Dada una array A[], encuentre un subconjunto de tamaño máximo en el que la suma de cada par de elementos sea un número primo. Imprime su longitud y el subconjunto. Considere muchas consultas para diferentes arrays y el valor máximo de un elemento como 100000.
Ejemplos:
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).
Hagamos algunas observaciones y luego pasemos al problema. La suma de dos números es par si y solo ambos números son pares o impares. Un número par no puede ser un número primo excepto 2. Ahora, si tomamos tres números a, b y c, dos de ellos deben ser pares o impares (teorema de Pigeonhole). Entonces, nuestra solución existe solo en dos casos: (Deje que el subconjunto sea B)
- Caso I : Cuando B contiene solo dos números enteros (> 1) cuya suma es un número primo.
- Caso II : Cuando B contiene un número de unos (1) y otro número X, donde X + 1 debe ser un número primo (solo es posible cuando X es un número par).
Primero cuente el número de unos en la array usando un bucle for.
- Si el conteo de 1 es mayor que 0, recorra todo el arreglo y verifique si [A[i] + 1] es un número primo y (A[i] != 1), si encuentra alguno, imprima el tamaño de subarreglo como (cuenta de 1s) +1 y todos los unos(1s) y el A[i] encontrado. Salga del programa.
- Si el paso anterior falla (es decir, no se encuentra A[i]), imprima todos los unos (1). Salga del programa.
- Si el paso anterior falla (es decir, cuenta de 1s = 0 ), Verifique que cada par de elementos en la array su suma sea un número primo. Imprime 2 y el par de enteros.
- De lo contrario Imprimir -1.
A continuación se muestra la implementación del enfoque anterior:
C++
// CPP program to find a subset in which sum of
// every pair in it is a prime
#include <bits/stdc++.h>
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 29
C#
// 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
<?php
// PHP program to find
// a subset in which
// sum of every pair in
// it is a prime
$MAX = 100001;
$isPrime = array();
for($i = 0;
$i < $MAX; $i++)
$isPrime[$i] = 0;
function sieve()
{
global $MAX, $isPrime;
for ($p = 2;
$p * $p < $MAX; $p++)
{
// If isPrime[p] is not
// changed, then it is
// a prime
if ($isPrime[$p] == 0)
{
// Update all
// multiples of p
for ($i = $p * 2;
$i < $MAX; $i += $p)
$isPrime[$i] = 1;
}
}
}
function findSubset($a, $n)
{
$cnt1 = 0;
global $MAX, $isPrime;
// Counting no.of
// ones in the array
for ($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 ($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");
// Print $all the ones(1s).
for ($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");
// Print $all ones(1s).
for ($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
<script>
// JavaScript program to find a
// subset in which sum of
// every pair in it is a prime
let MAX = 100001;
let isPrime = new Array(MAX);
for(let i=0;i<MAX;i++)
{
isPrime[i]=0;
}
function sieve()
{
for (let 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 (let i = p * 2;
i < MAX; i += p)
isPrime[i] = 1;
}
}
return -1;
}
function findSubset(a,n)
{
let cnt1 = 0;
// Counting no. of
// ones in the array
for (let 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 (let i = 0; i < n; i++)
{
// Find a[i], where
// a[i] + 1 is prime.
if ((a[i] != 1) &&
(isPrime[a[i] + 1] == 0))
{
document.write((cnt1 + 1)+"<br>");
// Print all
// the ones(1s).
for (let j = 0;
j < cnt1; j++)
document.write(1 + " ");
// print a[i].
document.write(a[i]+"<br>");
return 0;
}
}
}
// Case-II: array contains
// only ones(1s)
if (cnt1 >= 2)
{
document.write(cnt1);
// Print all ones(1s).
for (let i = 0;
i < cnt1; i++)
document.write(1 + " ");
document.write("<br>");
return 0;
}
// Case-III: array does
// not contain 1s
for (let i = 0; i < n; i++)
{
for (let j = i + 1;
j < n; j++)
{
// Find a pair of integers
// whose sum is prime
if (isPrime[a[i] + a[j]] == 0)
{
document.write(2+"<br>");
document.write(a[i] +
" " + a[j]+"<br>");
return 0;
}
}
}
// Array contains only
// a single element.
document.write(-1);
return -1;
}
// Driver Code
sieve();
let A=[2, 1, 1 ];
let n = A.length;
findSubset(A, n);
// This code is contributed by unknown2108
</script>
Producción:
3 1 1 2
Complejidad del tiempo : O(n 2 )
Publicación traducida automáticamente
Artículo escrito por Harsha_Mogali y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA