1101: numbers in reverse order (function topic)

Title Description

Enter a positive integer n, Calculation n In reverse order m, Output m And n And . Ask the program to define a inverse() Function and a main() function ,inverse() Function receives an integer value , Returns the result of the number in reverse order . for example , Given number 7631, The function returns 1367 . 
int inverse(int n) 
{ 
// Calculate and return n The reverse number of  
} 

Input

A positive integer n, The input guarantees that the calculation results are in int Within the scope of .

Output

Output an integer ,n And n The reverse order of the sum of numbers . Output takes up one line .

The sample input  Copy

1780

Sample output  Copy

2651

See the following for code and details ：

#include <stdio.h>
#include <stdlib.h>

int inverse(int n);

int main()
{
    int n;
    scanf ("%d",&n);
    int nn=inverse(n);//nn Is the number in reverse order of the input number 
    printf ("%d",n+nn); // The sum of the output number itself and the number in reverse order 
    return 0;
}
int inverse(int n)
 {

     int i=0;

    for (int m=n;m>0;i=i+1) // Judge the length of the input number 
    {
        m=m/10;

    }

    int s=0;
    for (int k=0;k<i;k++)
    {
       
        s=s+(int)(n/pow(10,(i-k-1)))*(int)(pow(10,(i-1))/pow(10,(i-k-1)));
        //pow The data type is double, Need to be converted to int type ,s Is an inverse ordinal number 
        int nn=(pow(10,(i-k-1)));
        n=n%nn;



    }

    return s;

 }