else:
        print("No Such Triplet exists between", 
                                  L, "and", R);
  
# Driver code
  
# finding possible Triplet 
# between 2 and 10
L = 2;
R = 10;
possibleTripletInRange(L, R);
  
# finding possible Triplet
# between 23 and 46
L = 23;
R = 46;
possibleTripletInRange(L, R);
  
# This code is contributed by mits

C＃

// C#  program to find a 
// non transitive co-prime 
// triplets between L and R 
using System;
  
public class GFG{
      
      
// Checks if any possible triplet 
// (a, b, c) satifying the condition 
// that (a, b) is coprime, (b, c) 
// is coprime but (a, c) isnt 
static void possibleTripletInRange(int L, 
                                int R) 
{ 
    bool flag = false; 
    int possibleA = 0, 
        possibleB = 0, 
        possibleC = 0; 
  
    int numbersInRange = (R - L + 1); 
  
    // Case 1 : Less than 3 
    // numbers between L and R 
    if (numbersInRange < 3) 
    { 
        flag = false; 
    } 
  
    // Case 2: More than 3 
    // numbers between L and R 
    else if (numbersInRange > 3) 
    { 
        flag = true; 
  
        // triplets should always 
        // be of form (2k, 2k + 1, 
        // 2k + 2) 
        if (L % 2 > 0) 
        { 
            L++; 
        } 
  
        possibleA = L; 
        possibleB = L + 1; 
        possibleC = L + 2; 
    } 
  
    else
    { 
        /* Case 3.1: Exactly 3 numbers 
                    in range of form 
                    (2k, 2k + 1, 2k + 2) */
        if (!(L % 2 > 0)) 
        { 
            flag = true; 
            possibleA = L; 
            possibleB = L + 1; 
            possibleC = L + 2; 
        } 
        else
        { 
            /* Case 3.2: Exactly 3 numbers 
                        in range of form 
                        (2k - 1, 2k, 2k + 1) */
            flag = false; 
        } 
    } 
  
    // flag = True indicates 
    // that a pair exists 
    // between L and R 
    if (flag == true) 
    { 
            Console.WriteLine("(" + possibleA + 
                        ", " + possibleB + 
                        ", " + possibleC + 
            ")" + " is one such possible" + 
                    " triplet between " + 
                        L + " and " + R ); 
    } 
    else { 
        Console.WriteLine("No Such Triplet" + 
                        " exists between " + 
                            L + " and " + R); 
    } 
} 
  
// Driver code 
      
static public void Main (){
      
    int L, R; 
    // finding possible Triplet 
    // between 2 and 10 
    L = 2; 
    R = 10; 
    possibleTripletInRange(L, R); 
    // finding possible Triplet 
    // between 23 and 46 
    L = 23; 
    R = 46; 
    possibleTripletInRange(L, R); 
    } 
} 
// This code is contributed by ajit 

输出：

(2, 3, 4) is one such possible triplet between 2 and 10
(24, 25, 26) is one such possible triplet between 24 and 46

该方法的时间复杂度为O(1)。