سوالات چالشی هفته دوم دومین دوره لیگ مای‌کد به همراه پاسخ‌های مورد انتظار تست کیس‌ها جهت آشنایی بیشتر شرکت‌کنندگان در لیگ منتشر می‌گردد.

پاسخ‌های تیم برنده جهت افزایش جنبه آموزشی  و استفاده سایر شرکت‌کنندگان پس از آزمون کامنت گذاری شده است.

دریافت
عنوان: نرمال سازی داده ها با min,max
حجم: 299 کیلوبایت

پاسخ تیم برنده

دریافت
عنوان: سوال چالشی الگوریتم Sleep Sort
حجم: 282 کیلوبایت

پاسخ تیم برنده 

دریافت
عنوان: سوال چالشی اعداد دوری
حجم: 607 کیلوبایت
پاسخ تیم برنده

تابع اصلی مرتبط با سوال اول

 
#Den stands for Denominator(Bottom part) and Num stands for Numerator(Upper part)

# Function to return the GCD (ک.م.م)
def gcd(a, b):
    while b > 0:
        a, b = b, a % b
    return a
# Main function to solve
def solve(user_input):
    # Getting input and splitting into a list of strings
    parts = user_input.split()
    
    # Converting strings to integers manually
    nums = []
    for item in parts:
        nums.append(int(item))
    # Nothing to return if 0
    if len(nums) == 0:
        return ""

    # Mins and Maxs (Could've used sort but oh well . . .)
    min_val = nums[0]
    max_val = nums[0]
    for n in nums:
        if n < min_val:
            min_val = n #Sets as new min
        if n > max_val:
            max_val = n #Sets as new max
    
    results = []
    
    # Special case (Rule)
    if min_val == max_val:
        for i in range(len(nums)):
            results.append("1/2")
    else:
        # Calculate the denominator 
        denominator = max_val - min_val
        #Calculate the numerator
        for x in nums:
            numerator = x - min_val
            #Special case (Rule)
            if numerator == 0:
                results.append("0")
            else:
                # Simplify the fraction using the GCD (ک.م.م)
                common_factor = gcd(numerator, denominator)
                simplified_num = numerator // common_factor
                simplified_den = denominator // common_factor
                
                # Print numerator if den == 1
                if simplified_den == 1:
                    results.append(str(simplified_num))
                else:
                    # Formatting 
                    fraction_string = str(simplified_num) + "/" + str(simplified_den)
                    results.append(fraction_string)
                    
    # Printing the list joined by spaces
    output = ""
    for i in range(len(results)):
        output += results[i]
        if i < len(results) - 1:
            output += " "
    return output

# Input and Output
input_file = open("input1.txt", "r")
output_file = open("output1.txt", "w")

for line in input_file:
    if line.strip():
        processed_line = solve(line)
        output_file.write(processed_line + "\n")

input_file.close()
output_file.close()

تابع اصلی مرتبط با سوال دوم

def solve(user_input):
    # Sppppliiiiit
    parts = user_input.split()
    # Convert to int
    all_numbers = [int(x) for x in parts]

    # The first number is k
    k = all_numbers[0]

    # The rest of the numbers are the sensor data
    sensor_data = all_numbers[1:]
    # Remove duplicates and then sort 
    unique_sorted_data = sorted(list(set(sensor_data)))
    # Check if k is larger than the number of unique items available
    
    if k > len(unique_sorted_data): # !Must be lower!
        return "UNKNOWN"
    else:
        # prints the smallest k
        return str(unique_sorted_data[k-1])

# Input and Output
input_file = open("input2.txt", "r")
output_file = open("output2.txt", "w")

for line in input_file:
    if line.strip():
        processed_line = solve(line)
        output_file.write(processed_line + "\n")

input_file.close()
output_file.close()

تابع اصلی مرتبط با سوال سوم

 
def solve_puzzle(line_text):
    # (1/17 through 16/17)
    valid_rotations = {}
    k_to_string = {}
    for k in range(1, 17):
        value = (k * (10**16)) // 17 #Set to 16 digits and divide by 17
        s_val = str(value).zfill(16) #Set to 16 digits , fill with 0s if needed
        valid_rotations[s_val] = k #For pos
        k_to_string[k] = s_val #For actual num
        
    user_input = line_text.split()
    ring_outer = user_input[0]
    ring_middle = user_input[1]
    
    if ring_outer in valid_rotations and ring_middle in valid_rotations:
        # Get the 'k' value (the numerator) for each input
        k1 = valid_rotations[ring_outer]
        k2 = valid_rotations[ring_middle]
        # Calculate the sum for the inner ring
        k3 = k1 + k2
        # If the resulting k is between 1 and 16, it's a valid cyclic rotation
        if k3 in k_to_string:
            return k_to_string[k3]
        else:
            # If the sum > 16, it's no longer a simple rotation (as in Case 3)
            return "INVALID"
    else:
        # If one of the inputs wasn't a valid rotation to begin with
        return "INVALID"

# Input and Output
input_file = open("input3.txt", "r")
output_file = open("output3.txt", "w")

for line in input_file:
    if line.strip():
        processed_line = solve_puzzle(line)
        output_file.write(processed_line + "\n")

input_file.close()
output_file.close()