A1061 Dating (20 分| 字符串处理，附详细注释，逻辑分析)

写在前面

• 思路分析

• 按照题目所给方法找到相等字符后判断，输出时间不足2位数要在前面添0，即用%02d输出

• 10分钟a题

测试用例

input:
3485djDkxh4hhGE 
2984akDfkkkkggEdsb 
s&hgsfdk 
d&Hyscvnm

output:
THU 14:04

ac代码

#include <iostream>
#include <cctype>

using namespace std;
int main()
{

    string a, b, c, d;
    cin >> a >> b >> c >> d;
    char t[2];
    int pos, i=0, j=0;
    while(i < a.length() && i<b.length())
    {
        if(a[i]==b[i] && (a[i]>= 'A' && a[i]<='G'))
        {
            t[0] = a[i];
            break;
        }
        i++;
    }
    i = i+1;
    while(i<a.length() && i<b.length())
    {
        if(a[i]==b[i] && ((a[i]>= 'A' && a[i]<='N') || isdigit(a[i])))
        {
            t[1] = a[i];
            break;
        }
        i++;
    }

    while(j<c.length() && j<d.length())
    {
        if(c[j]==d[j] && isalpha(c[j]))
        {
            pos = j;
            break;
        }
        j++;
    }
    string week[7] = {"MON ", "TUE ", "WED ", "THU ", "FRI ", "SAT ", "SUN "};
    int m = isdigit(t[1]) ? t[1]-'0' : t[1]-'A' + 10;
    cout << week[t[0]-'A'];
    printf("%02d:%02d", m, pos);

    return 0;
}