素数三元组
素数三元组是一组三个素数的形式( p，p+2，p+6 )或( p，p+4，p+6 )。这是三个素数最接近的分组，因为每三个连续奇数中有一个是三的倍数，因此除了(2，3，5)和(3，5，7)之外，不是素数(除了 3 本身)。
示例:
Input : n = 15
Output : 5 7 11
7 11 13
Input : n = 25
Output : 5 7 11
7 11 13
11 13 17
13 17 19
17 19 23
一个简单的解决方案是遍历从 1 到 n-6 的所有数字。对于每个数字，我检查我，我+2，我+6，还是我，我+4，我+6 是素数。如果是，打印三胞胎。
一个有效的解决方法是厄拉多塞的筛首先找到所有的素数，这样我们就可以快速检查一个数是否是素数。 以下是实施办法。
C++
// C++ program to find prime triplets smaller
// than or equal to n.
#include <bits/stdc++.h>
using namespace std;
// function to detect prime number
// here we have used sieve method
// https://www.geeksforgeeks.org/sieve-of-eratosthenes/
// to detect prime number
void sieve(int n, bool prime[])
{
for (int p = 2; p * p <= n; p++) {
// If prime[p] is not changed, then it is a prime
if (prime[p] == true) {
// Update all multiples of p
for (int i = p * 2; i <= n; i += p)
prime[i] = false;
}
}
}
// function to print prime triplets
void printPrimeTriplets(int n)
{
// Finding all primes from 1 to n
bool prime[n + 1];
memset(prime, true, sizeof(prime));
sieve(n, prime);
cout << "The prime triplets from 1 to "
<< n << "are :" << endl;
for (int i = 2; i <= n-6; ++i) {
// triplets of form (p, p+2, p+6)
if (prime[i] && prime[i + 2] && prime[i + 6])
cout << i << " " << (i + 2) << " " << (i + 6) << endl;
// triplets of form (p, p+4, p+6)
else if (prime[i] && prime[i + 4] && prime[i + 6])
cout << i << " " << (i + 4) << " " << (i + 6) << endl;
}
}
int main()
{
int n = 25;
printPrimeTriplets(n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find prime triplets
// smaller than or equal to n.
import java.io.*;
import java.util.*;
class GFG {
// function to detect prime number
// here we have used sieve method
// https://www.geeksforgeeks.org/sieve-of-eratosthenes/
// to detect prime number
static void sieve(int n, boolean prime[])
{
for (int p = 2; p * p <= n; p++) {
// If prime[p] is not changed,
//then it is a prime
if (prime[p] == true) {
// Update all multiples of p
for (int i = p * 2; i <= n; i += p)
prime[i] = false;
}
}
}
// function to print prime triplets
static void printPrimeTriplets(int n)
{
// Finding all primes from 1 to n
boolean prime[]=new boolean[n + 1];
Arrays.fill(prime,true);
sieve(n, prime);
System.out.println("The prime triplets"+
" from 1 to " + n + "are :");
for (int i = 2; i <= n-6; ++i) {
// triplets of form (p, p+2, p+6)
if (prime[i] && prime[i + 2] && prime[i + 6])
System.out.println( i + " " + (i + 2) +
" " + (i + 6));
// triplets of form (p, p+4, p+6)
else if (prime[i] && prime[i + 4] &&
prime[i + 6])
System.out.println(i + " " + (i + 4) +
" " + (i + 6));
}
}
public static void main(String args[])
{
int n = 25;
printPrimeTriplets(n);
}
}
/*This code is contributed by Nikita Tiwari.*/
Python 3
# Python 3 program to find
# prime triplets smaller
# than or equal to n.
# function to detect prime number
# using sieve method
# https://www.geeksforgeeks.org/sieve-of-eratosthenes/
# to detect prime number
def sieve(n, prime) :
p = 2
while (p * p <= n ) :
# If prime[p] is not changed
# , then it is a prime
if (prime[p] == True) :
# Update all multiples of p
i = p * 2
while ( i <= n ) :
prime[i] = False
i = i + p
p = p + 1
# function to print
# prime triplets
def printPrimeTriplets(n) :
# Finding all primes
# from 1 to n
prime = [True] * (n + 1)
sieve(n, prime)
print( "The prime triplets from 1 to ",
n , "are :")
for i in range(2, n - 6 + 1) :
# triplets of form (p, p+2, p+6)
if (prime[i] and prime[i + 2] and
prime[i + 6]) :
print( i , (i + 2) , (i + 6))
# triplets of form (p, p+4, p+6)
elif (prime[i] and prime[i + 4] and
prime[i + 6]) :
print(i , (i + 4) , (i + 6))
# Driver code
n = 25
printPrimeTriplets(n)
# This code is contributed by Nikita Tiwari.
C
// C# program to find prime
// triplets smaller than or
// equal to n.
using System;
class GFG
{
// function to detect
// prime number
static void sieve(int n,
bool[] prime)
{
for (int p = 2;
p * p <= n; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p] == false)
{
// Update all multiples of p
for (int i = p * 2;
i <= n; i += p)
prime[i] = true;
}
}
}
// function to print
// prime triplets
static void printPrimeTriplets(int n)
{
// Finding all primes
// from 1 to n
bool[] prime = new bool[n + 1];
sieve(n, prime);
Console.WriteLine("The prime triplets " +
"from 1 to " +
n + " are :");
for (int i = 2; i <= n - 6; ++i)
{
// triplets of form (p, p+2, p+6)
if (!prime[i] &&
!prime[i + 2] &&
!prime[i + 6])
Console.WriteLine(i + " " + (i + 2) +
" " + (i + 6));
// triplets of form (p, p+4, p+6)
else if (!prime[i] &&
!prime[i + 4] &&
!prime[i + 6])
Console.WriteLine(i + " " + (i + 4) +
" " + (i + 6));
}
}
// Driver Code
public static void Main()
{
int n = 25;
printPrimeTriplets(n);
}
}
// This code is contributed by mits
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find prime
// triplets smaller than or
// equal to n.
// function to print
// prime triplets
function printPrimeTriplets($n)
{
// Finding all primes
// from 1 to n
$prime = array_fill(0, $n + 1, true);
// to detect prime number
for ($p = 2; $p * $p <= $n; $p++)
{
// If prime[p] is not changed,
// then it is a prime
if ($prime[$p] == true)
{
// Update all multiples of p
for ($i = $p * 2; $i <= $n; $i += $p)
$prime[$i] = false;
}
}
echo "The prime triplets from 1 to " .
$n . " are :\n";
for ($i = 2; $i <= $n-6; ++$i)
{
// triplets of form (p, p+2, p+6)
if ($prime[$i] && $prime[$i + 2] &&
$prime[$i + 6])
echo $i. " " . ($i + 2) .
" " . ($i + 6) . "\n";
// triplets of form (p, p+4, p+6)
else if ($prime[$i] && $prime[$i + 4] &&
$prime[$i + 6])
echo $i. " " . ($i + 4) .
" " . ($i + 6) . "\n";
}
}
// Driver Code
$n = 25;
printPrimeTriplets($n);
// This code is contributed by mits.
?>
java 描述语言
<script>
// Javascript program to find prime
// triplets smaller than or
// equal to n.
// function to print
// prime triplets
function printPrimeTriplets(n)
{
// Finding all primes
// from 1 to n
let prime = new Array(n + 1).fill(true);
// to detect prime number
for (let p = 2; p * p <= n; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p] == true)
{
// Update all multiples of p
for (let i = p * 2; i <= n; i += p)
prime[i] = false;
}
}
document.write("The prime triplets from 1 to " + n + " are :<br>");
for (let i = 2; i <= n-6; ++i)
{
// triplets of form (p, p+2, p+6)
if (prime[i] && prime[i + 2] &&
prime[i + 6])
document.write(i + " " + (i + 2) +
" " + (i + 6) + "<br>");
// triplets of form (p, p+4, p+6)
else if (prime[i] && prime[i + 4] &&
prime[i + 6])
document.write(i + " " + (i + 4) +
" " + (i + 6) + "<br>");
}
}
// Driver Code
let n = 25;
printPrimeTriplets(n);
// This code is contributed by gfgking
</script>
输出:
The prime triplets from 1 to 25 are :
5 7 11
7 11 13
11 13 17
13 17 19
17 19 23
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