计算与它的反面相差 X 的三位数

给定一个整数X ，任务是计算与它的反面相差X的三位数的总数。如果不存在这样的数字，则打印 -1。

例子：

Input: X = 792
Output : 10
Explanation :
901 – 109 = 792
911 – 119 = 792
921 – 129 = 792
931 – 139 = 792
941 – 149 = 792
951 – 159 = 792
961 – 169 = 792
971 – 179 = 792
981 – 189 = 792
991 – 199 = 792

Input: X = 0
Output: 90

方法：根据以下观察可以解决给定的问题：

Let N = rpq 
Therefore, N = 100r + 10q + p 
Therefore, reverse of N = 100p + 10q + r 
Therefore, the problem reduces to solving (100r + 10q + p) – (r + 10q + 100p) = X 
-> 99(r – p) = X 
-> r – p = X / 99
Therefore, if given X is a multiple of 99, then solution exists. 
 

根据以上观察，按照以下步骤解决问题：

• 检查 X 是否是 99 的倍数。如果发现不正确，则打印 -1 作为不存在解决方案。
• 否则，计算X / 99 。使用数字 [1, 9] 生成所有对，对于每一对，检查它们的差值是否等于X / 99 。
• 如果发现任何对都为真，则将 count 增加 10，因为可以将中间数字置换为所获得对的 [0, 9] 范围内的任何值。
• 最后，打印获得的计数值。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to count three-digit
// numbers having difference x
// with its reverse
int Count_Number(int x)
{
    int ans = 0;
 
    // if x is not multiple of 99
    if (x % 99 != 0) {
 
        // No solution exists
        ans = -1;
    }
    else {
 
        int diff = x / 99;
 
        // Generate all possible pairs
        // of digits [1, 9]
        for (int i = 1; i < 10; i++) {
            for (int j = 1; j < 10; j++) {
 
                // If any pair is obtained
                // with difference x / 99
                if ((i - j) == diff) {
 
                    // Increase count
                    ans += 10;
                }
            }
        }
    }
 
    // Return the count
    return ans;
}
 
// Driver Code
int main()
{
    int x = 792;
    cout << Count_Number(x) << endl;
 
    return 0;
}

Java

// Java program to implement
// the above approach
import java.io.*;
import java.util.Arrays;
 
class GFG{
  
// Function to count three-digit
// numbers having difference x
// with its reverse
static int Count_Number(int x)
{
    int ans = 0;
   
    // If x is not multiple of 99
    if (x % 99 != 0)
    {
         
        // No solution exists
        ans = -1;
    }
    else
    {
        int diff = x / 99;
   
        // Generate all possible pairs
        // of digits [1, 9]
        for(int i = 1; i < 10; i++)
        {
            for(int j = 1; j < 10; j++)
            {
                 
                // If any pair is obtained
                // with difference x / 99
                if ((i - j) == diff)
                {
                     
                    // Increase count
                    ans += 10;
                }
            }
        }
    }
     
    // Return the count
    return ans;
}
  
// Driver Code
public static void main (String[] args)
{
    int x = 792;
     
    System.out.println(Count_Number(x));
}
}
 
// This code is contributed by sanjoy_62

Python3

# Python3 program to implement
# the above approach
 
# Function to count three-digit
# numbers having difference x
# with its reverse
def Count_Number(x):
   
    ans = 0;
 
    # If x is not multiple
    # of 99
    if (x % 99 != 0):
 
        # No solution exists
        ans = -1;
    else:
        diff = x / 99;
 
        # Generate all possible pairs
        # of digits [1, 9]
        for i in range(1, 10):
            for j in range(1, 10):
 
                # If any pair is obtained
                # with difference x / 99
                if ((i - j) == diff):
                    # Increase count
                    ans += 10;
 
    # Return the count
    return ans;
 
# Driver Code
if __name__ == '__main__':
   
    x = 792;
    print(Count_Number(x));
 
# This code is contributed by shikhasingrajput

C#

// C# program to implement
// the above approach 
using System;
 
class GFG{
  
// Function to count three-digit
// numbers having difference x
// with its reverse
static int Count_Number(int x)
{
    int ans = 0;
    
    // If x is not multiple of 99
    if (x % 99 != 0)
    {
         
        // No solution exists
        ans = -1;
    }
    else
    {
        int diff = x / 99;
    
        // Generate all possible pairs
        // of digits [1, 9]
        for(int i = 1; i < 10; i++)
        {
            for(int j = 1; j < 10; j++)
            {
                  
                // If any pair is obtained
                // with difference x / 99
                if ((i - j) == diff)
                {
                     
                    // Increase count
                    ans += 10;
                }
            }
        }
    }
      
    // Return the count
    return ans;
}
  
// Driver Code
public static void Main()
{
    int x = 792;
      
    Console.WriteLine(Count_Number(x));
}
}
 
// This code is contributed by code_hunt

Javascript

 
// Javascript program to implement
// the above approach
 
// Function to count three-digit
// numbers having difference x
// with its reverse
function Count_Number(x)
{
    let ans = 0;
    
    // If x is not multiple of 99
    if (x % 99 != 0)
    {
         
        // No solution exists
        ans = -1;
    }
    else
    {
        let diff = x / 99;
    
        // Generate all possible pairs
        // of digits [1, 9]
        for(let i = 1; i < 10; i++)
        {
            for(let j = 1; j < 10; j++)
            {
                 
                // If any pair is obtained
                // with difference x / 99
                if ((i - j) == diff)
                {
                      
                    // Increase count
                    ans += 10;
                }
            }
        }
    }
      
    // Return the count
    return ans;
}
 
// Driver code
let x = 792;
  
document.write(Count_Number(x));
 
// This code is contributed by splevel62
 

输出

10

时间复杂度： O(1)
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