题解 | #字母统计#

/*
描述
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
输入描述：
有多组数据，每组数据有两行，分别表示两个日期，形式为YYYYMMDD
输出描述：
每组数据输出一行，即日期差值
*/

#include<iostream>

using namespace std;

bool isLeapYear(int year) {
    if ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0)
        return true;
    else
        return false;
}


int main() {
    int monthDays[2][12] = {
    {31,28,31,30,31,30,31,31,30,31,30,31},            // 非闰年月份的天数
    {31,29,31,30,31,30,31,31,30,31,30,31}            // 闰年月份的天数
    };
    char date1[9];        // note: 多申请一个空间给 \n
    char date2[9];
    int y1, m1, d1, y2, m2, d2;
    int n1 = 0, n2 = 0;
    while (cin >> date1 >> date2) {
        n1 = n2 = 0;
        sscanf(date1, "%4d%2d%2d", &y1, &m1, &d1);
        sscanf(date2, "%4d%2d%2d", &y2, &m2, &d2);
        //cout << date1 << " " << date2 << endl;
        //cout << y1 << " " << m1 << " " << d1<<endl;
        //cout << y2 << " " << m2 << " " << d2<<endl;

        // year -> days
        for (int i = 0; i <= y1; ++i)
            n1 += isLeapYear(i) ? 366 : 365;
        for (int i = 0; i <= y2; ++i)
            n2 += isLeapYear(i) ? 366 : 365;
        // month -> days
        for (int j = 1; j < m1; ++j)
            n1 += monthDays[isLeapYear(y1)][j - 1];
        for (int j = 1; j < m2; ++j)
            n2 += monthDays[isLeapYear(y2)][j - 1];
        // days
        n1 += d1;
        n2 += d2;
        cout << abs(n1 - n2) + 1 << endl;
    }
    return 0;
}