方法有挺多的,比如说先枚举日期再判断是不是回文数,或者先枚举回文数再判断日期是否合法等等。
讲一下作者采用的方法。八位数,如果要是回文数,至少应该是ABCDDBCA,A有9种,B、C、D有10种,所以最多只要枚举9*10*10*10个数字就能出答案。
#include<iostream>
using namespace std;
bool DAY(int n) {
int year = n / 10000;
int month = (n % 10000) / 100;
if (month > 12 || month < 1)return false;
int day = n % 100;
if (day > 31 || day < 1)return false;
if (month == 2 || month == 4 || month == 6 || month == 9 || month == 11) {
if (day > 30)return false;
}
int flag = 0;
if (year % 100 != 0) {
if (year % 4 == 0)flag = 1;
}
else {
if (year % 400 == 0)flag = 1;
}
if (!flag) {
if (month == 2) {
if (day > 28)return false;
}
}
return true;
}
int main() {
int n;
cin >> n;
int a = n / 10000000 % 10;
int b = n / 1000000 % 10;
int c = n / 100000 % 10;
int d = n / 10000 % 10;
while (1) {
int sum = a * 10000000 + b * 1000000 + c * 100000 + d * 10000 + d * 1000 + c * 100 + b * 10 + a;
if (sum > n && DAY(sum)) {
cout << sum << endl;
break;
}
else {
d++;
if (d > 9) { d = 0; c++; }
if (c > 9) { c = 0; b++; }
if (b > 9) { b = 0; a++; }
}
}
while (1) {
int sum = a * 10000000 + b * 1000000 + a * 100000 + b * 10000 + b * 1000 + a * 100 + b * 10 + a;
if (sum > n && DAY(sum)) {
cout << sum << endl;
break;
}
b++;
if (b > 9) { b = 0; a++; }
}
return 0;
}