通过用它们的 GCD 替换对，将前 N 个自然数的序列修改为给定的数组

给定一个正整数N和阵列ARR []中，任务是检查通过选择任何一对（如果第一n个自然数的序列，即{1，2，3，… N}可以由等于给Arr [] i, j)来自序列并用i和j的 GCD 替换i和j 。如果可能，请打印“是” 。否则，打印“否” 。

例子：

Input: N = 4, arr[] = {1, 2, 3, 2}
Output: Yes
Explanation: For the pair (2, 4) in the sequence {1, 2, 3, 4}, GCD(2, 4) = 2. Now, the sequence modifies to {1, 2, 3, 2}, which is same as arr[].

Input: N = 3, arr[] = {1, 2, 2}
Output: No

方法：这个想法是基于这样一个事实，即两个数字的 GCD 介于1和两个数字的最小值之间。根据 gcd 的定义，它是将两者分开的最大数。因此，当且仅当存在某个作为其因数的数字时，才使索引处的数字变小。因此，可以得出结论，对于数组中的每个^第i^个索引，如果以下条件成立，则数组arr[]可以从前N 个自然数的序列中获得。

(i + 1) % arr[i] == 0

请按照以下步骤解决问题：

• 使用变量i遍历数组arr[] 。
• 对于每个^第i^个索引，检查 ( i + 1) % arr[i]是否等于0 。如果发现任何数组元素为 false，则打印“No” 。
• 否则，在完全遍历数组后，打印“Yes” 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to check if array arr[]
// can be obtained from first N
// natural numbers or not
void isSequenceValid(vector& B,
                     int N)
{
    for (int i = 0; i < N; i++) {
 
        if ((i + 1) % B[i] != 0) {
            cout << "No";
            return;
        }
    }
 
    cout << "Yes";
}
 
// Driver Code
int main()
{
    int N = 4;
    vector arr{ 1, 2, 3, 2 };
 
    // Function Call
    isSequenceValid(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
class GFG{
      
// Function to check if array arr[]
// can be obtained from first N
// natural numbers or not
static void isSequenceValid(int[] B,
                            int N)
{
    for(int i = 0; i < N; i++)
    {
        if ((i + 1) % B[i] != 0)
        {
            System.out.print("No");
            return;
        }
    }
    System.out.print("Yes");
}
  
// Driver code
public static void main(String[] args)
{
    int N = 4;
    int[] arr = { 1, 2, 3, 2 };
     
    // Function Call
    isSequenceValid(arr, N);
}
}
  
// This code is contributed by sanjoy_62

Python3

# Python3 program for the above approach
 
# Function to check if array arr[]
# can be obtained from first N
# natural numbers or not
def isSequenceValid(B, N):
     
    for i in range(N):
        if ((i + 1) % B[i] != 0):
            print("No")
            return
         
    print("Yes")
     
# Driver Code
N = 4
arr = [ 1, 2, 3, 2 ]
  
# Function Call
isSequenceValid(arr, N)
 
# This code is contributed by susmitakundugoaldanga

C#

// C# program for the above approach
using System;
 
class GFG{
       
// Function to check if array arr[]
// can be obtained from first N
// natural numbers or not
static void isSequenceValid(int[] B,
                            int N)
{
    for(int i = 0; i < N; i++)
    {
        if ((i + 1) % B[i] != 0)
        {
            Console.WriteLine("No");
            return;
        }
    }
    Console.WriteLine("Yes");
}
   
// Driver code
public static void Main()
{
    int N = 4;
    int[] arr = { 1, 2, 3, 2 };
      
    // Function Call
    isSequenceValid(arr, N);
}
}
 
// This code is contributed by code_hunt

Javascript

// Javascript program to implement
// the above approach
 
// Function to check if array arr[]
// can be obtained from first N
// natural numbers or not
function isSequenceValid(B, N)
{
    for(let i = 0; i < N; i++)
    {
        if ((i + 1) % B[i] != 0)
        {
            document.write("No");
            return;
        }
    }
    document.write("Yes");
}
 
// Driver code
 
    let N = 4;
    let arr = [ 1, 2, 3, 2 ];
      
    // Function Call
    isSequenceValid(arr, N);
     
    // This code is contributed by souravghosh0416.

输出：

Yes

时间复杂度： O(N)
辅助空间： O(1)