0705.py

new0705.py

#coding:utf8
import sys

## prepare
d_arr = (
    (0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31),
    (0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31)
)
ds_arr = (
    (None, 0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334, 365),
    (None, 0, 31, 60, 91, 121, 152, 182, 213, 244, 274, 305, 335, 366),
)

## prepare end

## cache is leap year
def is_leap_year_old(y):
    """闰年判断 这样最快"""
    return (y % 4 == 0) and (y % 100 != 0 or y % 400 == 0)


def leap_count_old(year):
    """从0年开始的闰年数量(不包含0年)"""
    return year // 4 - year // 100 + year // 400


is_leap_array = [is_leap_year_old(y) for y in range(10000)]
### leap_count_array[i]是i年1月1日的day_id 所以要计算截止前一年的闰年数量
leap_count_array = [leap_count_old(y-1)+365*y for y in range(10000)]
## cache is leap year end

def solve(y, m, d, n):
    ny, nm = y + n // 12, m + n % 12

    if nm > 12:
        ny += 1
        nm -= 12
    is_leap_ny = is_leap_array[ny]
    is_leap_y = is_leap_array[y]

    nd = d_arr[is_leap_ny][nm]

    if d < nd:
        nd = d
    return leap_count_array[ny] + nd + ds_arr[is_leap_ny][nm] \
        - (leap_count_array[y] + d + ds_arr[is_leap_y][m])