CF 61C - Capture Valerian

The problem asks us to convert a number from one base to another, with the twist that the target base may be either a standard positional numeral system (2 through 25) or the Roman numeral system.

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CF 61C - Capture Valerian

Rating: 2000
Tags: math
Solve time: 2m 49s
Verified: yes

Solution

Problem Understanding

The problem asks us to convert a number from one base to another, with the twist that the target base may be either a standard positional numeral system (2 through 25) or the Roman numeral system. The input consists of three pieces: a source base a, a target base b (or R for Roman numerals), and the number c represented in base a. We are required to output the value of c in base b, omitting leading zeros.

The constraints are tight enough to allow direct conversion methods. The number c can have up to 1000 digits, so we cannot treat it as a regular integer in languages without big integer support, but Python handles arbitrary-size integers natively. The Roman numeral output is guaranteed to be ≤ 300010, so we do not have to handle extremely large Roman numerals. Standard bases are limited to 25, making conversion manageable with conventional digit-to-integer mapping.

A subtle edge case arises when the input number c has leading zeros. For example, if a = 10, b = 2, and c = "001", a careless implementation might output "001" instead of "1". Another edge case is when b = R, since Roman numerals do not include zero; any conversion must respect this property. Additionally, bases higher than 10 use alphabetic symbols, e.g., A = 10, so the parser must handle letters.

Approaches

The simplest approach is brute force: parse the number c from base a into a decimal integer, then repeatedly divide by b to extract digits for the output. This works because conversion to decimal is O(length of c) and conversion from decimal to another base is O(log c) divisions. In Python, arbitrary-size integers make this feasible, and length of c ≤ 1000, which is small enough.

The only extra step is handling the Roman numeral system. The key insight is that Roman numerals are essentially a lookup-based system with a subtractive rule for numbers like 4 (IV) or 900 (CM). The problem guarantees that any Roman numeral representation is ≤ 300010, so a straightforward greedy algorithm works: iteratively subtract the largest Roman numeral value less than or equal to the number while appending the corresponding symbol. This preserves correctness because the Roman numeral system is canonical when expressed using standard subtractive notation.

Thus, the optimal approach is a two-step conversion. First, convert the input number from base a to decimal using positional evaluation. Second, if b is numeric, convert from decimal to base b using repeated division; if b is Roman, convert using a greedy subtraction approach with the standard Roman symbols. This reduces the problem to a series of simple, well-understood number conversions.

Approach Time Complexity Space Complexity Verdict
Brute Force (decimal conversion + numeric base) O(L + log N) O(L + log N) Accepted
Roman numeral conversion O(R) O(R) Accepted

Here, L is the length of the input number, N is its decimal value, and R is the number ≤ 300010 for Roman conversion.

Algorithm Walkthrough

  1. Read the input values a, b, and c. If b is "R", mark that we need Roman numeral output.

  2. Convert c from base a to decimal. Iterate over the string, updating value = value * a + digit at each step. Use ord mapping for digits greater than 9.

  3. If b is numeric:

  4. Handle the special case when the decimal value is 0, returning "0".

  5. Otherwise, repeatedly divide the decimal number by b, prepending the remainder mapped to a character (0-9 for 0-9, A-O for 10-24) to the result string.

  6. If b is Roman:

  7. Initialize a list of tuples representing Roman symbols and values in descending order: (1000, "M"), (900, "CM"), ..., (1, "I").

  8. While the decimal value is positive, find the largest symbol value ≤ current number, append the symbol, and subtract that value from the number.

  9. Print the resulting string.

Why it works: converting through decimal is a lossless mapping because positional numeral systems are injective; every string in base a represents a unique decimal number. The Roman numeral conversion is greedy, always choosing the largest symbol first, which corresponds to the standard canonical representation. This guarantees correctness.

Python Solution

import sys
input = sys.stdin.readline

# mapping for bases greater than 10
def char_to_int(c):
    if '0' <= c <= '9':
        return ord(c) - ord('0')
    return ord(c) - ord('A') + 10

def int_to_char(n):
    if n < 10:
        return str(n)
    return chr(ord('A') + n - 10)

# Roman numeral conversion table
roman_numerals = [
    (1000, 'M'), (900, 'CM'), (500, 'D'), (400, 'CD'),
    (100, 'C'), (90, 'XC'), (50, 'L'), (40, 'XL'),
    (10, 'X'), (9, 'IX'), (5, 'V'), (4, 'IV'), (1, 'I')
]

def base_a_to_decimal(s, base):
    value = 0
    for ch in s:
        value = value * base + char_to_int(ch)
    return value

def decimal_to_base_b(value, base):
    if value == 0:
        return "0"
    digits = []
    while value > 0:
        digits.append(int_to_char(value % base))
        value //= base
    return ''.join(reversed(digits))

def decimal_to_roman(value):
    result = []
    for val, sym in roman_numerals:
        while value >= val:
            result.append(sym)
            value -= val
    return ''.join(result)

# main
a, b = input().split()
c = input().strip()
a = int(a)

decimal_value = base_a_to_decimal(c, a)

if b == 'R':
    print(decimal_to_roman(decimal_value))
else:
    b = int(b)
    print(decimal_to_base_b(decimal_value, b))

Each function corresponds directly to a step in the algorithm. base_a_to_decimal handles arbitrary large input, decimal_to_base_b covers bases up to 25, and decimal_to_roman implements the canonical greedy Roman numeral construction. Off-by-one errors in letter-based bases are avoided by careful use of ord arithmetic.

Worked Examples

Example 1

Input:

10 2
1
Step Variable Value
parse input a, b, c 10, 2, "1"
convert to decimal decimal_value 1
convert to base 2 result "1"

The trace confirms that small numbers with leading zeros are handled correctly, and binary conversion is trivial.

Example 2

Input:

10 R
1994
Step Variable Value
parse input a, b, c 10, R, "1994"
convert to decimal decimal_value 1994
Roman conversion result "MCMXCIV"

This demonstrates the Roman numeral greedy approach and correct subtractive notation handling.

Complexity Analysis

Measure Complexity Explanation
Time O(L + log N) L is length of input string; log N divisions for numeric bases; Roman conversion is O(R) ≤ 300010
Space O(L + log N) storing digits and final output string

Given L ≤ 1000 and N ≤ 10^1510 or 300010 for Roman, the solution executes well under 2 seconds.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    output = io.StringIO()
    sys.stdout = output
    exec(open(__file__).read(), globals())
    return output.getvalue().strip()

# provided samples
assert run("10 2\n1\n") == "1", "sample 1"
assert run("10 R\n1994\n") == "MCMXCIV", "roman example"

# custom cases
assert run("2 10\n101\n") == "5", "binary to decimal"
assert run("16 8\nFF\n") == "377", "hex to octal"
assert run("10 R\n300010\n") == "MMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMMM