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Nth Decimal Place

Find the digit at the Nth decimal place when dividing X by Y

Understand the Problem

Problem Statement

Nth Decimal Place: The program must accept three integers X, Y, and N as the input. The program must print the digit present in the Nth decimal place right to the decimal point when X is divided by Y as the output.

Note: If the number of digits in the decimal part of the result is less than N, then print 0 as the output.

Constraints

  • 1 ≤ X, Y ≤ 10^6
  • 1 ≤ N ≤ 10^6
  • Division result should be calculated with sufficient precision to determine the Nth decimal place

Examples

Example 1
Input
22 7 12
Output
7
Explanation

22/7 = 3.142857142857... The 12th decimal digit is 7.

Example 2
Input
4 2 1
Output
0
Explanation

4/2 = 2.0, so there are no decimal places. The 1st decimal digit is 0.

Example 3
Input
1 3 5
Output
3
Explanation

1/3 = 0.333333... The 5th decimal digit is 3.

Solution

#include <stdio.h>

int main() {
    int x, y, n;
    scanf("%d %d %d", &x, &y, &n);
    
    // Get the remainder after integer division
    int remainder = x % y;
    
    // If no decimal part exists
    if (remainder == 0) {
        printf("0\n");
        return 0;
    }
    
    // Find the Nth decimal digit
    for (int i = 0; i < n; i++) {
        remainder *= 10;
        int digit = remainder / y;
        remainder = remainder % y;
        
        // If we've reached the end of decimal expansion
        if (remainder == 0) {
            if (i == n - 1) {
                printf("%d\n", digit);
            } else {
                printf("0\n");
            }
            return 0;
        }
        
        // If this is the Nth decimal place
        if (i == n - 1) {
            printf("%d\n", digit);
            return 0;
        }
    }
    
    return 0;
}
Time:O(N) - We perform N iterations to reach the Nth decimal place
Space:O(1) - Only using a constant amount of extra space
Approach:

The C solution uses modular arithmetic to find the Nth decimal digit:

  1. Read input values X, Y, and N
  2. Calculate the remainder after integer division (X % Y)
  3. If remainder is 0, there's no decimal part, so output 0
  4. Loop N times to reach the Nth decimal place:
    • Multiply remainder by 10
    • The quotient when dividing by Y gives the current decimal digit
    • Update remainder to (remainder * 10) % Y
    • If remainder becomes 0 during the process, we've reached the end of decimal expansion
  5. Output the digit found at the Nth position

This approach avoids floating-point precision issues and works efficiently.

Visual Explanation

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