日期差值

Posted 2020-11-27 kalicener

样例输入

20130101

20130105

样例输出

5

#include<stdio.h>
#include<iostream>


using namespace std;

int main() {
    int month[12] = { 31,28,31,30,31,30,31,31,30,31,30,31 };
    long int i = 0,sum1,sum2;
    int all1,all2,year1, year2, month1, month2, day1, day2;
    cin >> all1 >> all2;//获取两个出生日期
    year1 = all1 / 10000;//得到出生年份
    year2 = all2 / 10000;
    month1 = (all1 - year1 * 10000) / 100;//出生的月份
    month2 = (all2 - year2 * 10000) / 100;
    day1 = all1 - year1 * 10000 - month1 * 100;//出生的日期
    day2 = all2 - year2 * 10000 - month2 * 100;
    sum1 = (year1 - 1) / 4 + (year1 - 1) * 365;//年份是从公元01年开始的
    for (; i < month1-1; i++)
    {
        if (year1 % 2 == 0 && i == 1)
            sum1 += 29;//如果该年为闰年，2月有29天
        else
            sum1 += month[i];
    }
    sum1 += day1;//把日期加上，大功告成！
    sum2 = (year2 - 1) / 4 + (year2 - 1) * 365;
    for (i=0; i < month2 - 1; i++)
    {
        if (year2 % 2 == 0 && i == 1)
            sum2 += 29;
        else
            sum2 += month[i];
    }
    sum2 += day2;
    if (sum1 > sum2)
        cout << sum1 - sum2;//输出两个人出生日期间差的天数。
    else
        cout << sum2 - sum1;
}