C: How to convert a string where each char is followed by its occurrence into absolute string?

Question

I need help in writing the code where I enter the input in which each char is followed by its occurrence and output should be absolute string.

Example
Input a10b5c4
Output aaaaaaaaaabbbbbcccc

Answer 1

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

int main()
{
    char input[100] = "", c, x;
    int i = 0, num = 0;

    printf("Enter the input (char followed by its occurrence)\n");
    scanf("%s", input);

    while ((c = input[i++])) {
        if ((c >= 'a' && c <='z') || (c >= 'A' && c <= 'Z')) {
            x = input[i];
            while(isdigit(x)) {
                num = (num * 10) + (x - '0');
                x = input[++i];
            }
            while(num--)
                printf("%c", c);
            num = 0;
       }
    }

    printf("\n");
    return 0;
}

Here is the expected output for given input

input: a10b5c4
output: aaaaaaaaaabbbbbcccc

Answer 2

#include<stdio.h>
int main()
{
    char inputstring[100],outputstring[1000];
    int i,j,temp;

    printf("Enter the string\n");
    scanf("%s",inputstring);

    i=1;
    j=0;

    char ch=inputstring[0];

    while(inputstring[i]!=NULL)
    {
        if(inputstring[i]>='0'&&inputstring[i]<='9')
        {
            temp=0;
            while(inputstring[i]>='0'&&inputstring[i]<='9')
            {
                temp=temp*10+inputstring[i]-48;
                i++;
            }
            while(temp--)
                outputstring[j++]=ch;
        }
        else
        {
            ch=inputstring[i];
            i++;
        }
    }

    outputstring[j]='\0';
    printf("\n%s",outputstring);

    return 0;
}