codeup 1928

问题 A: 日期差值

时间限制: 1 Sec 内存限制: 32 MB
提交: 2590 解决: 640
[提交][状态][讨论版][命题人:外部导入]

题目描述

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

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

每组数据输出一行，即日期差值

样例输入

20130101
20130105

样例输出

5

#include<iostream>
using namespace std;
int main() {
    int date1, date2;
 
    while (cin >> date1 >> date2) {
        int month[13] = { 0,31,28,31,30,31,30,31,31,30,31,30,31 };
        int year1, year2, month1, month2, day1, day2;
        int sum = 0;
        if (date2 <date1) {
            int t = date1;
            date1 =date2;
            date2 = t;
        }
        year1 = date1 / 10000;
               month1 =date1 / 100 % 100; 
               day1 = num1 % 100;
        year2 = date2 / 10000;
                month2 = num2 / 100 % 100; 
                day2 = date2 % 100;
        for (int i = year1 + 1; i < year2; i++) {
            if ((i % 4 == 0 && i % 100 != 0) || i % 400 == 0) {
                sum += 366;
            }
            else sum += 365;
        }
        if ((year1 % 4 == 0 && year1 % 100 != 0) || year1 % 400 == 0) {
            month[2] = 29;
        }
        if (year1 == year2) {
            for (int i = month1 + 1; i < month2; i++) {
                sum += month[i];
            }
            if (month1 == month2) {
                sum += day2 - day1;
            }
            else {
                sum += month[month1] - day1;
                sum += day2;
            }
        }
        else {
            for (int i = month1 + 1; i <= 12; i++) {
                sum += month[i];
            }
            sum += month[month1] - day1;
            if ((year2 % 4 == 0 && year2 % 100 != 0) || year2 % 400 == 0) {
                month[2] = 29;
            }
            else month[2] = 28;
            for (int i = 1; i < month2; i++) {
                sum += month[i];
            }
            sum += day2;
        }
        cout << ++sum << endl;
    }
    return 0;
}