给定一个由N个正整数组成的数组arr [] ，任务是通过交换对(arr [i]，arr [j])使得i！= j(1≤i，j≤ n)和GCD (arr [i]，arr [j])等于数组中存在的最小元素。
例子：
Input: arr[] = {4, 3, 6, 6, 2, 9}
Output: Yes
Explanation:
Minimum array element = 2.
Swap arr[0] and arr[2], since gcd(4, 6) = 2. Array modifies to {6, 3, 4, 6, 2, 9}.
Swap arr[0] and arr[4], since gcd(6, 2) = 2.
Therefore, the modified array {2, 3, 4, 6, 6, 9} is non-decreasing.
Input: arr[] = {7, 5, 2, 2, 4}
Output: No
方法：
- 首先，遍历数组以找到最小元素。将所有这些元素存储在另一个数组中并对该数组进行排序。
- 对于每个数组元素，检查其是否在正确的位置。如果发现为真，则继续进行下一个元素。否则，请检查它是否可以被数组的最小元素整除，因为只有这些元素将具有GCD，而其他元素的GCD等于数组的最小元素。
- 如果任何数组元素不在其正确位置并且该元素不能被数组的最小元素整除，则打印“否”。
下面是上述方法的实现：
C++
// C++ program for above approach
#include
using namespace std;
// Function to check if the array
// can be made non-decreasing
bool check(int a[], int n)
{
int b[n];
int minElement = INT_MAX;
// Iterate till N
for(int i = 0; i < n; i++)
{
// Find the minimum element
b[i] = a[i];
minElement = min(minElement, a[i]);
}
// Sort the array
sort(b, b + n);
int k = 1;
// Iterate till N
for(int i = 0; i < n; i++)
{
// Check if the element is
// at its correct position
if (a[i] != b[i] &&
a[i] % minElement != 0)
{
k = 0;
break;
}
}
// Return the answer
return k == 1 ? true : false;
}
// Driver Code
int main()
{
int a[] = { 4, 3, 6, 6, 2, 9 };
int n = sizeof(a) / sizeof(a[0]);
// Print the answer
if (check(a, n) == true)
cout << "Yes \n";
else
cout<<"No \n";
return 0;
}
// This code is contributed by akhilsaini Java
// Java program for above approach
import java.io.*;
import java.util.*;
import java.math.*;
class GFG {
// Function to check if the array
// can be made non-decreasing
public static boolean check(int[] a, int n)
{
int[] b = new int[n];
int minElement = Integer.MAX_VALUE;
// Iterate till N
for (int i = 0; i < n; i++) {
// Find the minimum element
b[i] = a[i];
minElement
= Math.min(minElement, a[i]);
}
// Sort the array
Arrays.sort(b);
int k = 1;
// Iterate till N
for (int i = 0; i < n; i++) {
// Check if the element is
// at its correct position
if (a[i] != b[i]
&& a[i] % minElement != 0) {
k = 0;
break;
}
}
// Return the answer
return k == 1 ? true : false;
}
// Driver Code
public static void main(String[] args)
{
int[] a = { 4, 3, 6, 6, 2, 9 };
int n = a.length;
// Print the answer
if (check(a, n) == true) {
System.out.println("Yes");
}
else {
System.out.println("No");
}
}
}Python3
# Python3 program for above approach
import sys
# Function to check if the array
# can be made non-decreasing
def check(a, n):
b = [None] * n
minElement = sys.maxsize
# Iterate till N
for i in range(0, n):
# Find the minimum element
b[i] = a[i]
minElement = min(minElement, a[i])
# Sort the array
b.sort()
k = 1
# Iterate till N
for i in range(0, n):
# Check if the element is
# at its correct position
if ((a[i] != b[i]) and
(a[i] % minElement != 0)):
k = 0
break
# Return the answer
if k == 1:
return True
else:
return False
# Driver Code
if __name__ == "__main__":
a = [ 4, 3, 6, 6, 2, 9 ]
n = len(a)
# Print the answer
if check(a, n) == True:
print("Yes")
else:
print("No")
# This code is contributed by akhilsainiC#
// C# program for above approach
using System;
class GFG{
// Function to check if the array
// can be made non-decreasing
static bool check(int[] a, int n)
{
int[] b = new int[n];
int minElement = int.MaxValue;
// Iterate till N
for(int i = 0; i < n; i++)
{
// Find the minimum element
b[i] = a[i];
minElement = Math.Min(minElement, a[i]);
}
// Sort the array
Array.Sort(b);
int k = 1;
// Iterate till N
for(int i = 0; i < n; i++)
{
// Check if the element is
// at its correct position
if (a[i] != b[i] &&
a[i] % minElement != 0)
{
k = 0;
break;
}
}
// Return the answer
return k == 1 ? true : false;
}
// Driver Code
static public void Main()
{
int[] a = { 4, 3, 6, 6, 2, 9 };
int n = a.Length;
// Print the answer
if (check(a, n) == true)
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by akhilsaini输出：
Yes
时间复杂度： O(N logN)
辅助空间： O(N)