题目描述

题目分析

对于输入的abc我们可以以a为年也可以以c为年，将abc,cab,cba这三种情况进行判断合法性即可，注意需要排序去重，所以考虑使用set

此处为纯模拟的写法，但使用循环代码会更加简洁。

方法一：

#include<bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
set<string> st;
string s[N];
int cnt;
int m[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
bool is_ren(int x)
{if((x % 4 == 0 && x % 100 != 0)||(x % 400 == 0))return true;return false;
}
int main()
{int a, b, c;scanf("%d/%d/%d",&a, &b, &c);//以a为年 //abcif(a <= 59){//abcif(b <= 12 && b > 0){int x = 2000 + a;if(is_ren(x))m[2] = 29;else m[2] = 28;if(c <= m[b] && c > 0){cnt ++;s[cnt] = "20";if(a < 10)s[cnt] += "0";s[cnt] += (to_string(a) + "-" );if(b < 10)s[cnt] += "0";s[cnt] += (to_string (b) + "-");if(c < 10)s[cnt] +="0";s[cnt] += to_string(c);st.insert(s[cnt]);}}}else if(a > 59){//abcif(b <= 12 && b > 0){int x = 1900 + a;if(is_ren(x))m[2] = 29;else m[2] = 28;if(c <= m[b] && c > 0){cnt ++;s[cnt] = "19";if(a < 10)s[cnt] += "0";s[cnt] += to_string(a) + "-" ;if(b < 10)s[cnt] += "0";s[cnt] += to_string (b) + "-";if(c < 10)s[cnt] +="0";s[cnt] += to_string(c);st.insert(s[cnt]);}}}//以c为年 //cab, cba if(c <= 59){//cabif(a <= 12 && a > 0){int x = 2000 + c;if(is_ren(x))m[2] = 29;else m[2] = 28;if(b <= m[a] && b > 0){s[++ cnt] = "20";if(c < 10)s[cnt] += "0";s[cnt] += to_string(c) + "-" ;if(a < 10)s[cnt] += "0";s[cnt] += to_string (a) + "-";if(b < 10)s[cnt] +="0";s[cnt] += to_string(b);st.insert(s[cnt]);}}//cbaif(b <= 12 && b > 0){int x = 2000 + c;if(is_ren(x))m[2] = 29;else m[2] = 28;if(a <= m[b] && a > 0){cnt ++;s[cnt] = "20";if(c < 10)s[cnt] += "0";s[cnt] += to_string(c) + "-" ;if(b < 10)s[cnt] += "0";s[cnt] += to_string (b) + "-"; if(a < 10)s[cnt] +="0";s[cnt] += to_string(a);st.insert(s[cnt]);}	}}else if(c > 59){//cabif(a <= 12 && a > 0){int x = 1900 + c;if(is_ren(x))m[2] = 29;else m[2] = 28;if(b <= m[a] && b > 0){cnt ++;s[cnt] = "19";if(c < 10)s[cnt] += "0";s[cnt] += to_string(c) + "-" ;if(a < 10)s[cnt] += "0";s[cnt] += to_string (a) + "-";if(b < 10)s[cnt] +="0";s[cnt] += to_string(b);st.insert(s[cnt]);}}//cbaif(b <= 12 && b > 0){int x = 1900 + c;if(is_ren(x))m[2] = 29;else m[2] = 28;if(a <= m[b] && a > 0){cnt ++;s[cnt] = "20";if(c < 10)s[cnt] += "0";s[cnt] += to_string(c) + "-" ;if(b < 10)s[cnt] += "0";s[cnt] += to_string (b) + "-";if(a < 10)s[cnt] +="0";s[cnt] += to_string(a);st.insert(s[cnt]);}}}for(auto i : st){cout << i << '\n';}return 0;
}

方法二：

#include<bits/stdc++.h>
using namespace std;
int days[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
bool check_valid(int year, int month, int day)
{if(month == 0 || month > 12)return false;if(day == 0)return false;if(month != 2){if(day > days[month])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 ++){int year = date / 10000, month = date % 10000 / 100, day = date % 100;if(check_valid(year, month, day)){if(year % 100 == a && month == b && day == c ||month == a && day == b && year % 100 == c ||day == a && month == b && year % 100 == c)printf("%d-%02d-%02d\n", year, month, day);}}return 0;
}