检查前N个自然数的串联是否可被3整除

给定整数N ，任务是检查前N个自然数的串联是否可被3整除。如果是可分割的，则打印是；否则，则打印“否” 。

例子：

Input: N = 3 
Output: Yes 
Explanation: 
The concatenated number = 123 
Since it is divisible by 3, the output is Yes
Input: N = 7 
Output: No 
Explanation: The concatenated number = 1234567 
Since it is not divisible by 3, the output is No. 
 

天真的方法：
最简单的方法是连接前N个自然数，并计算结果数的位数之和，然后检查该数是否可被3整除。
时间复杂度： O(N)
辅助空间： O(1)
高效方法：
为了优化上述方法，我们可以观察到一种模式。第一N的自然数的级联是不被3整除以下系列1，4，7，10，13，16，19，等等。该级数的第N^个项由公式3×n +1给出。因此，如果(N – 1)不能被3整除，则结果数可以被3整除，因此打印Yes 。否则，打印编号。
下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
 
using namespace std;
 
// Function that returns True if
// concatenation of first N natural
// numbers is divisible by 3
bool isDivisible(int N)
{
    // Check using the formula
    return (N - 1) % 3 != 0;
}
 
 
// Driver Code
int main()
{
    // Given Number
    int N = 6;
     
    // Function Call
    if (isDivisible(N))
        cout << ("Yes");
     
    else
        cout << ("No");
     
    return 0;
}
 
// This code is contributed by Mohit Kumar

Java

// Java program for the above approach
class GFG{
  
// Function that returns True if
// concatenation of first N natural
// numbers is divisible by 3
static boolean isDivisible(int N)
{
    // Check using the formula
    return (N - 1) % 3 != 0;
}
  
  
// Driver Code
public static void main(String[] args)
{
    // Given Number
    int N = 6;
      
    // Function Call
    if (isDivisible(N))
        System.out.println("Yes");
      
    else
        System.out.println("No");
}
}
 
// This code is contributed by Ritik Bansal

Python 3

# Python program for the above approach
 
# Function that returns True if
# concatenation of first N natural
# numbers is divisible by 3
def isDivisible(N):
 
    # Check using the formula
    return (N - 1) % 3 != 0
 
 
# Driver Code
if __name__ == "__main__":
 
    # Given Number
    N = 6
 
    # Function Call
    if (isDivisible(N)):
        print("Yes")
 
    else:
        print("No")

C#

// C# program for the above approach
using System;
class GFG{
 
// Function that returns True if
// concatenation of first N natural
// numbers is divisible by 3
static bool isDivisible(int N)
{
    // Check using the formula
    return (N – 1) % 3 != 0;
}
 
// Driver Code
public static void Main()
{
    // Given Number
    int N = 6;
     
    // Function Call
    if (isDivisible(N))
    Console.Write("Yes");
     
    else
    Console.Write("No");
}
}
 
// This code is contributed by Code_Mech

Javascript

// javascript program for the above approach
 
// Function that returns True if
// concatenation of first N natural
// numbers is divisible by 3
function isDivisible(N)
{
    // Check using the formula
    return (N - 1) % 3 != 0;
}
  
// Driver Code
 
// Given Number
var N = 6;
  
// Function Call
if (isDivisible(N))
    document.write("Yes");
else
    document.write("No");
 
// This code is contributed by Princi Singh.

输出：

Yes

时间复杂度： O(1)
辅助空间： O(1)