C Exercises: Count the total number of digit 1 appearing in all positive integers less than or equal to a given integer n

C Programming Mathematics: Exercise-11 with Solution

Write a C program to count the total number of digit 1 appearing in all positive integers less than or equal to a given integer n.

For example:
Input n = 12,
Return 5, because digit 1 occurred 5 times in the following numbers: 1, 10, 11, 12.

Example:
Input:
n = 12
n = 30
Output:
Total number of digit 1 appearing in 12 (less than or equal) is 5.
Total number of digit 1 appearing in 30 (less than or equal) is 13.

Pictorial Presentation:

Sample Solution:

C Code:

#include <stdio.h>

static int count_DigitOne(int n) {
        int m = 0, k = 0, x = 0, base = 1;
        while (n > 0) {
            k = n % 10;
            n = n / 10;

            if (k > 1) { x += (n+1)*base; }
            else if (k < 1) { x += n*base; }
            else { x += n*base+m+1; }

            m += k*base;
            base *= 10;
        }
        return x;
    }

int main(void)
{
    int n = 12;
    printf("\nTotal number of digit 1 appearing in %d (less than or equal) is %d.", n, count_DigitOne(n));
    n = 30;
    printf("\nTotal number of digit 1 appearing in %d (less than or equal) is %d.", n, count_DigitOne(n));
    return 0;
}

Sample Output:

Total number of digit 1 appearing in 12 (less than or equal) is 5.
Total number of digit 1 appearing in 30 (less than or equal) is 13.

Flowchart:

C Programming Code Editor:

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