素数三元组，由两个元素之差最大为 N 的值组成，等于第三个
原文:https://www . geesforgeks . org/prime-triples-高达-n-两个元素之间有差异-等于三分之一/
给定一个正整数 N ，任务是找到所有的素数三元组 {P，Q，R} ，使得P = R–Q和 P 、 Q 和 R 小于 N 。
示例:
输入: N = 8 输出: 2 3 5 2 5 7* 解释:* 满足给定条件的仅有 2 个素数三元组是:
- {2，3，5}: P = 2，Q = 3，R = 5。因此，P，Q 和 R 是素数，P = R–Q。
- {2，5，7}: P = 2，Q = 5，R = 7。因此，P，Q 和 R 是素数，P = R–Q。
输入:N = 5 T5输出:** 2 3 5
方法:给定的问题可以基于以下观察来解决:
- 通过重新排列给定的方程，可以观察到 P + Q = R ，两个奇数之和为偶数，一个奇数和一个偶数之和为奇数。
- 因为只有一个偶数素数，即 2 。让 P 为奇素数， Q 为奇素数那么 R 永远不可能是素数，所以 P 必须永远是 2 也就是偶素数， Q 就是奇素数那么 R 应该是奇素数。所以有必要找质数 Q 这样 Q > 2 和 R = P + Q ≤ N(其中 P = 2)，R 应该是质数。
按照以下步骤解决问题:
- 使用厄拉多塞的筛预计算范围【1，N】内的所有质数。
- 初始化一个向量的向量，比如 V ，来存储所有的结果三元组。
- 使用一个变量迭代范围【3，N】，比如说 i 。如果 i 和 (i + 2) 为素数和 2 + i ≤ N ，则将当前三元组存储在向量 V 中。
- 完成以上步骤后，打印 V[] 中存储的所有三胞胎。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Stores 1 and 0 at indices which
// are prime and non-prime respectively
bool prime[100000];
// Function to find all prime
// numbers from the range [0, N]
void SieveOfEratosthenes(int n)
{
// Consider all numbers to prime initially
memset(prime, true, sizeof(prime));
// Iterate over the range [2, sqrt(N)]
for (int p = 2; p * p <= n; p++) {
// If p is a prime
if (prime[p] == true) {
// Update all tultiples
// of p as false
for (int i = p * p;
i <= n; i += p) {
prime[i] = false;
}
}
}
}
// Function to find all prime triplets
// satisfying the given conditions
void findTriplets(int N)
{
// Generate all primes up to N
SieveOfEratosthenes(N);
// Stores the triplets
vector<vector<int> > V;
// Iterate over the range [3, N]
for (int i = 3; i <= N; i++) {
// Check for the condition
if (2 + i <= N && prime[i]
&& prime[2 + i]) {
// Store the triplets
V.push_back({ 2, i, i + 2 });
}
}
// Print all the stored triplets
for (int i = 0; i < V.size(); i++) {
cout << V[i][0] << " "
<< V[i][1] << " "
<< V[i][2] << "\n";
}
}
// Driver Code
int main()
{
int N = 8;
findTriplets(N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.util.*;
class GFG{
// Stores 1 and 0 at indices which
// are prime and non-prime respectively
static boolean[] prime = new boolean[100000];
static void initialize()
{
for(int i = 0; i < 100000; i++)
prime[i] = true;
}
// Function to find all prime
// numbers from the range [0, N]
static void SieveOfEratosthenes(int n)
{
// Iterate over the range [2, sqrt(N)]
for(int p = 2; p * p <= n; p++)
{
// If p is a prime
if (prime[p] == true)
{
// Update all tultiples
// of p as false
for(int i = p * p; i <= n; i += p)
{
prime[i] = false;
}
}
}
}
// Function to find all prime triplets
// satisfying the given conditions
static void findTriplets(int N)
{
// Generate all primes up to N
SieveOfEratosthenes(N);
// Stores the triplets
ArrayList<ArrayList<Integer>> V = new ArrayList<ArrayList<Integer>>();
// List<List<int> > V = new List<List<int>>();
// Iterate over the range [3, N]
for(int i = 3; i <= N; i++)
{
// Check for the condition
if (2 + i <= N && prime[i] && prime[2 + i])
{
// Store the triplets
ArrayList<Integer> a1 = new ArrayList<Integer>();
a1.add(2);
a1.add(i);
a1.add(i + 2);
V.add(a1);
}
}
// Print all the stored triplets
for(int i = 0; i < V.size(); i++)
{
System.out.println(V.get(i).get(0) + " " +
V.get(i).get(1) + " " +
V.get(i).get(2));
}
}
// Driver Code
public static void main(String args[])
{
initialize();
int N = 8;
findTriplets(N);
}
}
// This code is contributed by ipg2016107
Python 3
# Python3 program for the above approach
from math import sqrt
# Stores 1 and 0 at indices which
# are prime and non-prime respectively
prime = [True for i in range(100000)]
# Function to find all prime
# numbers from the range [0, N]
def SieveOfEratosthenes(n):
# Iterate over the range [2, sqrt(N)]
for p in range(2, int(sqrt(n)) + 1, 1):
# If p is a prime
if (prime[p] == True):
# Update all tultiples
# of p as false
for i in range(p * p, n + 1, p):
prime[i] = False
# Function to find all prime triplets
# satisfying the given conditions
def findTriplets(N):
# Generate all primes up to N
SieveOfEratosthenes(N)
# Stores the triplets
V = []
# Iterate over the range [3, N]
for i in range(3, N + 1, 1):
# Check for the condition
if (2 + i <= N and prime[i] and prime[2 + i]):
# Store the triplets
V.append([2, i, i + 2])
# Print all the stored triplets
for i in range(len(V)):
print(V[i][0], V[i][1], V[i][2])
# Driver Code
if __name__ == '__main__':
N = 8
findTriplets(N)
# This code is contributed by bgangwar59.
C
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Stores 1 and 0 at indices which
// are prime and non-prime respectively
static bool[] prime = new bool[100000];
static void initialize()
{
for(int i = 0; i < 100000; i++)
prime[i] = true;
}
// Function to find all prime
// numbers from the range [0, N]
static void SieveOfEratosthenes(int n)
{
// Iterate over the range [2, sqrt(N)]
for(int p = 2; p * p <= n; p++)
{
// If p is a prime
if (prime[p] == true)
{
// Update all tultiples
// of p as false
for(int i = p * p; i <= n; i += p)
{
prime[i] = false;
}
}
}
}
// Function to find all prime triplets
// satisfying the given conditions
static void findTriplets(int N)
{
// Generate all primes up to N
SieveOfEratosthenes(N);
// Stores the triplets
List<List<int>> V = new List<List<int>>();
// Iterate over the range [3, N]
for(int i = 3; i <= N; i++)
{
// Check for the condition
if (2 + i <= N && prime[i] ==
true && prime[2 + i])
{
// Store the triplets
List<int> a1 = new List<int>();
a1.Add(2);
a1.Add(i);
a1.Add(i + 2);
V.Add(a1);
}
}
// Print all the stored triplets
for(int i = 0; i < V.Count; i++)
{
Console.WriteLine(V[i][0] + " " +
V[i][1] + " " +
V[i][2]);
}
}
// Driver Code
public static void Main()
{
initialize();
int N = 8;
findTriplets(N);
}
}
// This code is contributed by SURENDRA_GANGWAR
java 描述语言
<script>
// Javascript program for the above approach
// Stores 1 and 0 at indices which
// are prime and non-prime respectively
var prime = new Array(100000);
// Function to find all prime
// numbers from the range [0, N]
function SieveOfEratosthenes(n)
{
// Consider all numbers to prime initially
prime.fill(true);
// Iterate over the range [2, sqrt(N)]
for(var p = 2; p * p <= n; p++)
{
// If p is a prime
if (prime[p] == true)
{
// Update all tultiples
// of p as false
for(var i = p * p;
i <= n; i += p)
{
prime[i] = false;
}
}
}
}
// Function to find all prime triplets
// satisfying the given conditions
function findTriplets(N)
{
// Generate all primes up to N
SieveOfEratosthenes(N);
// Stores the triplets
var V = [];
// Iterate over the range [3, N]
for(var i = 3; i <= N; i++)
{
// Check for the condition
if (2 + i <= N && prime[i] == true &&
prime[2 + i] == true)
{
// Store the triplets
var a1 = [2, i, i + 2];
V.push(a1);
}
}
// Print all the stored triplets
for(var i = 0; i < V.length; i++)
{
document.write(V[i][0] + " " +
V[i][1] + " " +
V[i][2] + "<br>");
}
}
// Driver code
N = 8;
findTriplets(N);
// This code is contributed by SoumikMondal
</script>
Output:
2 3 5
2 5 7
时间复杂度: O(Nlog(log(N))) 辅助空间:* O(1)
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