1.题目描述
有两个日期,求两个日期之间的天数,如果是连续的两个日期,则天数为两天
2.样例
20110412
20110422
11
3.代码思想:构造日期结构体, 预处理,将日期减去元年得到一个固定的日期差值,每个日期对应一个差值,计算两个日期的差值即将它们互减,时间复杂度较低
#include<stdio.h>
#define ISYEAR(x) x % 100 != 0 && x % 4 ==0 || x % 400 ==0 ? 1 :0
int dayofmonth[13][2]={0,0,31,31,28,29,31,31,30,30,31,31,30,30,31,31,31,31,30,30,31,31,30,30,31,31};
typedef struct Date{
int Day;
int Month;
int yesr;
};
int buf[5001][13][32];
int Abs(int x)
{
return x > 0 ? x : -x;
}
int main(){
Date tmp;
int cnt = 0;
tmp.Day=1;
tmp.Month = 1;
tmp.year = 0;
while(tmp.year != 5001){ //预处理,形成一个以年月份为下标的数组
bufday[tmp.year][tmp.Month][tmp.Day] = cnt;
tmp.Day ++;
if(tmp.Day >dayofmonth[tmp.Month][ISYEAR(tmp.year)]){
tmp.Day = 1;
tmp.Month ++;
if(tmp.Month>12){
tmp.Month = 1;
tmp.year ++;
}
}
cnt ++;
}
int d1,m1,y1;
int d2,m2,y2;
while(scanf("%4d%2d%2d",&y1,&m1,&d1)!= EOF){
scanf("%4d%2d%2d",&y2,&m2,&d2);
printf("%d\n",Abs(bufday[y1][m1][d1]-bufday[y2][m2][d2])+1);
}
return 0;
}