直线数

给定数字N ，任务是检查N是否为直线数。如果N是直线数，则打印“是”，否则打印“否” 。

A straight-line Number is a number greater than 99 in which the digits form an Arithmetic Progression.

例子：

Input: N = 135
Output: Yes
Explanation:
The digits of the number N are 1, 3, and 5.
1 3 5 is an AP with d = 2

Input: N = 136
Output: No

方法：想法是将给定数字N转换为字符串，并检查连续数字之间的差异是否相同。如果所有对应数字之间的差异相同，则打印“是”，否则打印“否” 。

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 N is a
// Straight Line number or not
bool isStraighLineNum(int N)
{
    // N must be > 99
    if (N <= 99)
        return false;
  
    string str = to_string(N);
  
    // Difference between consecutive
    // digits must be same
    int d = str[1] - str[0];
  
    for (int i = 2; i < str.length(); i++)
        if (str[i] - str[i - 1] != d)
            return false;
  
    return true;
}
  
// Driver Code
int main()
{
    // Given Number N
    int N = 135;
  
    // Function Call
    if (isStraighLineNum(n))
        cout << "Yes";
    else
        cout << "No";
}

Java

// Java program for the above approach
class GFG{ 
  
// Function to check if N is a
// Straight Line number or not
static boolean isStraighLineNum(int N)
{
      
    // N must be > 99
    if (N <= 99)
        return false;
  
    String s = Integer.toString(N);
  
    // Difference between consecutive
    // digits must be same
    int d = s.charAt(1) - s.charAt(0);
  
    for(int i = 2; i < s.length(); i++)
       if (s.charAt(i) - s.charAt(i - 1) != d)
           return false;
  
    return true;
}
  
// Driver code 
public static void main(String[] args) 
{ 
      
    // Given Number N
    int n = 135;
  
    // Function Call
    if (isStraighLineNum(n))
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
} 
} 
  
// This code is contributed by Pratima Pandey

Python3

# Python3 program for the above approach
  
# Function to check if N is a
# Straight Line number or not
def isStraighLineNum(N):
  
    # N must be > 99
    if (N <= 99):
        return False;
  
    str1 = str(N);
  
    # Difference between consecutive
    # digits must be same
    d = int(str1[1]) - int(str1[0]);
  
    for i in range(2, len(str1)):
        if (int(str1[i]) - 
            int(str1[i - 1]) != d):
            return False;
  
    return True;
  
# Driver Code
  
# Given Number N
N = 135;
  
# Function Call
if (isStraighLineNum(N)):
    print("Yes");
else:
    print("No");
  
# This code is contributed by Code_Mech

C#

// C# program for the above approach
using System;
class GFG{ 
  
// Function to check if N is a
// Straight Line number or not
static bool isStraighLineNum(int N)
{
      
    // N must be > 99
    if (N <= 99)
        return false;
  
    string s = N.ToString();
  
    // Difference between consecutive
    // digits must be same
    int d = s[1] - s[0];
  
    for(int i = 2; i < s.Length; i++)
       if (s[i] - s[i - 1] != d)
           return false;
  
    return true;
}
  
// Driver code 
public static void Main() 
{ 
      
    // Given Number N
    int n = 135;
  
    // Function Call
    if (isStraighLineNum(n))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
} 
} 
  
// This code is contributed by Code_Mech

输出：

Yes

时间复杂度： O(log ₁₀ N)