# 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 digits 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.

**Visual Presentation:**

**Sample Solution:**

**C Code:**

```
#include <stdio.h>
// Function to count the total number of occurrences of digit 1 in numbers up to the given limit
static int count_DigitOne(int n) {
int m = 0, k = 0, x = 0, base = 1;
// Loop to iterate through each digit of the number
while (n > 0) {
k = n % 10; // Get the current digit
n = n / 10; // Update the number by removing the last digit
// Count occurrences of digit 1 based on the current digit
if (k > 1) {
x += (n + 1) * base; // Increment occurrences if the digit is greater than 1
} else if (k < 1) {
x += n * base; // Increment occurrences if the digit is less than 1
} else {
x += n * base + m + 1; // Increment occurrences if the digit is equal to 1
}
m += k * base; // Update the count of occurrences so far
base *= 10; // Update the base for the next digit
}
return x; // Return the total count of digit 1 occurrences
}
// Main function to test the count_DigitOne function with different values
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.

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