标签:九度
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
有多组数据,每组数据有两行,分别表示两个日期,形式为YYYYMMDD
每组数据输出一行,即日期差值
20110412 20110422
11
#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; }
标签:九度
原文地址:http://blog.csdn.net/vintionnee/article/details/39303821