给定整数N ，任务是计算范围[1，N]中不同素数三元组(a，b，c)的数量，以使a 且a + b = c 。
注意：如果两个素元组中存在的素数中的至少一个不同，则它们是不同的。
例子：
Input: N = 6
Output: 1
Explanation: Among numbers in the range [1, 6], the only prime triplet is (2, 3, 5) (Since 2 + 3 = 5).
Input: N = 10
Output: 2
Explanation: The distinct prime triplets satisfying the condition are (2, 3, 5), (2, 5, 7).
方法：可以根据以下观察到的问题解决问题：
观察：
For every prime number p from 1 to N, it is a part of a triplet if and only if it can be represented as a sum of two prime numbers.
Since a prime number is an odd number, it must be equal to the sum of an even number and an odd number.
Hence the only even prime is 2. Therefore, for a prime number p to constitute a unique tuple (2, p-2, p), the number p – 2 must be a prime number.
请按照以下步骤解决问题：
- 初始化一个变量，例如count = 0 ，以存储主要三元组的数量。
- 从1到N进行迭代，对于每个数字p ，检查此数字p和p – 2是否为质数。
- 如果它们是素数，则将count加1 。
- 打印count的值。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to check if a
// number is a prime or not
bool isPrime(int N)
{
if (N <= 1)
return false;
for (int i = 2; i <= sqrt(N); i++) {
if (N % i == 0)
return false;
}
return true;
}
// Function to count the number
// of valid prime triplets
void countPrimeTuples(int N)
{
// Stores the count
// of prime triplets
int count = 0;
// Iterate from 2 to N and check for each
// p, whether p & (p - 2) are prime or not
for (int i = 2; i <= N; i++) {
if (isPrime(i) && isPrime(i - 2))
count++;
}
// Print the count obtained
cout << count;
}
// Driver Code
int main()
{
int N = 6;
countPrimeTuples(N);
return 0;
} Java
// Java program for the above approach
import java.io.*;
class GFG
{
// Function to check if a
// number is a prime or not
static boolean isPrime(int N)
{
if (N <= 1)
return false;
for (int i = 2; i <= Math.sqrt(N); i++)
{
if (N % i == 0)
return false;
}
return true;
}
// Function to count the number
// of valid prime triplets
static void countPrimeTuples(int N)
{
// Stores the count
// of prime triplets
int count = 0;
// Iterate from 2 to N and check for each
// p, whether p & (p - 2) are prime or not
for (int i = 2; i <= N; i++)
{
if (isPrime(i) && isPrime(i - 2))
count++;
}
// Print the count obtained
System.out.println(count);
}
// Driver Code
public static void main (String[] args)
{
int N = 6;
countPrimeTuples(N);
}
}
// This code is contributed by Dharanendra L V.Python3
# Python3 program for the above approach
import math
# Function to check if a
# number is a prime or not
def isPrime(N) :
if (N <= 1) :
return False
for i in range(2, int(math.sqrt(N) + 1)):
if (N % i == 0) :
return False
return True
# Function to count the number
# of valid prime triplets
def countPrimeTuples(N) :
# Stores the count
# of prime triplets
count = 0
# Iterate from 2 to N and check for each
# p, whether p & (p - 2) are prime or not
for i in range(2, N + 1):
if (isPrime(i) and isPrime(i - 2)) :
count += 1
# Prthe count obtained
print(count)
# Driver Code
N = 6
countPrimeTuples(N)
# This code is contributed by susmitakundugoaldanga.C#
// C# program for the above approach
using System;
public class GFG
{
// Function to check if a
// number is a prime or not
static bool isPrime(int N)
{
if (N <= 1)
return false;
for (int i = 2; i <= Math.Sqrt(N); i++)
{
if (N % i == 0)
return false;
}
return true;
}
// Function to count the number
// of valid prime triplets
static void countPrimeTuples(int N)
{
// Stores the count
// of prime triplets
int count = 0;
// Iterate from 2 to N and check for each
// p, whether p & (p - 2) are prime or not
for (int i = 2; i <= N; i++)
{
if (isPrime(i) && isPrime(i - 2))
count++;
}
// Print the count obtained
Console.WriteLine(count);
}
// Driver Code
static public void Main ()
{
int N = 6;
countPrimeTuples(N);
}
}
// This code is contributed by Dharanendra L V.1时间复杂度： O(N 3/2 )
辅助空间： O(1)