EOJ Monthly 2020.1 A. 回文时间

左手的ㄟ右手 2023-06-30 15:57 10阅读 0赞

[https://acm.ecnu.edu.cn/contest/247/problem/A/][https_acm.ecnu.edu.cn_contest_247_problem_A]

**因为是对称的，只考虑前七位**  
**2020012 | 2100202**

0123   45   6
    2020 | 01 | 2 
    年   | 月 | 日
    0位和2位的取值范围：{ "0","1","2","3","4","5","6","7","8","9"}
    1位和3位的取值范围： { "0","1","2","3","4","5"}
    4-5位的取值范围： {  "01","02","10","11","12" }
    6位的取值范围 ： { "1","2"}
    从第六位开始+1，一次向前进位，类似字符串加法

**code**

#include <iostream>
    #include <sstream>
    #include <vector>
    #include <algorithm>
    #include <stdio.h>
    
    using namespace std;
    
    typedef long long LL;
    
    #define debug(x) cout<<#x<<": "<<x<<endl;
    
    string LL_str(LL a){ 
        stringstream ss;
        ss<<a;
        return ss.str();
    }
    
    
    int main()
    { 
        vector<string> nian0 = { "0","1","2","3","4","5","6","7","8","9"};
        vector<string> nian1 = { "0","1","2","3","4","5"};
        vector<string> nian2 = { "0","1","2","3","4","5","6","7","8","9"};
        vector<string> nian3 = { "0","1","2","3","4","5"};
        vector<string> yue = {  "01","02","10","11","12" };
        vector<string> ri0 = { "1","2"};
    
        int indexnian0 = 2;
        int indexnian1 = 0;
        int indexnian2 = 2;
        int indexnian3 = 0;
        int indexyue = 0;
        int indexri0 = 1;
    
        int K = 200;
        scanf("%d",&K);
        for(int k=0;k<K;k++){ 
    
            indexri0 += 1;
    
            if( indexri0 == ri0.size()  ){ 
                indexri0 = 0;
                indexyue += 1;
            }
    
            if( indexyue == yue.size()  ){ 
                indexyue = 0;
                indexnian3 += 1;
            }
    
            if( indexnian3 == nian3.size()  ){ 
                indexnian3 = 0;
                indexnian2 += 1;
            }
    
            if( indexnian2 == nian2.size()  ){ 
                indexnian2 = 0;
                indexnian1 += 1;
            }
    
            if( indexnian1 == nian1.size()  ){ 
                indexnian1 = 0;
                indexnian0 += 1;
            }
    
            if( indexnian0 == nian0.size()  ){ 
                indexnian0 = 0;
            }
    
    
        }
        string ret =
            nian0[indexnian0] +
            nian1[indexnian1] +
            nian2[indexnian2] +
            nian3[indexnian3] +
            yue[indexyue]+
            ri0[indexri0];
            //debug(ret)
            cout<<ret;
            reverse(ret.begin(),ret.end());
            cout<<ret<<endl;
    
        return 0;
    }

![在这里插入图片描述][20200118185644618.png]

[https_acm.ecnu.edu.cn_contest_247_problem_A]: https://acm.ecnu.edu.cn/contest/247/problem/A/
[20200118185644618.png]: https://img-blog.csdnimg.cn/20200118185644618.png