第一题：36进制转10进制（不能使用转换函数，用例中出现非36进制需要返回0，有范围限制）

def f(num):
    dic = {'a':10,'b':11,'c':12,'d':13,'e':14,'f':15,'g':16,
          'h':17,'i':18,'j':19,'k':20,'l':21,'m':22,'n':23,
          'o':24,'p':25,'q':26,'r':27,'s':28,'t':29,'u':30,
          'v':31,'w':32,'x':33,'y':34,'z':35}
    re = 0
    if not num[0].isdigit() and num[0]!='+' and num[0]!='-' and not dic.get(num[0]):
        return 0
    tmp = 1
    if num[0] == '+':
        tmp = 1
        num = num[1:]
    if num[0] == '-':
        tmp = -1
        num = num[1:]
    n = len(num)
    j = 0
    for i in range(n-1,-1,-1):
        if num[i].isdigit():
            re += 36**j*int(num[i])
        else:
            if not dic.get(num[i]):
                return 0
            re += 36**j*dic.get(num[i])
        j += 1
    return re*tmp

s = input().strip()
re = f(s)
if re>9223372036854775807:
    re = 9223372036854775807
if re<-9223372036854775807:
    re = -9223372036854775807
print(re)

第二题：连续n个正整数最多可以构造多少个不同形状的二叉搜索树

n = int(input().strip())
if n==0:
    print(0)
else:
    dp = [0]*(n+1)
    dp[0]=1
    dp[1]=1
    for i in range(2,n+1):
        for j in range(1,i+1):
            dp[i]+=dp[j-1]*dp[i-j]
    print(dp[-1])