九度题目1096：日期差值

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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：

2011041220110422

样例输出：

#include<stdio.h>#include<algorithm>#include<iostream>#include<stack>#include<vector>#include<string.h>#include<limits.h>#include<stdlib.h>#define ABS(x) ((x)>=0?(x):(-(x)))using namespace std;static int month[12]={0,31,28,31,30,31,30,31,31,30,31,30};int days_from(int year){    int y = year/10000;    int m = (year - y*10000)/100;    int d = year%100;    int result = 0;    int i;    for(i=1000;i<y;i++)    {        if((i%4==0&&i%100!=0)||i%400==0)            result+=366;        else            result+=365;    }    for(i=1;i<m;i++)    {        if(i==2)        {            if((y%4==0&&y%100!=0)||y%400==0)                result += month[i]+1;            else                result += month[i];        }        else            result += month[i];    }    return result+d;}int main(){    freopen("test.in","r",stdin);    freopen("test.out","w",stdout);    int year1, year2;    int days1, days2;    while(cin>>year1>>year2)    {        days1 = days_from(year1);         days2 = days_from(year2);         cout<<ABS(days1-days2)+1<<endl;    }    fclose(stdin);    fclose(stdout);    return 0;}

闰年的判断 ((y%4==0&&y%100!=0)||y%400==0)

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