Reciprocal Cycles

// https://projecteuler.net/problem=26

// A unit fraction contains 1 in the numerator. The decimal representation of the unit fractions with denominators 2 to 10 are given:

// 1/2 = 0.5
// 1/3 = 0.(3)
// 1/4 = 0.25
// 1/5 = 0.2
// 1/6 = 0.1(6)
// 1/7 = 0.(142857)
// 1/8 = 0.125
// 1/9 = 0.(1)
// 1/10 = 0.1

// Where 0.1(6) means 0.1666..., and has a 1-digit recurring cycle. It can be seen that 1/7 has a 6-digit recurring cycle.

// Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its decimal fraction part.
//
// Answer: 983
//
// Authored by: Tim Varley 💘
// Assisted-by: Grok Code Fast via Crush 💘 <crush@charm.land>

#include <iostream>
#include <map>

int longest_reciprocal_cycle(int max_d) {
    int max_cycle = 0;
    int max_d_val = 0;
    for (int d = 2; d < max_d; ++d) {
        std::map<int, int> remainder_positions;
        int remainder = 1;
        int position = 0;
        while (remainder != 0 && remainder_positions.find(remainder) == remainder_positions.end()) {
            remainder_positions[remainder] = position;
            remainder *= 10;
            remainder %= d;
            ++position;
        }
        if (remainder != 0) {
            int cycle_length = position - remainder_positions[remainder];
            if (cycle_length > max_cycle) {
                max_cycle = cycle_length;
                max_d_val = d;
            }
        }
    }
    return max_d_val;
}

#if ! defined UNITTEST_MODE
int main(int argc, char const *argv[]) {
    std::cout << longest_reciprocal_cycle(1000) << std::endl;
}
#endif // UNITTEST_MODE

package org.tvarley.euler.solutions;

import org.tvarley.euler.Solution;

public class Solution026 implements Solution {
  public String solve() {
    int maxCycle = 0;
    int result = 0;

    for (int d = 2; d < 1000; d++) {
      int len = cycleLength(d);
      if (len > maxCycle) {
        maxCycle = len;
        result = d;
      }
    }

    return Integer.toString(result);
  }

  private int cycleLength(int d) {
    int[] remainders = new int[d];
    int remainder = 1;
    int position = 0;

    while (remainder != 0) {
      if (remainders[remainder] != 0) {
        return position - remainders[remainder];
      }
      remainders[remainder] = position;
      remainder = (remainder * 10) % d;
      position++;
    }

    return 0;
  }
}

function cycleLength(d) {
  const remainders = new Map();
  let remainder = 1;
  let position = 0;

  while (remainder !== 0) {
    if (remainders.has(remainder)) {
      return position - remainders.get(remainder);
    }
    remainders.set(remainder, position);
    remainder = (remainder * 10) % d;
    position++;
  }

  return 0;
}

module.exports = {
  answer: () => {
    let maxCycle = 0;
    let result = 0;

    for (let d = 2; d < 1000; d++) {
      const len = cycleLength(d);
      if (len > maxCycle) {
        maxCycle = len;
        result = d;
      }
    }

    return result;
  }
};

def solve():
    """
    Reciprocal cycles
    A unit fraction contains 1 in the numerator. The decimal representation of the unit
    fractions with denominators 2 to 10 are given. Where 0.1(6) means 0.166666..., the
    recurring part is in parentheses. Find the value of d < 1000 for which 1/d contains
    the longest recurring cycle in its decimal fraction part.
    https://projecteuler.net/problem=26
    """
    def cycle_length(d):
        remainders = {}
        r = 1
        pos = 0
        while r != 0:
            if r in remainders:
                return pos - remainders[r]
            remainders[r] = pos
            r = (r * 10) % d
            pos += 1
        return 0

    best_d, best_len = 1, 0
    for d in range(2, 1000):
        length = cycle_length(d)
        if length > best_len:
            best_len = length
            best_d = d
    return best_d

def cycle_length(d)
  seen = {}
  remainder = 1
  position = 0
  while remainder != 0 && !seen[remainder]
    seen[remainder] = position
    remainder *= 10
    remainder %= d
    position += 1
  end
  remainder == 0 ? 0 : position - seen[remainder]
end

def euler026_solution
  max_length = 0
  max_d = 0
  (2..999).each do |d|
    len = cycle_length(d)
    if len > max_length
      max_length = len
      max_d = d
    end
  end
  max_d
end

puts euler026_solution if __FILE__ == $PROGRAM_NAME

package main

import "fmt"

func cycleLength(d int) int {
    remainders := make(map[int]int)
    r := 1
    pos := 0
    for r != 0 {
        if prev, ok := remainders[r]; ok {
            return pos - prev
        }
        remainders[r] = pos
        r = (r * 10) % d
        pos++
    }
    return 0
}

func main() {
    best, bestLen := 0, 0
    for d := 2; d < 1000; d++ {
        l := cycleLength(d)
        if l > bestLen {
            bestLen = l
            best = d
        }
    }
    fmt.Println(best)
}

// https://projecteuler.net/problem=26
//
// A unit fraction contains 1 in the numerator. The decimal representation of the unit
// fractions with denominators 2 to 10 are given:
// 1/6 = 0.1(6), where the cycle length is 1.
// 1/7 = 0.(142857), where the cycle length is 6.
//
// Find the value of d < 1000 for which 1/d contains the longest recurring cycle in its
// decimal fraction part.
//
// Answer: 983

fn cycle_length(d: u32) -> usize {
    let mut remainders = vec![0usize; d as usize];
    let mut r = 1usize;
    let mut pos = 0usize;
    loop {
        r %= d as usize;
        if r == 0 {
            return 0;
        }
        if remainders[r] != 0 {
            return pos - remainders[r] + 1;
        }
        remainders[r] = pos + 1;
        r *= 10;
        pos += 1;
    }
}

pub fn longest_recurring_cycle(limit: u32) -> u32 {
    let mut best_d = 0u32;
    let mut best_len = 0usize;
    for d in 2..limit {
        let len = cycle_length(d);
        if len > best_len {
            best_len = len;
            best_d = d;
        }
    }
    best_d
}

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn euler_026() {
        assert_eq!(longest_recurring_cycle(1000), 983);
    }
}