# BFRAC - Editorial

Contest

Author: Rahul Sharma

Tester: Sumukh Bhardwaj

Editorialist: Rahul Sharma

Easy

# PREREQUISITES:

Basic programming, modulo, math

# PROBLEM:

Given numerator Num and denominator Den of a fraction, a non-negative integer K and a positive integer N. A is given as decimal part of fraction of K length. You need to tell the value of A \mod N.

# EXPLANATION:

rem = Num \mod Den
sum = 0

To get first decimal digit
digit = (rem*10)/Den
rem = (rem*10) \mod Den

Since we want to take the modulo N of the decimal part we will use Modular Arithmetic to update the sum
sum = ((sum \mod N * 10 \mod N ) \mod N + digit \mod N) \mod N
This will ensure that we never overflow.
Repeating this K times will give the answer which will be present in sum

# COMPLEXITIES:

Time Complexity: O(K) for each test case
Space Complexity: O(1) for each test case

# SOLUTIONS:

Setter's Solution
``````#include<bits/stdc++.h>
using namespace std;

int main()
{
long long int a, b, k, n, sum = 0, add = 0, rem;

cin >> a >> b >> k >> n;

rem = a % b;

long long int i = 0;

while(i < k)
{