[L, R] 范围内的整数计数不与 [1, R] 中的数字形成任何三角形对

给定整数L和R ，任务是找到范围 [L, R]（比如 X）内的整数个数，它们不与范围 [1, R] 内的任何其他整数形成三角形对。

注意：如果 GCD(X, Y)、X/GCD(X, Y) 和 Y/GCD(X, Y) 可以构成三角形的边，则一对整数 {X, Y} 称为三角形对。

例子：

Input: L = 1, R = 5
Output: 3
Explanation: In range [1, 5] there are 3 elements 1, 3 and 5 which does not form any triangular pair.
Other integers 2 and 4 can form triangular pairs. the pair(2, 4) is a triangular pair.
GCD(2, 4) = 2. So the triplet formed is (2, 4/2, 2/2) = (2, 2, 1).
This can be sides of a valid triangle. So only 3 elements are there.

Input: L = 2, R = 6
Output: 2

方法：可以根据以下观察解决问题：

From 1 to N, only 1 and the prime numbers in the range √N to N does not form triangular pairs with any other integers in the range 1 to N

上述观察可以证明如下：

Proof :

For composite numbers: All the composite numbers will form triangular pair with atleast 1 integer.

• Suppose x is a composite number, then x = y * z (y ≥ z, y ≥ 2, z ≥ 2).
• Let a be another number such that a = ((y-1)*z).
• So gcd(a, x) = z, a/gcd(a, x) = y – 1 and x/gcd(a, x) = y.

The triplet {y, y-1, z} would obviously form the sides of a triangle.

For prime numbers: Consider any prime number p in the range [1, N] and any non-coprime number of p, say a.

• So, gcd(a, p) = p, a/gcd(a, p) = a/p and p/gcd(a, p) = 1.
• Now, for the pair {a, p} to form a triangular pair, {1, p, a/p} must form a triangle.
• For that, (p+1) > a/p and 1 + a/p > p, which is possible only when p = a/p or p = √a.

That means, prime number p in range 1 to N can form triangular pair with another integer if p ≤ √N.

So, it can be concluded that in the range [1, N], the integer 1 and prime numbers larger than √N cannot form triangular pairs with any other integers in the range.

So the numbers in range [L, R] which does not form any triangular pair can be calculated by finding such numbers in range [1, L) (say C1) and in range [1, R] (say C2) and then subtracting C1 from C2.

按照下面提到的步骤来实现上述观察：

• 使用 eratosthenes 筛将质数的数量存储到从 1 到 R 的每个整数。
• 如上述观察所示计算 C1 和 C2。
• 从 C2 中减去 C1（比如count ）。
• 返回计数作为最终答案。

下面是上述方法的实现：

C++

// C++ code to implement the above approach
 
#include 
using namespace std;
const int M = 1e5;
 
// Array to store primes upto each
// integer from 1 to N
int p[M + 1];
 
// Array to store whether each integer from
// 1 to N is prime  or not
bool isPrime[M + 1];
 
// Function to count integers in given
// range which does not form a triangle
// with any other integer in the same
// range for all  elements of array
int countNum(int L, int R)
{
    memset(isPrime, true, sizeof(isPrime));
    isPrime[0] = false, isPrime[1] = false;
 
    // Standard sieve method to store
    // which integer is prime
    for (int i = 2; i * i <= R; i++) {
        if (isPrime[i] == true) {
            for (int j = i * i; j <= R;
                 j += i) {
                isPrime[j] = false;
            }
        }
    }
 
    // Prefix sum method to store total primes
    // upto any integer till M
    p[0] = 0;
    for (int i = 1; i <= R; i++) {
        p[i] = p[i - 1];
        if (isPrime[i]) {
            p[i]++;
        }
    }
 
    // Though 2 is prime but it is even also
    p[1] = 1;
    int count = p[R] - p[L - 1];
 
    // Returning answer
    return count;
}
 
// Driver Code
int main()
{
    int L = 1, R = 5;
 
    // Function call
    int answer = countNum(L, R);
    cout << answer;
    return 0;
}

Java

// Java code to implement the above approach
import java.io.*;
 
class GFG
{
 
  // Function to count integers in given
  // range which does not form a triangle
  // with any other integer in the same
  // range for all  elements of array
  public static int countNum(int L, int R)
  {
 
    // Array to store primes upto each
    // integer from 1 to N
    int p[] = new int[100001];
 
    // Array to store whether each integer from
    // 1 to N is prime  or not
    boolean isPrime[] = new boolean[100001];
    for (int i = 0; i < 100001; i++)
      isPrime[i] = true;
 
    isPrime[0] = false;
    isPrime[1] = false;
 
    // Standard sieve method to store
    // which integer is prime
    for (int i = 2; i * i <= R; i++) {
      if (isPrime[i] == true) {
        for (int j = i * i; j <= R; j += i) {
          isPrime[j] = false;
        }
      }
    }
 
    // Prefix sum method to store total primes
    // upto any integer till M
    p[0] = 0;
    for (int i = 1; i <= R; i++) {
      p[i] = p[i - 1];
      if (isPrime[i] == true) {
        p[i]++;
      }
    }
 
    // Though 2 is prime but it is even also
    p[1] = 1;
    int count = p[R] - p[L - 1];
 
    // Returning answer
    return count;
  }
 
  public static void main(String[] args)
  {
    int L = 1, R = 5;
 
    // Function call
    int answer = countNum(L, R);
    System.out.print(answer);
  }
}
 
// This code is contributed by Rohit Pradhan.

Python3

# python3 code to implement the above approach
import math
 
M = int(1e5)
 
# Array to store primes upto each
# integer from 1 to N
p = [0 for _ in range(M + 1)]
 
# Array to store whether each integer from
# 1 to N is prime or not
isPrime = [True for _ in range(M + 1)]
 
# Function to count integers in given
# range which does not form a triangle
# with any other integer in the same
# range for all elements of array
def countNum(L, R):
    global isPrime
 
    isPrime[0], isPrime[1] = False, False
 
    # Standard sieve method to store
    # which integer is prime
    for i in range(2, int(math.sqrt(R)) + 1):
        if (isPrime[i] == True):
            for j in range(i*i, R+1, i):
                isPrime[j] = False
 
        # Prefix sum method to store total primes
        # upto any integer till M
    p[0] = 0
 
    for i in range(1, R+1):
        p[i] = p[i - 1]
        if (isPrime[i]):
            p[i] += 1
 
        # Though 2 is prime but it is even also
    p[1] = 1
    count = p[R] - p[L - 1]
 
    # Returning answer
    return count
 
# Driver Code
if __name__ == "__main__":
 
    L, R = 1, 5
 
    # Function call
    answer = countNum(L, R)
    print(answer)
 
    # This code is contributed by rakeshsahni

C#

// C# code to implement the above approach
using System;
class GFG
{
 
  static int M = 100000;
 
  // Array to store primes upto each
  // integer from 1 to N
  static int[] p = new int[M + 1];
 
  // Array to store whether each integer from
  // 1 to N is prime  or not
  static bool[] isPrime = new bool[M + 1];
 
  // Function to count integers in given
  // range which does not form a triangle
  // with any other integer in the same
  // range for all  elements of array
  static int countNum(int L, int R)
  {
    for (int i = 0; i < M + 1; i++) {
      isPrime[i] = true;
    }
 
    isPrime[0] = false;
    isPrime[1] = false;
 
    // Standard sieve method to store
    // which integer is prime
    for (int i = 2; i * i <= R; i++) {
      if (isPrime[i] == true) {
        for (int j = i * i; j <= R; j += i) {
          isPrime[j] = false;
        }
      }
    }
 
    // Prefix sum method to store total primes
    // upto any integer till M
    p[0] = 0;
    for (int i = 1; i <= R; i++) {
      p[i] = p[i - 1];
      if (isPrime[i]) {
        p[i]++;
      }
    }
 
    // Though 2 is prime but it is even also
    p[1] = 1;
    int count = p[R] - p[L - 1];
 
    // Returning answer
    return count;
  }
 
  // Driver Code
  public static void Main()
  {
    int L = 1, R = 5;
 
    // Function call
    int answer = countNum(L, R);
    Console.Write(answer);
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

    // JavaScript code for the above approach
    let M = 1e5;
 
    // Array to store primes upto each
    // integer from 1 to N
    let p = new Array(M + 1);
 
    // Array to store whether each integer from
    // 1 to N is prime  or not
    let isPrime = new Array(M + 1).fill(true);
 
    // Function to count integers in given
    // range which does not form a triangle
    // with any other integer in the same
    // range for all  elements of array
    function countNum(L, R)
    {
        isPrime[0] = false, isPrime[1] = false;
 
        // Standard sieve method to store
        // which integer is prime
        for (let i = 2; i * i <= R; i++) {
            if (isPrime[i] == true) {
                for (let j = i * i; j <= R;
                    j += i) {
                    isPrime[j] = false;
                }
            }
        }
 
        // Prefix sum method to store total primes
        // upto any integer till M
        p[0] = 0;
        for (let i = 1; i <= R; i++) {
            p[i] = p[i - 1];
            if (isPrime[i]) {
                p[i]++;
            }
        }
 
        // Though 2 is prime but it is even also
        p[1] = 1;
        let count = p[R] - p[L - 1];
 
        // Returning answer
        return count;
    }
 
    // Driver Code
    let L = 1, R = 5;
 
    // Function call
    let answer = countNum(L, R);
    document.write(answer);
 
    // This code is contributed by Potta Lokesh

输出

3

时间复杂度： O(M*log(log(M)))，其中 M 为 1e5
辅助空间： O(M)