给定一个由N个正整数组成的数组arr [] ，任务是找到一个大于1的整数，该整数与所有给定的数组元素互质。
例子：
Input: arr[ ] = {10,13,17,19}
Output: 23
Explanation:
The GCD of 23 with every array element is 1. Therefore, 23 is coprime with all the given array elements.
Input: arr[] = {13, 17, 23, 24, 50}
Output: 53
Explanation:
The GCD of 53 with every array element is 1. Therefore, 53 is coprime with all the given array elements.
方法：想法是利用一个事实，即大于最大数组元素的质数将与所有给定的数组元素互质。因此，只需找到大于数组中存在的最大元素的质数即可。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find an element which
// is coprime with all array elements
int find_X(int arr[], int N)
{
// Stores maximum array element
int R = INT_MIN;
for (int i = 0; i < N; i++)
R = max(R, arr[i]);
// Stores if index of an array is prime or not
bool prime[1000001];
for (int i = 0; i < 1000001; i++)
prime[i] = true;
int p = 2;
while (p * p <= 1000002)
{
// 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 < 1000001; i += p)
{
prime[i] = false;
}
}
// Increment p by 1
p = p + 1;
}
prime[0] = false;
prime[1] = false;
// Traverse the range [R, 10000000 + 1]
for (int i = R; i < 1000001; i++) {
// If i is greater than R and prime
if (i > R and prime[i] == true) {
// Return i
return i;
}
}
// Dummy value to omit return error
return -1;
}
// Driven Program
int main()
{
// Given array
int arr[] = { 10, 13, 17, 19 };
// stores the length of array
int N = sizeof(arr) / sizeof(arr[0]);
// Function call
cout << find_X(arr, N);
return 0;
}
// This code is contributed by Kingash. Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG {
// Function to find an element which
// is coprime with all array elements
static int find_X(int arr[])
{
// Stores maximum array element
int R = Integer.MIN_VALUE;
for (int i = 0; i < arr.length; i++)
R = Math.max(R, arr[i]);
// Stores if index of an array is prime or not
boolean prime[] = new boolean[1000001];
Arrays.fill(prime, true);
int p = 2;
while (p * p <= 1000002) {
// 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 < 1000001; i += p) {
prime[i] = false;
}
}
// Increment p by 1
p = p + 1;
}
prime[0] = false;
prime[1] = false;
// Traverse the range [R, 10000000 + 1]
for (int i = R; i < 1000001; i++) {
// If i is greater than R and prime
if (i > R && prime[i] == true) {
// Return i
return i;
}
}
// Dummy value to omit return error
return -1;
}
// Driver Code
public static void main(String[] args)
{
// Given array
int arr[] = { 10, 13, 17, 19 };
// Function call
System.out.println(find_X(arr));
}
}
// This code is contributed by Kingash.Python3
# Python3 program for the above approach
import math
# Function to find an element which
# is coprime with all array elements
def find_X(arr):
# Stores maximum array element
R = max(arr)
# Stores if index of an array is prime or not
prime = [True for i in range(0, 1000001)]
p = 2
while (p * p <= 1000002):
# If prime[p] is not changed,
# then it is a prime
if (prime[p] == True):
# Update all multiples of p
for i in range(p * 2, 1000001, p):
prime[i] = False
# Increment p by 1
p = p + 1
prime[0] = False
prime[1] = False
# Traverse the range [R, 10000000 + 1]
for i in range(R, 1000001):
# If i is greater than R and prime
if i > R and prime[i] == True:
# Return i
return i
# Driver Code
arr = [10, 13, 17, 19]
print(find_X(arr))C#
// C# program for the above approach
using System;
class GFG{
// Function to find an element which
// is coprime with all array elements
static int find_X(int[] arr)
{
// Stores maximum array element
int R = Int32.MinValue;
for(int i = 0; i < arr.Length; i++)
R = Math.Max(R, arr[i]);
// Stores if index of an array is prime or not
bool[] prime = new bool[1000001];
for(int i = 0; i < 1000001; i++)
{
prime[i] = true;
}
int p = 2;
while (p * p <= 1000002)
{
// 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 < 1000001; i += p)
{
prime[i] = false;
}
}
// Increment p by 1
p = p + 1;
}
prime[0] = false;
prime[1] = false;
// Traverse the range [R, 10000000 + 1]
for(int i = R; i < 1000001; i++)
{
// If i is greater than R and prime
if (i > R && prime[i] == true)
{
// Return i
return i;
}
}
// Dummy value to omit return error
return -1;
}
// Driver Code
public static void Main(String []args)
{
// Given array
int[] arr = { 10, 13, 17, 19 };
// Function call
Console.WriteLine(find_X(arr));
}
}
// This code is contributed by souravghosh0416输出：
23时间复杂度： O(N * log(N))
辅助空间： O(N)