题目

分析

代码中巧妙的用到3重循环，完美的解决了输出的顺序问题【题目要求从小到大】

需要注意的是2月的值，在不同的年份中应该更新2月的值

还有就是最后的3个if【T^T，我一开始写的3个if语句】，要么改成else if，要么把判定的条件改成用“||”连接【害我看半天T^T】，不然用3个if回导致重复输出

代码

#include <iostream>
#include <vector>
#include <set>
#include <string>
#include <algorithm>
#include <math.h>
#include <queue>
#include <climits>  // 包含INT_MAX常量
#include <cctype>
using namespace std;int month[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};int check(int x) {return (x % 4 == 0 && x % 100 != 0) || (x % 400 == 0);
}int main() {int a, b, c;scanf("%d/%d/%d", &a, &b, &c);//循环从早到晚，巧妙满足了题目要求的从早到晚排列for (int year = 1960; year <= 2059; year++) {if (check(year))month[2] = 29;elsemonth[2] = 28;for (int mth = 1; mth <= 12; mth++) {for (int day = 1; day <= month[mth]; day++) {if (a == year % 100 && b == mth && day == c)printf("%d-%02d-%02d\n", year, mth, day);else if (a == mth && b == day && c == year % 100 )printf("%d-%02d-%02d\n", year, mth, day);else if (a == day && b == mth &&  c == year % 100 )printf("%d-%02d-%02d\n", year, mth, day);}}}return 0;
}