给定一个大小为N的正整数的数组arr [] ，任务是将所有特殊素数按其相对位置排序(不影响其他元素的位置)。特殊质数是质数，可以表示为其他两个质数之和。
例子：
Input: arr[] = {31, 5, 2, 1, 7}
Output: 5 7 2 1 31
The special primes are 31 (29 + 2), 5 (2 + 3), 7 (5 + 2).
Sort them, we get 5, 7 and 31.
So, the original array becomes {5, 7, 2, 1, 31}
Input: arr[] = {3, 13, 11, 43, 2, 19, 17}
Output: 3 13 11 19 2 43 17
方法：
- 开始遍历数组，对于每个元素arr [i] ，如果arr [i]是一个特殊的质数，则将其存储在向量中并更新arr [i] = -1
- 将所有特殊质数存储在列表中之后，对更新后的列表进行排序。
- 再次遍历数组，对于每个元素，
- 如果arr [i] = -1，则打印之前未打印过的列表中的第一个元素。
- 否则，打印arr [i] 。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function for the Sieve of Eratosthenes
void sieveOfEratosthenes(bool prime[], int n)
{
prime[0] = prime[1] = false;
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 greater than or
// equal to the square of it
// numbers which are multiple of p and are
// less than p^2 are already been marked.
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
}
// Function to sort the special primes
// in their relative positions
void sortSpecialPrimes(int arr[], int n)
{
// Maximum element from the array
int maxVal = *max_element(arr, arr + n);
// prime[i] will be true if i is a prime
bool prime[maxVal + 1];
memset(prime, true, sizeof(prime));
sieveOfEratosthenes(prime, maxVal);
// To store the special primes
// from the array
vector list;
for (int i = 0; i < n; i++) {
// If current element is a special prime
if (prime[arr[i]] && prime[arr[i] - 2]) {
// Add it to the ArrayList
// and set arr[i] to -1
list.push_back(arr[i]);
arr[i] = -1;
}
}
// Sort the special primes
sort(list.begin(), list.end());
int j = 0;
for (int i = 0; i < n; i++) {
// Position of a special prime
if (arr[i] == -1)
cout << list[j++] << " ";
else
cout << arr[i] << " ";
}
}
// Driver code
int main()
{
int arr[] = { 31, 5, 2, 1, 7 };
int n = sizeof(arr) / sizeof(int);
sortSpecialPrimes(arr, n);
return 0;
} Java
// Java implementation of the above approach
import java.util.*;
class GFG{
// Function for the Sieve of Eratosthenes
static void sieveOfEratosthenes(boolean prime[],
int n)
{
prime[0] = prime[1] = false;
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
// greater than or equal to the
// square of it numbers which are
// multiple of p and are less than
// p^2 are already been marked.
for(int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
}
// Function to sort the special primes
// in their relative positions
static void sortSpecialPrimes(int arr[], int n)
{
// Maximum element from the array
int maxVal = Arrays.stream(arr).max().getAsInt();
// prime[i] will be true if i is a prime
boolean []prime = new boolean[maxVal + 1];
for(int i = 0; i < prime.length; i++)
prime[i] = true;
sieveOfEratosthenes(prime, maxVal);
// To store the special primes
// from the array
Vector list = new Vector();
for(int i = 0; i < n; i++)
{
// If current element is a special prime
if (prime[arr[i]] && prime[arr[i] - 2])
{
// Add it to the ArrayList
// and set arr[i] to -1
list.add(arr[i]);
arr[i] = -1;
}
}
// Sort the special primes
Collections.sort(list);
int j = 0;
for(int i = 0; i < n; i++)
{
// Position of a special prime
if (arr[i] == -1)
System.out.print(list.get(j++) + " ");
else
System.out.print(arr[i] + " ");
}
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 31, 5, 2, 1, 7 };
int n = arr.length;
sortSpecialPrimes(arr, n);
}
}
// This code is contributed by PrinciRaj1992 C#
// C# implementation of the above approach
using System;
using System.Collections.Generic;
using System.Linq;
class GFG{
// Function for the Sieve of Eratosthenes
static void sieveOfEratosthenes(bool []prime,
int n)
{
prime[0] = prime[1] = false;
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
// greater than or equal to the
// square of it numbers which are
// multiple of p and are less than
// p^2 are already been marked.
for(int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
}
// Function to sort the special primes
// in their relative positions
static void sortSpecialPrimes(int []arr, int n)
{
// Maximum element from the array
int maxVal = arr.Max();
// prime[i] will be true if i is a prime
bool []prime = new bool[maxVal + 1];
for(int i = 0; i < prime.Length; i++)
prime[i] = true;
sieveOfEratosthenes(prime, maxVal);
// To store the special primes
// from the array
List list = new List();
for(int i = 0; i < n; i++)
{
// If current element is a special prime
if (prime[arr[i]] && prime[arr[i] - 2])
{
// Add it to the List
// and set arr[i] to -1
list.Add(arr[i]);
arr[i] = -1;
}
}
// Sort the special primes
list.Sort();
int j = 0;
for(int i = 0; i < n; i++)
{
// Position of a special prime
if (arr[i] == -1)
Console.Write(list[j++] + " ");
else
Console.Write(arr[i] + " ");
}
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 31, 5, 2, 1, 7 };
int n = arr.Length;
sortSpecialPrimes(arr, n);
}
}
// This code is contributed by PrinciRaj1992 输出：
5 7 2 1 31