摘要
Title: 3498. 日期差值
Tag: 日期
Memory Limit: 64 MB
Time Limit: 1000 ms

3498. 日期差值

题意

有两个日期，求两个日期之间的天数，如果两个日期是连续的我们规定他们之间的天数为两天。
思路
1. 用Python时调用datetime和timedelta库，直接进行计算即可，但也有限制，年数不能太大，会超限制
2. 正常写法

代码

调库

'''
Author: NEFU AB-IN
Date: 2023-05-18 17:20:27
FilePath: \Acwing\3498\3498.py
LastEditTime: 2023-05-18 18:31:46
'''
# import
import sys, math
from collections import Counter, deque
from heapq import heapify, heappop, heappush, nlargest, nsmallest
from bisect import bisect_left, bisect_right
from datetime import datetime, timedelta
from string import ascii_lowercase, ascii_uppercase


class sa:
    def __init__(self, x, y):
        self.x = x
        self.y = y

    def __lt__(self, x):
        pass


# Final
N = int(1e3 + 10)
INF = int(2e9)

# Define
sys.setrecursionlimit(INF)
input = lambda: sys.stdin.readline().rstrip("\r\n")  # Remove when Mutiple data
read = lambda: map(int, input().split())
letterTonumber = lambda x: ord(x.upper()) - 64

# —————————————————————Division line ——————————————————————

while True:
    try:
        s = input()
        e = input()
        if s > e:
            s, e = e, s
        d1 = datetime(year=int(s[:4]), month=int(s[4:6]), day=int(s[6:]))
        d2 = datetime(year=int(e[:4]), month=int(e[4:6]), day=int(e[6:]))

        print((d2 - d1).days + 1)
    except:
        break

正常写法

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int months[] = {
    0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31
};

int is_leap(int year)
{
    if (year % 100 && year % 4 == 0 || year % 400 == 0)
        return 1;
    return 0;
}

int get_month_days(int year, int month)
{
    int res = months[month];
    if (month == 2) res += is_leap(year);
    return res;
}

int get_total_days(int y, int m, int d)
{
    int res = 0;
    for (int i = 1; i < y; i ++ )
        res += 365 + is_leap(i);

    for (int i = 1; i < m; i ++ )
        res += get_month_days(y, i);

    return res + d;
}

int main()
{
    int y1, m1, d1, y2, m2, d2;
    while (scanf("%04d%02d%02d", &y1, &m1, &d1) != -1)
    {
        scanf("%04d%02d%02d", &y2, &m2, &d2);
        printf("%d\n", abs(get_total_days(y1, m1, d1) - get_total_days(y2, m2, d2)) + 1);
    }

    return 0;
}

3498. 日期差值

题意

思路

代码