题解 | #数制转换#

#本题蕴含思想可用于所有进制转换，即“若任意情况，则以10进制为中介，转10则sum+pow求和，10转则取余并整除”
def zs(s):  # 字母转数字
    return ord(s) - 55


def sz(n):  # 数字转字母
    return chr(n + 55)


def z10(a, n):  # a进制n  转   10进制
    sum = 0
    n = list(str(n))
    for k in range(len(n)):
        if ord(n[k]) in range(97, 123):
            n[k] = n[k].upper()
    for i in range(len(n)):
        if ord(n[i]) in range(65, 91):
            n[i] = str(zs(n[i]))
    n.reverse()
    for j in range(len(n)):
        sum = sum + pow(int(a), j) * int(n[j])
    return sum


def tz(b, n):  # 10进制的n  转   b进制
    a = []
    n = int(n)
    b = int(b)
    while True:
        a.append(n % b)
        n = n // b
        if n == 0:
            break
    for i in range(len(a)):
        if a[i] > 9:
            a[i] = sz(a[i])
    a.reverse()
    s = ""
    for j in range(len(a)):
        s = s + str(a[j])
    return s


while True:
    try:
        s = list(input().split())
        a = s[0]
        n = s[1]
        b = s[2]
        print(tz(b, z10(a, n)))
    except:
        break