1010 Radix

Posted 2021-02-16 d-i-p

1010 Radix

注意点

1. 如111 1 1 10类似情况下，若n为个位数，如果本身比另一个数小，则多少的进制都是没有用的(可能会造成空循环而超时)，不过好像没有这么一个测试用例
2. 进制应该比最少数据中的最大的数要大一，如8最少是9进制的数，z最少是36进制的数
3. 注意算法中的超时问题如9999999999 11 1 10很容易超时，注意匹配的算法
4. 如果用c等强类型语言写，注意溢出问题，int类型肯定是不够的

python3代码

def getNum(num, radix):
    sum = 0
    count = 0
    for i in num[::-1]:
        if i <= '9':
            sum += int(i)*pow(radix, count)
        else:
            sum += (ord(i)-87)*pow(radix, count)
        count += 1
    return sum

num0 = 0
num1 = 0  
data_list = input().split(" ")

if data_list[2] == '1':
    num0 = getNum(data_list[0], int(data_list[3]))
    num1 = data_list[1]
else:
    num0 = getNum(data_list[1], int(data_list[3]))
    num1 = data_list[0]

if len(num1) == 1 and getNum(num1, 36) < num0:
    print("Impossible")
    exit()

max_c = max(num1)
if max_c > '9':
    max_c = ord(max_c) - 87
else:
    max_c = int(max_c)

radix = max_c + 1
result = getNum(num1, radix)

basic = 5
while result < num0:
    radix *= basic
    result = getNum(num1, radix)
    if result > num0:
        radix //= basic
        basic = radix // 2
        result = getNum(num1, radix)
        while result < num0:
            if basic == 0:
                break
            radix += basic
            result = getNum(num1, radix)
            if result > num0:
                radix -= basic
                result = getNum(num1, radix)
                basic //= 2

if result == num0:
    print(radix)
else:
    print("Impossible")