九度oj 题目1096：日期差值

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：

20110412
20110422

样例输出：

 1 #include <cstdio>
 2 #include <cstdlib>
 3 #include <cstring>
 4 #include <string>
 5 #include <cmath>
 6 #include <algorithm>
 7 
 8 #define MAX 10
 9 using namespace std;
10 
11 char date1c[MAX];
12 char date2c[MAX];
13 char tempc[MAX];
14 
15 int date1[MAX];
16 int date2[MAX];
17 
18 int year1,month1,day1;
19 int year2,month2,day2;
20 
21 int month[][13] = {{0,31,28,31,30,31,30,31,31,30,31,30,31},{0,31,29,31,30,31,30,31,31,30,31,30,31}};
22 int dayCount[] = {365,366};
23 
24 int isR(int year) {
25     if(((year % 4 == 0) && (year % 100!= 0))||(year % 400 == 0)) {
26         return 1;
27     }
28     return 0;
29 }
30 int main(int argc, char const *argv[])
31 {
32     while(scanf("%s %s",date1c,date2c) != EOF) {
33         if(strcmp(date1c,date2c) > 0) {
34             sscanf(date1c,"%s",tempc);
35             sscanf(date2c,"%s",date1c);
36             sscanf(tempc,"%s",date2c);
37         }
38         for(int i = 0; i < 8; i++) {
39             date1[i] = date1c[i] - '0';
40             date2[i] = date2c[i] - '0';
41         }
42         year1 = date1[0]*1000+date1[1]*100+date1[2]*10+date1[3];
43         month1 = date1[4]*10 + date1[5];
44         day1 = date1[6]*10+date1[7];
45         //printf("%d %d %d
",year1,month1,day1);
46 
47         year2 = date2[0]*1000+date2[1]*100+date2[2]*10+date2[3];
48         month2 = date2[4]*10 + date2[5];
49         day2 = date2[6]*10+date2[7];
50 
51         //printf("%d %d %d
",year2,month2,day2);
52 
53         int yd1 = 0,yd2 = 0;
54             
55         for(int i = 1; i < month1; i++) {
56             yd1 = yd1 + month[isR(year1)][i];
57         }
58             yd1 = yd1 + day1;
59             //printf("%d
",yd1);
60         for(int i = year1; i < year2; i++) {
61             yd2 = yd2 + dayCount[isR(i)];
62         }    
63         
64         for(int i = 1; i < month2; i++) {
65             yd2 = yd2 + month[isR(year2)][i];
66         }
67         //printf("%d
",yd2);
68             yd2 = yd2 + day2;
69         int ans = yd2 - yd1 + 1;
70         printf("%d
",ans);
71     }
72     return 0;
73 }

用到第一年的第一天的时间差表示二者的时间差，自我感觉月份天数的表示法有些巧妙