solution1（通过50%）

#include<stdio.h>
void f(int a){
	int t = a;
	while(a){
		printf("%d", a % 10);
		a /= 10;
	}
	if(t < 10) printf("0");
}
int isLeap(int n){
	if(n % 400 == 0 || (n % 4 == 0 && n % 100 != 0)) return 1;
	return 0;
}
int main(){
	int t, d, m, y, dp, mp, ab[12] = {01011010, 10100101, 11111111, 20200202, 21211212, 30300303, 40400404, 50500505, 60600606, 70700707, 80800808, 90900909};
	int rd[50] = {1, 2, 3, 10, 11, 12, 13, 20, 21, 22, 30, 31, 32, 40, 41, 42, 50, 51, 52, 60, 61, 62, 70, 71, 72, 80, 81, 82, 90, 91, 92};
	int rm[12] = {1, 10, 11, 20, 21, 30, 40, 50, 70, 80, 90};
	scanf("%d", &t);
	d = t / 1000000;
	m = t / 10000 % 100;
	for(mp = 0; mp < 12; mp++){
		if(rm[mp] > m) {
			m = rm[mp];
			break;
		}
	}
	y = d * 100 + m;
	if(m == 20 && isLeap(y) && (d == 3 || d == 13)) m = 21;
	else if(m == 20 && !isLeap(y) && (d == 92 || d == 3 || d == 31)) m = 21;
	else if(mp == 12){
		m = 1;
		for(int i = 0; i < 50; i++){
			if(rd[i] > d){
				d = rd[i];
				break;
			}
		}
	}
	printf("%02d%02d", d, m);
	f(m); 
	f(d);
	printf("\n");
	for(int i = 0; i < 50; i++){
		if(ab[i] > t){
			printf("%d\n", ab[i]);
			break;
		}
	}
	return 0;
}

solution2

给出的日期范围是10⁸左右，再加上判断肯定超时，类似之前的思想先用某个限制条件缩小范围。无论是回文串还是满足ABABBABA的数字，共同特点都是需要保证回文，满足该条件的数据量会大大降低。进一步判断该回文数是否满足日期要求

• 1<= 天数<=31
• 1 <=月数<=12
• 闰年的二月<=29天，平年的二月<=28天
• 4，6，9，11月份<=30天

#include<stdio.h>
int isLeap(int y){
	if(y % 400 == 0 || (y % 4 == 0 && y % 100 != 0)) return 1;
	return 0;
}
int check(int y, int m, int d){
	if(d > 31 || d < 1) return 0;
	if(m > 12 || m < 1) return 0;
	if(m == 2 && isLeap(y) && d > 29) return 0;
	if(m == 2 && !isLeap(y) && d > 28) return 0;
	if(d > 30 && (m == 4 || m == 6 || m == 9 || m == 11)) return 0;
	return 1;
}
int main(){
	int t, flag = 1, over = 1;
	scanf("%d", &t);
	for(int i = t + 1; i <= 90900909 && over; i++){
		int a[8] = {0}, temp = i, k, num = 0;
		do{
			a[num++] = temp % 10;
			temp /= 10;
		}while(temp);
		for(k = 0; k < 4; k++){
			if(a[k] != a[7 - k]) break;
		}
		if(k == 4){
			if(check(i / 10000, i / 100 % 100, i % 100)){
				if(flag){
					printf("%d\n", i);
					flag = 0;
				}
				if(a[0] == a[2] && a[2] == a[5] && a[5] == a[7] && a[1] == a[3] && a[3] == a[4] && a[4] == a[6]){
					printf("%d", i);
					over = 0;
				}
			}
		}
	}
	return 0;
}