PATA1061题解

//
//  main.cpp
//  PATA1061
//
//  Created by Phoenix on 2018/2/16.
//  Copyright © 2018年 Phoenix. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <cstring>
char day[7][5] = {"MON", "TUE", "WED", "THU", "FRI", "SAT", "SUN"};

int main(int argc, const char * argv[]) {
    char str[4][65];
    for(int i = 0; i < 4; i++) {
        scanf("%s", str[i]);
    }
    int k = 0;
    while(k < strlen(str[0]) && k < strlen(str[1])) {
        if(str[0][k] == str[1][k] && str[0][k] >= 'A' && str[0][k] <= 'G')
            break;
        else k++;
    }
    printf("%s ", day[str[0][k] - 'A']);
    k++;
    while(k < strlen(str[0]) && k < strlen(str[1])) {
        if(str[0][k] == str[1][k] && ((str[0][k] >= 'A' && str[0][k] <= 'N' )|| (str[0][k] >= '0' && str[0][k] <= '9')))
            break;
        else k++;
    }
    if(str[0][k] >= '0' && str[0][k] <= '9') printf("%02d:", str[0][k] - '0');
    else printf("%d:", str[0][k] - 'A' + 10);
    k = 0;
    while(k < strlen(str[2]) && k < strlen(str[3])) {
        if(str[2][k] == str[3][k] && ((str[2][k] >= 'a' && str[2][k] <= 'z') || (str[2][k] >= 'A' && str[2][k] <= 'Z'))) break;
        else k++;
    }
    printf("%02d\n", k);
    return 0;
}