美好日子2021.12.2
据说2021年12月2日是一个美好日子，因为这是一个完全对称日！这里认为一个美好日子是一个共8位数字的完全对称日（例如20211202），其中年份占4位，月份、日份都是2位。对于给定的年份，请判断该年是否存在美好日子。
输入格式: 输出格式: 输入样例: 代码 #include int main() { int T,i; int year, m, d; scanf("%d", &T);//获取年数 for(i=0;i { scanf("%d", &year); m = year % 10 * 10 + year / 10 % 10; //取出月 d = year / 100 % 10 * 10 + year / 1000;//取出日 if (year % 4 == 0 && year % 100 != 0 || year % 400 == 0)//为闰年 { if (m == 1 && d > 0 && d <= 9) printf("%d0%d0%d", year, m, d); else if (m == 1 && d > 10 && d <= 31) printf("%d0%d%d", year, m, d); else if (m == 2 && d > 0 && d <= 9) printf("%d0%d0%d", year, m, d); else if (m == 2 && d > 10 && d <= 29) printf("%d0%d%d", year, m, d); else if (m == 3 && d > 0 && d <= 9) printf("%d0%d0%d", year, m, d); else if (m == 3 && d > 10 && d <= 31) printf("%d0%d%d", year, m, d); else if (m == 4 && d > 0 && d <= 9) printf("%d0%d0%d", year, m, d); else if (m == 4 && d > 10 && d <= 30) printf("%d0%d%d", year, m, d); else if (m == 5 && d > 0 && d <= 9) printf("%d0%d0%d", year, m, d); else if (m == 5 && d > 10 && d <= 31) printf("%d0%d%d", year, m, d); else if (m == 6 && d > 0 && d <= 9) printf("%d0%d0%d", year, m, d); else if (m == 6 && d > 10 && d <= 30) printf("%d0%d%d", year, m, d); else if (m == 7 && d > 0 && d <= 9) printf("%d0%d0%d", year, m, d); else if (m == 7 && d > 10 && d <= 31) printf("%d0%d%d", year, m, d); else if (m == 8 && d > 0 && d <= 9) printf("%d0%d0%d", year, m, d); else if (m == 8 && d > 10 && d <= 31) printf("%d0%d%d", year, m, d); else if (m == 9 && d > 0 && d <= 9) printf("%d0%d0%d", year, m, d); else if (m == 9 && d > 10 && d <= 30) printf("%d0%d%d", year, m, d); else if (m == 10 && d > 0 && d <= 9) printf("%d%d0%d", year, m, d); else if (m == 10 && d > 10 && d <= 31) printf("%d%d%d", year, m, d); else if (m == 11 && d > 0 && d <= 9) printf("%d%d0%d", year, m, d); else if (m == 11 && d > 10 && d <= 30) printf("%d%d%d", year, m, d); else if (m == 12 && d > 0 && d <= 9) printf("%d%d0%d", year, m, d); else if (m == 12 && d > 10 && d <= 31) printf("%d%d%d", year, m, d); else printf("none"); } else //平年 { if (m == 1 && d > 0 && d <= 9) printf("%d0%d0%d", year, m, d); else if (m == 1 && d > 10 && d <= 31) printf("%d0%d%d", year, m, d); else if (m == 2 && d > 0 && d <= 9) printf("%d0%d0%d", year, m, d); else if (m == 2 && d > 10 && d <= 28) printf("%d0%d%d", year, m, d); else if (m == 3 && d > 0 && d <= 9) printf("%d0%d0%d", year, m, d); else if (m == 3 && d > 10 && d <= 31) printf("%d0%d%d", year, m, d); else if (m == 4 && d > 0 && d <= 9) printf("%d0%d0%d", year, m, d); else if (m == 4 && d > 10 && d <= 30) printf("%d0%d%d", year, m, d); else if (m == 5 && d > 0 && d <= 9) printf("%d0%d0%d", year, m, d); else if (m == 5 && d > 10 && d <= 31) printf("%d0%d%d", year, m, d); else if (m == 6 && d > 0 && d <= 9) printf("%d0%d0%d", year, m, d); else if (m == 6 && d > 10 && d <= 30) printf("%d0%d%d", year, m, d); else if (m == 7 && d > 0 && d <= 9) printf("%d0%d0%d", year, m, d); else if (m == 7 && d > 10 && d <= 31) printf("%d0%d%d", year, m, d); else if (m == 8 && d > 0 && d <= 9) printf("%d0%d0%d", year, m, d); else if (m == 8 && d > 10 && d <= 31) printf("%d0%d%d", year, m, d); else if (m == 9 && d > 0 && d <= 9) printf("%d0%d0%d", year, m, d); else if (m == 9 && d > 10 && d <= 30) printf("%d0%d%d", year, m, d); else if (m == 10 && d > 0 && d <= 9) printf("%d%d0%d", year, m, d); else if (m == 10 && d > 10 && d <= 31) printf("%d%d%d", year, m, d); else if (m == 11 && d > 0 && d <= 9) printf("%d%d0%d", year, m, d); else if (m == 11 && d > 10 && d <= 30) printf("%d%d%d", year, m, d); else if (m == 12 && d > 0 && d <= 9) printf("%d%d0%d", year, m, d); else if (m == 12 && d > 10 && d <= 31)
首先输入一个正整数T（<8000）表示测试数据的组数，然后输入T组测试数据。对于每组测试数据，输入一个年份y（2020
对于每组测试数据，输出一行。若年份y存在美好日子，则输出该日期，否则输出“none”。引号不必输出。
2
2021
2022
输出样例:
20211202
none
PS:
2021年12月2日，顺利接娃放学，晚饭获中评。谨以此题纪念这个美好日子。
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