PAT 1014 B Holmes dating (C language), 124 test points errors and solutions

1014 Holmes appointment (20 minutes)

Sherlock Holmes received a strange note: we meet it! 3485djDkxh4hhGE 2984akDfkkkkggEdsb s & hgsfdk d & Hyscvnm. Detective soon realized that actually strange garbled appointments on time Thursday 14:04 note, because the front two in the first string for the same capital letters (case sensitive distinction) is the first four letters D representative Thursday; 2 is the E of the same character, it is the first five letters of the alphabet, the representative day of 14 hours (so one is 0:00 to 23:00 by the numbers 0 to 9, and the capital letters a to N shown); behind the first two for the same string of letters s appear in the 4th position (counting from 0), the representative of the first 4 minutes. Now given two pairs of strings, please help Holmes decoded appointment time.

Input format:
Input line 4 are given in the four non-empty, no spaces, and the length of the string 60 is not exceeded.

Output formats:
Output appointment time in a row, the format DAY HH: MM, which is a week DAY 3 character abbreviation that MON = Monday, TUE for Tuesday, WED for Wednesday, THU expressed Thursday, FRI means that on Friday, SAT for Saturday, SUN for Sunday. Enter the topic guarantee the existence of a unique solution for each test.

Sample input:

3485djDkxh4hhGE

2984akDfkkkkggEdsb

s&hgsfdk

d&Hyscvnm

Sample output:

THU 14:04

124 test points wrong answer, looked a long time, and finally changed out, the original is guilty of a stupid mistake, not the digital exchange value of the ASCII code numbers (count hours of code, line 25) ¯ □ ¯ ||
This many pits questions, pay attention to the point I do not want to talk about, seen a lot of big brothers code, feel written very lengthy, it needs to be raised, can pointer familiar with the pointer, it can greatly simplify the amount of code.

#include <stdio.h>
int main()
{
    int d,b=0,m,h;
    char a[4][61];
    for(int i=0;i<4;i++)
        gets(a[i]);
    for(int j=0;j<60;j++)
    {
        if(a[0][j]>='A' && a[0][j]<='G' && a[1][j]==a[0][j] && b==0)
        {    
            d=a[0][j]-'A'+1;
            b++;
            continue;
        }
        if(b==1 && a[1][j]==a[0][j] )
        {    
            if(a[0][j]>='A' && a[0][j]<='N')
            {
                h=a[0][j]-'A'+10;
                break;
            }
            if(a[0][j]>='0' && a[0][j]<='9')
            {
                h=a[0][j]-'0';
                break;
            }
        }
    }
    for(int k=0;k<60;k++)
    {
        if( (a[2][k]>='A' && a[2][k]<='Z') || (a[2][k]>='a' && a[2][k]<='z') && a[2][k]==a[3][k])
        {
            m=k;
            break;
        }
    }
    switch(d)
    {
        case 1:printf("MON ");break;
        case 2:printf("TUE ");break;
        case 3:printf("WED ");break;
        case 4:printf("THU ");break;
        case 5:printf("FRI ");break;
        case 6:printf("SAT ");break;
        case 7:printf("SUN ");break;
    }
    printf("%02d:%02d\n",h,m);
	return 0;
}

/*测试点124答案错误，看了半天，终于改出来了，原来是犯了低级错误，没把数字的ASCII码换回数字的数值（算小时的代码里，第25行）￣□￣｜｜*/
/*注意的点我不想讲了，看了很多大佬的代码，感觉自己写得很冗长，有待提高吧，对指针熟悉的可以用指针，可以大大简化代码量。*/