给定一个由N个正整数组成的数组arr [] ，任务是确定最小的正整数K ，使得没有一个数组元素不能被K整除。如果没有这样的整数
例子：
Input: arr[] = {3, 2, 6, 9, 2}
Output: 4
Explanation: None of the array elements is divisible by 4( smallest positive ).
Input: arr[] = {3, 5, 1, 19, 11}
Output: 2
方法：请按照以下步骤解决问题：
- 找到给定数组的最大元素，例如maxE 。
- 使用变量i遍历[1，maxE + 1]的范围，并检查给定数组中是否有可被i整除的整数。如果发现为真，则检查该范围内的下一个整数。
- 否则，打印当前数字i 。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the smallest number
// which doesn't divides any integer in
// the given array arr[]
void smallestNumber(int arr[], int len)
{
int maxi = 0;
// Traverse the array arr[]
for (int i = 0; i < len; i++) {
// Maximum array elemnt
maxi = std::max(maxi, arr[i]);
}
// Initialize variable
int ans = -1;
// Traverse from 2 to max
for (int i = 2; i < maxi + 2; i++) {
// Stores if any such
// integer is found or not
bool flag = true;
for (int j = 0; j < len; j++) {
// If any array element
// is divisible by j
if (arr[j] % i == 0) {
flag = false;
break;
}
}
if (flag) {
// Smallest integer
ans = i;
break;
}
}
// Print the answer
cout << ans;
}
// Driver Code
int main()
{
int arr[] = { 3, 2, 6, 9, 2 };
int N = sizeof(arr)
/ sizeof(arr[0]);
// Function Call
smallestNumber(arr, N);
return 0;
} Java
// Java program for the above approach
class GFG
{
// Function to find the smallest number
// which doesn't divides any integer in
// the given array arr[]
static void smallestNumber(int arr[], int len)
{
int maxi = 0;
// Traverse the array arr[]
for (int i = 0; i < len; i++)
{
// Maximum array elemnt
maxi = Math.max(maxi, arr[i]);
}
// Initialize variable
int ans = -1;
// Traverse from 2 to max
for (int i = 2; i < maxi + 2; i++)
{
// Stores if any such
// integer is found or not
boolean flag = true;
for (int j = 0; j < len; j++)
{
// If any array element
// is divisible by j
if (arr[j] % i == 0)
{
flag = false;
break;
}
}
if (flag)
{
// Smallest integer
ans = i;
break;
}
}
// Print the answer
System.out.print(ans);
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 3, 2, 6, 9, 2 };
int N = arr.length;
// Function Call
smallestNumber(arr, N);
}
}
// This code is contributed by shikhasingrajputPython3
# Python3 program for the above approach
# Function to find the smallest number
# which doesn't divides any integer in
# the given array arr[]
def smallestNumber(arr, lenn):
maxi = 0
# Traverse the array arr[]
for i in range(lenn):
# Maximum array elemnt
maxi = max(maxi, arr[i])
# Initialize variable
ans = -1
# Traverse from 2 to max
for i in range(2, maxi + 2):
# Stores if any such
# integer is found or not
flag = True
for j in range(lenn):
# If any array element
# is divisible by j
if (arr[j] % i == 0):
flag = False
break
if (flag):
# Smallest integer
ans = i
break
# Prthe answer
print (ans)
# Driver Code
if __name__ == '__main__':
arr = [3, 2, 6, 9, 2]
N = len(arr)
#Function Call
smallestNumber(arr, N)
# This code is contributed by mohit kumar 29.C#
// C# program for the above approach
using System;
public class GFG
{
// Function to find the smallest number
// which doesn't divides any integer in
// the given array arr[]
static void smallestNumber(int []arr, int len)
{
int maxi = 0;
// Traverse the array arr[]
for (int i = 0; i < len; i++)
{
// Maximum array elemnt
maxi = Math.Max(maxi, arr[i]);
}
// Initialize variable
int ans = -1;
// Traverse from 2 to max
for (int i = 2; i < maxi + 2; i++)
{
// Stores if any such
// integer is found or not
bool flag = true;
for (int j = 0; j < len; j++)
{
// If any array element
// is divisible by j
if (arr[j] % i == 0)
{
flag = false;
break;
}
}
if (flag)
{
// Smallest integer
ans = i;
break;
}
}
// Print the answer
Console.WriteLine(ans);
}
// Driver Code
public static void Main(string[] args)
{
int []arr = { 3, 2, 6, 9, 2 };
int N = arr.Length;
// Function Call
smallestNumber(arr, N);
}
}
// This code is contributed by AnkThon输出：
4时间复杂度： O(N * max)其中max是给定数组中的最大元素
辅助空间： O(N)