Bluebridge cup real topic 2020 palindrome date simulation construction provincial competition

Title Description

2020 During the Spring Festival , There was a special date that caught everyone's attention ：2020 year 2 month 2 Japan . Because if you press this date “yyyymmdd” In the form of 8 The number of digits is 20200202, It happens to be a palindrome number . We call such dates palindrome dates .

Someone said 20200202 yes “ Once in a thousand years ” The special day of . Xiao Ming does not agree with this , Because not to 2 The date of the next palindrome will be after five years ：20211202 namely 2021 year 12 month 2 Japan .

Also expressed 20200202 It's not just a palindrome date , Or a ABABBABA Type palindrome date . Xiao Ming doesn't agree with this either , Because about 100 I'll meet the next one in two years ABABBABA Type palindrome date ：21211212 namely 2121 year 12 month 12 Japan . Not worth calculating “ Once in a thousand years ”, At best “ Once in a thousand years ”.

Given a 8 Number of dates , Please calculate the next reply date after that date and the next ABABBABA When are the dates of the palindrome .

Input description

The input contains an eight digit integer NN, Indicates the date .

For all profiling use cases ,10000101 \leq N \leq 8999123110000101≤N≤89991231, Guarantee NN It's a legal date 8 Digit representation .

Output description

Output two lines , Each row 1 Eight digits . The first line indicates the date of the next palindrome , The second line represents the next ABABBABA Type palindrome date .

I/o sample

Example

Input
20200202

Output
20211202
21211212

Operation limit

Maximum operation time ：1s
Maximum running memory : 256M

Code

#include <iostream>
#include<string>
using namespace std;
int year, mon, day;// year   month   Japan 
int maxday[13] = {
     0,31,28,31,30,31,30,31,31,30,31,30,31 };// Record the number of days per month 
bool isLeapYear()// Determine if it's a leap year 
{
    
	if ((year % 4 == 0) && (year % 100 != 0) || (year % 400 == 0))
	{
    
		return true;
	}
	return false;
}
int main()
{
    
	int x;
	cin >> x;
	string s;
	int y, z;
	y = z = 0;
	bool flag1 = true;
	year = x / 10000;
	mon = x % 10000 / 100;
	day = x % 100;
	while (1)
	{
    
		day++;
		if (isLeapYear())// Leap year's 2 Month is 29 God 
		{
    
			maxday[2] = 29;
		}
		else
		{
    
			maxday[2] = 28;
		}
		if (day > maxday[mon])// It's more than a month , Add one month 
		{
    
			mon++;
			day = 1;
		}
		if (mon > 12)// It's been more than a year , Add one year 
		{
    
			year++;
			mon = 1;
		}
		bool flag;
		flag = true;
		s = to_string(year * 10000 + mon * 100 + day);// Convert to string form for easy comparison 
		if (s[0] != s[7] || s[1] != s[6] || s[2] != s[5] || s[3] != s[4])
		{
    
			flag = false;
		}
		if (flag && flag1)
		{
    
			cout << s << endl;
			flag1 = false;// Record whether a palindrome year has been found 
		}
		int a = s[0];
		int b = s[1];
		if (s[2] == a && s[5] == a && s[7] == a && s[3] == b && s[4] == b && s[6] == b && a != b)
		{
    
			cout << s << endl;
			break;
		}
	}//  Please enter your code here 
	return 0;
}

The main points of

When writing this question, I didn't consider the correctness of the date at the beginning , Just simply judge the two palindromes and output , Discovery cannot pass . After reading the questions carefully, I found that the date needs to be legal , So we set up an array to store the days of each month , Then judge whether it is a leap year , Then change it 2 The number of days in the month is ok . There is no way to increase one by one in the cycle judgment , Instead, it adopts the legal form of date addition , In this way, you can not judge illegal numbers . Because the title limits the length of the input number , So when judging palindromes, you can also use direct comparison , It is more direct than circular judgment .