1229.発行日

アイデア：すべての状況を列挙してから、法的状況コードをフィルタリングし
ます。

# include<iostream>
# include<cstdio>
# include<cstring>
# include<algorithm>
using namespace std;

int days[13] = {
    
    0,31,28,31,30,31,30,31,31,30,31,30,31};

bool check(int year,int mouth,int day)//判断日子是否合法
{
    
    
    if(mouth == 0 && mouth > 12) return false;
    if(day == 0) return false;
    if(mouth != 2)
    {
    
    
        if(day > days[mouth]) return false;
    }
    else
    {
    
    
        int leap = year % 100 && year % 4 == 0 || year % 400 == 0;
        if(day > 28 + leap)
        {
    
    
            return false;
        }
    }
    return true;
}

int main()
{
    
    
    int a,b,c;
    scanf("%d/%d/%d",&a,&b,&c);
    for(int date = 19600101;date <= 20591231;date++)//枚举所有情况
    {
    
    
        int year = date / 10000;
        int mouth = date % 10000 / 100;
        int day = date % 100;
        if(check(year,mouth,day))
        {
    
    
            //判断枚举的所有情况中和输入数据相同的结果
            if(year % 100 == a && mouth == b && day == c ||
             mouth  == a && day == b && year % 100 == c ||
             day == a && mouth == b && year % 100 == c)
             {
    
    
                 printf("%d-%02d-%02d\n",year,mouth,day);
             }
        }
    }
    
    return 0;
}