检查一个数字的所有前缀是否可以被剩余的位数整除

给定一个数N，任务是检查对于i ( 0 <= i <= len )的每个值，一个数的前i位是否可以被(len – i + 1)整除，其中len是N中的位数。如果发现是真的，则打印“是” 。否则，打印“否”。

例子：

Input: N = 52248.
Output: Yes
Explanation: 

• 5 is divisible by 5
• 52 is divisible by 4
• 522 is divisible by 3
• 5224 is divisible by 2
• 52248 is divisible by 1

Input: N = 59268
Output : No
 

方法：想法是遍历给定数字的所有前缀，对于每个前缀，检查它是否满足条件。

请按照以下步骤解决问题：

• 初始化一个变量，比如i为1，以保持(len – i + 1 ) 的值。
• 在n大于0 时迭代。
  • 如果N不能被i 整除，则返回false。
  • 否则，将N除以10并将i的值增加1 。
• 最后，返回真。

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 all prefixes
// of a number is divisible by
// remaining count of digits or not
bool prefixDivisble(int n)
{
    int i = 1;
 
    while (n > 0) {
 
        // Traverse and check divisibility
        // for each updated number
        if (n % i != 0)
            return false;
 
        // Update the original number
        n = n / 10;
 
        i++;
    }
 
    return true;
}
 
// Driver Code
int main()
{
    // Given Input
    int n = 52248;
 
    // Function Call
    if (prefixDivisble(n))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
 
    return 0;
}

Java

// Java program  for the above approach
 
import java.io.*;
 
class GFG{
     
// Function to check if all prefixes
// of a number is divisible by
// remaining count of digits or not
public static boolean prefixDivisble(int n)
{
    int i = 1;
 
    while (n > 0)
    {
         
        // Traverse and check divisibility
        // for each updated number
        if (n % i != 0)
            return false;
 
        // Update the original number
        n = n / 10;
 
        i++;
    }
    return true;
}
 
// Driver Code
public static void main (String[] args)
{
 
    // Given Input
    int n = 52248;
     
    // Function Call
    if (prefixDivisble(n))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
 
// This code is contributed by lokeshpotta20.

Python3

# Python3 program for the above approach
 
# Function to check if all prefixes of
# a number is divisible by remaining count
# of digits or not
def prefixDivisble(n):
     
    i = 1
     
    while n > 0:
         
        # Traverse and check divisibility
        # for each updated number
        if n % i != 0:
            return False
         
        # Update the original number
        n = n // 10
        i += 1
     
    return True
 
# Driver Code
 
# Given Input
n = 52248
 
# Function Call
if (prefixDivisble(n)):
   print("Yes")
else:
   print("No")
 
# This code is contributed by abhivick07

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
  
// Function to check if all prefixes
// of a number is divisible by
// remaining count of digits or not
static bool prefixDivisble(int n)
{
    int i = 1;
 
    while (n > 0)
    {
         
        // Traverse and check divisibility
        // for each updated number
        if (n % i != 0)
            return false;
 
        // Update the original number
        n = n / 10;
 
        i++;
    }
    return true;
}
 
// Driver Code
public static void Main()
{
     
    // Given Input
    int n = 52248;
 
    // Function Call
    if (prefixDivisble(n))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
 
// This code is contributed by ipg2016107

Javascript

 
// Javascript program for the above approach
 
// Function to check if all prefixes
// of a number is divisible by
// remaining count of digits or not
function prefixDivisble(n)
{
    let i = 1;
     
    while (n > 0)
    {
         
        // Traverse and check divisibility
        // for each updated number
        if (n % i != 0)
            return false;
 
        // Update the original number
        n = parseInt(n / 10);
 
        i++;
    }
    return true;
}
 
// Driver Code
 
// Given Input
let n = 52248;
 
// Function Call
if (prefixDivisble(n) == true)
    document.write("Yes");
else
    document.write("No");
 
// This code is contributed by lokeshpotta20.
 

输出

Yes

时间复杂度： O(len)，其中 len 是N中的位数。
辅助空间： O(1)