1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define isSat(x) ((x) >= 1 &&(x) <= 12)//判断是否为合法的月份
 4 #define isRun(x) ((x%4 == 0 && x%100 != 0) || (x%400 == 0))//判断是否为闰年
 5
 6 char data[2][8];
 7 int monthDay[2][13]={{0,31,28,31,30,31,30,31,31,30,31,30,31}, //0 : 平年
 8                      {0,31,29,31,30,31,30,31,31,30,31,30,31}};//1 : 闰年
 9
10 int Transform(int a,int b,int d){//将字符串对应的年、月、日转化成数子
11     int num=0;
12     for(int i=a;i <= b;++i)
13         num=num*10+(data[d][i]-'0');
14     return num;
15 }
16 void Search(int year,char *s){
17     int index=0;
18     do
19     {
20         s[index++]=year%10+'0';
21         year /= 10;
22     }while(year != 0);
23 }
24 void Solve()
25 {
26     int year1=Transform(0,3,0),month1=Transform(4,5,0),day1=Transform(6,7,0);
27     int year2=Transform(0,3,1),month2=Transform(4,5,1),day2=Transform(6,7,1);
28     int res=0;
29     int pre=0;
30     for(int i=year1;i <= year2;++i)
31     {
32         char s[4];
33         Search(i,s);//找到将当前的年份的回文
34         int month=(s[0]-'0')*10+(s[1]-'0');//当前年份的回文对应的月
35         int day=(s[2]-'0')*10+(s[3]-'0');//日
36         if(!isSat(month) || day > monthDay[isRun(i)][month])
37             continue;
38         if(year1 == year2)
39         {
40             if(month > month1 || (month == month1 && day >= day1))
41             {
42                 if(month < month2 || (month == month2 && day <= day2))
43                     res++;
44             }
45         }
46         else
47         {
48             if(i == year1)
49                 res += (month > month1 || (month == month1 && day >= day1) ? 1:0);
50             else if(i == year2)
51                 res += (month < month2 || (month == month2 && day <= day2) ? 1:0);
52             else//如果year1 != year2 ，且 i != year1 && i != year2 ，则当前合法的日期一定在date1和date2之间
53                 res++;
54         }
55     }
56     printf("%d\n",res);
57 }
58 int main()
59 {
60     scanf("%s%s",data[0],data[1]);
61     Solve();
62 }

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int date1,date2;
 4 int monthDay[13]={0,31,29,31,30,31,30,31,31,30,31,30,31};
 5
 6 void Solve()
 7 {
 8     int res=0;
 9     for(int i=1;i <= 12;++i)//枚举月份
10         for(int j=1;j <= monthDay[i];++j)//枚举每月对应的日
11         {
12             int year=j%10*1000+j/10*100+i%10*10+i/10;//当前月，日对应的回文年份
13             int cnt=year*10000+i*100+j;//回文日期对应的八位数
14             res += (cnt >= date1 && cnt <= date2 ? 1:0);
15         }
16     printf("%d\n",res);
17 }
18 int main()
19 {
20     scanf("%d%d",&date1,&date2);
21     Solve();
22 }