思路：每年都只有一个回文串例如1234年 就是12344321，所以根据前4位年份去构建对应的回文串，判断月份、日份是否在合法范围内即可。ABABBABA型同理只有一个，可以根据前2位构建这种类型的回文串。
n=int(input())+1
n1=n
#用字典存储每个月对应的天数 xx为闰年时2月的天数
dict={"01":31,"02":28,"03":31,"04":30,"05":31,"06":30,"07":31,"08":31,"09":30,"10":31,"11":30,"12":31,"xx":29}
while(1):
s1=str(n)
s2=str(int(n/10000))+s1[3]+s1[2]+s1[1]+s1[0] #构建最近的回文数
temp=s2[4]+s2[5] #计算月份
year=int(s2[0]+s2[1]+s2[2]+s2[3])
if ((year%4==0 and year%100!=0) or year%400==0) and temp=="02": #将闰年的2月月份改为xx
temp="xx"
if temp not in dict.keys(): #判断月份是否在字典中
n+=10000
continue
else:
if dict.get(temp)s1[0]): #如果初始为12340123 则直接构建的话 应该是12211221<12340123 所以提前把第二位+1 13311331符合AB型条件
n1 += 1000000
while(1):
s1 = str(n1)
s2 = str(int(n1 / 1000000)) + s1[0] + s1[1] + s1[1] + s1[0] + s1[1] + s1[0]
temp = s2[4] + s2[5]
year = int(s2[0] + s2[1] + s2[2] + s2[3])
if (year % 4 == 0 and year % 100 != 0) or year % 400 == 0:
temp = "xx"
if temp not in dict.keys():
n1 += 1000000
continue
else:
if dict.get(temp) < int(s2[6] + s2[7]):
n1 += 1000000
continue
else:
print(s2)
break
