c - 在 C 编程中设置循环限制并计算回文

Question

所以这段代码的目标是将用户号转换为回文。如果该数不是回文数，则计算该数+反转数，直到成为回文数。如果仍然不是回文，则取最后一个数字的总和并将其反转并将其相加，最多尝试 10 次。

int main()
{
    int n;
    int reverse = 0;    //initial value will be 0
    int temp;           //temporary variable

//Check if number is a palindrome===============================

    printf("Enter an integer: ");
    scanf_s("%d", &n);
    temp = n;           //Make input number have a temporary variable

    while (temp != 0)
    {
        reverse = reverse * 10;
        reverse = reverse + temp % 10;
        temp = temp / 10;
    }

//Check if number entered by user and it's reverse number is equal
if (reverse == n)
    printf("%d\t1, reverse is %d\n", n, reverse);    //Is a palindrome
else
    printf("%d\t0, reverse is %d\n", n, reverse);   //Not a palindrome
//==========================================================================

//Keep adding numbers until it reaches a palindrome
    int sum;

    while (n /= reverse)
    {
        sum = n + reverse;
        n++;
    }

if (reverse == sum)
    printf("%d it works", sum, reverse);
else
    ("%d didn't work", sum, reverse);

我还没有达到极限。但我的问题是，如何将逆序和用户号码相加，然后求和+其逆序？另外，什么样的循环最适合此类问题？

Answer 1

你可以做的就是在另一个 while 循环(考虑为循环 1)中编写用于生成数字相反的 while 循环(考虑为循环 2)，当数字是回文时，该循环会中断(终止)。如果给定的数字不是回文，那么您将数字更改为 number=number+reverse。如果您想尝试 10 次，那么您可以通过使用计数器为循环 1 添加另一个条件。下面是我写的代码。

#include<stdio.h>
int main()
{
    int n;
    int count=0;         //taking counter for number of trials
    int reverse=0;      //initial value will be 0
    int temp;           //temporary variable

    printf("Enter an integer: ");
    scanf("%d", &n);

    while (count<10)
    //count less than 10 condition makes the max. no. of trials equal to 10   
    {   
        temp=n;   //assigning a temporary variable to n
        reverse=0; 
        //Finding the reverse of the number n===============================
        while (temp != 0)    
        {
            reverse = reverse * 10;
            reverse = reverse + temp % 10;
            temp = temp / 10;
        }

        //Check if number entered by user and it's reverse number is equal
        if (reverse == n){
            printf("\n%d\t1, reverse is %d\n", n, reverse);
            //Is a palindrome
            break; 
            /*break condition makes the loop 1 to terminate if the number 'n' is palindrome*/  
        }
        else
            printf("\n%d\t0, reverse is %d\n", n, reverse);   
            //Not a palindrome
        //================================================================

        //Keep adding numbers until it reaches a palindrome
        n=n+ reverse;  
        count++;   
         }
 }