题解 | #计算日期到天数转换#

根据输入的日期，计算是这一年的第几天。
保证年份为4位数且日期合法。
进阶：时间复杂度：O(n)\O(n) ，空间复杂度：O(1)\O(1) 
输入描述：输入一行，每行空格分割，分别是年，月，日
输出描述：输出是这一年的第几天
输入：2012 12 31
输出：366
输入：1982 3 4
输出：63

def exam(instr):
    ins = instr
    lis = list(map(int,ins.split(' ')))
    # [2012, 12, 31]
    year = lis[0]
    month = lis[1]
    day = lis[2]
    cnt = []
        
    if year % 4 != 0:
        dic = {1:31,2:28,3:31,4:30,5:31,6:30,7:31,8:31,9:30,10:31,11:30,12:31}
        for i in range(1,month):
            cnt.append(dic[i])
            # print(i,'===',cnt)
            # 1 === [31]
            # 2 === [31, 28]
            # 11 === [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30]

        cnt.append(day)
        print(sum(cnt))
        
    if year % 4 == 0 and year != 1900:
        dic_run = {1:31,2:29,3:31,4:30,5:31,6:30,7:31,8:31,9:30,10:31,11:30,12:31}
        for i in range(1,month):
            cnt.append(dic_run[i])
            # print(i,'===',cnt)
            # 2 === [31, 29]
            # 11 === [31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30]
        cnt.append(day)
        print(sum(cnt))
        # 366
        
    # 特殊情况单独处理，没有百度，你就什么都不知道，哎
    if year == 1900 :
        dic_run = {1:31,2:28,3:31,4:30,5:31,6:30,7:31,8:31,9:30,10:31,11:30,12:31}
        for i in range(1,month):
            cnt.append(dic_run[i])
        cnt.append(day)
        print(sum(cnt))
        # 365
    
instr = input().strip()
exam(instr=instr)