以每对之和为素数的最大子集
给定一个数组 A[]，找到一个最大大小的子集，其中每对元素的和都是质数。打印它的长度和子集。考虑对不同数组的许多查询，并且元素的最大值为 100000。
例子 ：
Input : A[] = {2, 1, 2}
Output : 2
1 2
Explanation :
Here, we can only form subsets with size 1 and 2.
maximum sized subset = {1, 2}, 1 + 2 = 3, which
is prime number.
So, Answer = 2 (size), {1, 2} (subset)
Input : A[] = {2, 1, 1}
Output : 3
1 1 2
Explanation :
Maximum subset = {2, 1, 2}, since 1 + 2 = 3,
1 + 1 = 2, both are prime numbers.
Answer = 3 (size), {2, 1, 1} (subset).让我们做一些观察，然后转向问题。两个数之和是偶数，且只有这两个数是奇数或偶数。除了 2 之外，偶数不能是素数。现在，如果我们取三个数 a、b 和 c，其中两个应该是奇数或偶数（鸽洞定理）。所以，我们的解决方案只存在于两种情况——（设子集为 B）
- 情况 I ：当 B 仅包含两个整数（>1）时，其和为素数。
- 情况二：当 B 包含一些个数（1）和另一个数 X 时，其中 X + 1 应该是素数（仅当 X 是偶数时才有可能）。
首先使用 for 循环计算数组中的个数。
- 如果 1 的计数大于 0，则遍历整个数组并检查 [A[i] + 1] 是否为质数和 (A[i] != 1)，如果找到，则打印子数组为(count of 1s) +1 和所有的 (1s) 和找到的 A[i]。退出程序。
- 如果上述步骤失败（即未找到 A[i]），则打印所有的（1）。退出程序。
- 如果上述步骤失败（即1s = 0 的计数），请检查数组中的每一对元素是否为素数。打印 2 和这对整数。
- 否则打印-1。
以下是上述方法的实现：
C++
// CPP program to find a subset in which sum of
// every pair in it is a prime
#include
using namespace std;
#define MAX 100001
bool isPrime[MAX] = { 0 };
int sieve()
{
for (int p = 2; p * p < MAX; p++)
{
// If isPrime[p] is not changed, then it
// is a prime
if (isPrime[p] == 0)
{
// Update all multiples of p
for (int i = p * 2; i < MAX; i += p)
isPrime[i] = 1;
}
}
}
int findSubset(int a[], int n)
{
int cnt1 = 0;
// Counting no.of ones in the array
for (int i = 0; i < n; i++)
if (a[i] == 1)
cnt1++;
// Case-I: count of ones(1s) > 0 and
// an integer > 1 is present in the array
if (cnt1 > 0)
{
for (int i = 0; i < n; i++)
{
// Find a[i], where a[i] + 1 is prime.
if ((a[i] != 1) and (isPrime[a[i] + 1] == 0))
{
cout << cnt1 + 1 << endl;
// Print all the ones(1s).
for (int j = 0; j < cnt1; j++)
cout << 1 << " ";
cout << a[i] << endl; // print a[i].
return 0;
}
}
}
// Case-II: array contains only ones(1s)
if (cnt1 >= 2)
{
cout << cnt1 << endl;
// Print all ones(1s).
for (int i = 0; i < cnt1; i++)
cout << 1 << " ";
cout << endl;
return 0;
}
// Case-III: array does not contain 1s
for (int i = 0; i < n; i++)
{
for (int j = i + 1; j < n; j++)
{
// Find a pair of integers whose sum is prime
if (isPrime[a[i] + a[j]] == 0)
{
cout << 2 << endl;
cout << a[i] << " " << a[j] << endl;
return 0;
}
}
}
// Array contains only a single element.
cout << -1 << endl;
}
// Driver function
int main()
{
sieve();
int A[] = { 2, 1, 1 };
int n = sizeof(A) / sizeof(A[0]);
findSubset(A, n);
return 0;
} Java
// Java program to find a
// subset in which sum of
// every pair in it is a prime
import java.io.*;
class GFG
{
static int MAX = 100001;
static int []isPrime = new int[MAX];
static int sieve()
{
for (int p = 2;
p * p < MAX; p++)
{
// If isPrime[p] is
// not changed, then
// it is a prime
if (isPrime[p] == 0)
{
// Update all
// multiples of p
for (int i = p * 2;
i < MAX; i += p)
isPrime[i] = 1;
}
}
return -1;
}
static int findSubset(int []a, int n)
{
int cnt1 = 0;
// Counting no. of
// ones in the array
for (int i = 0; i < n; i++)
if (a[i] == 1)
cnt1++;
// Case-I: count of
// ones(1s) > 0 and
// an integer > 1 is
// present in the array
if (cnt1 > 0)
{
for (int i = 0; i < n; i++)
{
// Find a[i], where
// a[i] + 1 is prime.
if ((a[i] != 1) &&
(isPrime[a[i] + 1] == 0))
{
System.out.println(cnt1 + 1);
// Print all
// the ones(1s).
for (int j = 0;
j < cnt1; j++)
System.out.print(1 + " ");
System.out.println(a[i]); // print a[i].
return 0;
}
}
}
// Case-II: array contains
// only ones(1s)
if (cnt1 >= 2)
{
System.out.println(cnt1);
// Print all ones(1s).
for (int i = 0;
i < cnt1; i++)
System.out.print(1 + " ");
System.out.println();
return 0;
}
// Case-III: array does
// not contain 1s
for (int i = 0; i < n; i++)
{
for (int j = i + 1;
j < n; j++)
{
// Find a pair of integers
// whose sum is prime
if (isPrime[a[i] + a[j]] == 0)
{
System.out.println(2);
System.out.println(a[i] +
" " + a[j]);
return 0;
}
}
}
// Array contains only
// a single element.
System.out.println(-1);
return -1;
}
// Driver Code
public static void main(String args[])
{
sieve();
int []A = new int[]{ 2, 1, 1 };
int n = A.length;
findSubset(A, n);
}
}
// This code is contributed by
// Manish Shaw(manishshaw1)Python3
# Python3 program to find a subset in which
# sum of every pair in it is a prime
import math as mt
MAX = 100001
isPrime = [0 for i in range(MAX)]
def sieve():
for p in range(2, mt.ceil(mt.sqrt(MAX))):
# If isPrime[p] is not changed,
# then it is a prime
if (isPrime[p] == 0) :
# Update all multiples of p
for i in range(2 * p, MAX, p):
isPrime[i] = 1
def findSubset(a, n):
cnt1 = 0
# Counting no.of ones in the array
for i in range(n):
if (a[i] == 1):
cnt1+=1
# Case-I: count of ones(1s) > 0 and
# an integer > 1 is present in the array
if (cnt1 > 0):
for i in range(n):
# Find a[i], where a[i] + 1 is prime.
if ((a[i] != 1) and
(isPrime[a[i] + 1] == 0)):
print(cnt1 + 1)
# Print all the ones(1s).
for j in range(cnt1):
print("1", end = " ")
print(a[i])
return 0
# Case-II: array contains only ones(1s)
if (cnt1 >= 2):
print(cnt1)
# Print all ones(1s).
for i in range(cnt1):
print("1", end = " ")
print("\n")
return 0
# Case-III: array does not contain 1s
for i in range(n):
for j in range(i + 1, n):
# Find a pair of integers whose
# sum is prime
if (isPrime[a[i] + a[j]] == 0):
print(2)
print(a[i], " ", a[j])
# Array contains only a single element.
print(-1)
# Driver Code
sieve()
A =[ 2, 1, 1]
n =len(A)
findSubset(A, n)
# This code is contributed
# by Mohit kumar 29C#
// C# program to find a subset
// in which sum of every pair
// in it is a prime
using System;
class GFG
{
static int MAX = 100001;
static int []isPrime = new int[MAX];
static int sieve()
{
for (int p = 2;
p * p < MAX; p++)
{
// If isPrime[p] is
// not changed, then
// it is a prime
if (isPrime[p] == 0)
{
// Update all
// multiples of p
for (int i = p * 2;
i < MAX; i += p)
isPrime[i] = 1;
}
}
return -1;
}
static int findSubset(int []a, int n)
{
int cnt1 = 0;
// Counting no. of
// ones in the array
for (int i = 0; i < n; i++)
if (a[i] == 1)
cnt1++;
// Case-I: count of
// ones(1s) > 0 and
// an integer > 1 is
// present in the array
if (cnt1 > 0)
{
for (int i = 0; i < n; i++)
{
// Find a[i], where
// a[i] + 1 is prime.
if ((a[i] != 1) &&
(isPrime[a[i] + 1] == 0))
{
Console.WriteLine(cnt1 + 1);
// Print all the ones(1s).
for (int j = 0; j < cnt1; j++)
Console.Write(1 + " ");
Console.WriteLine(a[i]); // print a[i].
return 0;
}
}
}
// Case-II: array contains
// only ones(1s)
if (cnt1 >= 2)
{
Console.WriteLine(cnt1);
// Print all ones(1s).
for (int i = 0; i < cnt1; i++)
Console.Write(1 + " ");
Console.WriteLine();
return 0;
}
// Case-III: array does
// not contain 1s
for (int i = 0; i < n; i++)
{
for (int j = i + 1; j < n; j++)
{
// Find a pair of integers
// whose sum is prime
if (isPrime[a[i] + a[j]] == 0)
{
Console.WriteLine(2);
Console.WriteLine(a[i] + " " + a[j]);
return 0;
}
}
}
// Array contains only
// a single element.
Console.WriteLine(-1);
return -1;
}
// Driver Code
static void Main()
{
sieve();
int []A = new int[]{ 2, 1, 1 };
int n = A.Length;
findSubset(A, n);
}
}
// This code is contributed by
// Manish Shaw(manishshaw1)PHP
0 and
// an integer > 1 is
// present in the array
if ($cnt1 > 0)
{
for ($i = 0; $i < $n; $i++)
{
// Find a[i], where
// a[i] + 1 is prime.
if (($a[$i] != 1) and
($isPrime[$a[$i] + 1] == 0))
{
echo (($cnt1 + 1) . "\n");
// Pr$all the ones(1s).
for ($j = 0;
$j < $cnt1; $j++)
{
echo ("1 ");
}
echo ($a[$i] . "\n");
return 0;
}
}
}
// Case-II: array contains
// only ones(1s)
if ($cnt1 >= 2)
{
echo (cnt1 . "\n");
// Pr$all ones(1s).
for ($i = 0;
$i < $cnt1; $i++)
echo ("1 ");
echo ("\n");
return 0;
}
// Case-III: array does
// not contain 1s
for ($i = 0; $i < $n; $i++)
{
for ($j = $i + 1;
$j < $n; $j++)
{
// Find a pair of integers
// whose sum is prime
if ($isPrime[$a[$i] +
$a[$j]] == 0)
{
echo (2 . "\n");
echo ($a[$i] . " " .
$a[$j] . "\n");
return 0;
}
}
}
// Array contains only
// a single element.
echo (-1 . "\n");
}
// Driver Code
sieve();
$A = array(2, 1, 1);
$n = count($A);
findSubset($A, $n);
// This code is contributed by
// Manish Shaw(manishshaw1)
?>Javascript
输出：
3
1 1 2时间复杂度： O(n 2 )