1061. Dating

Sherlock Holmes received a note with some strange strings: "Let's date!" 3485djdkxh4hhge 2984AKDFKKKKGGEDSB s&hgsfdk d&hyscvnm ". It took him only a minute to figure out that those strange strings is actually referring to the coded time "Thursday 14:0 4 "--since the first common capital 中文版 (case sensitive) shared by the first and strings is the 4th capital Le Tter ' D ', representing the 4th day in a week; The second common character is the 5th capital letter ' E ', representing the 14th hour (hence the hours from 0 to + a D Ay is represented by the numbers from 0 to 9 and the capital letters from A to N, respectively); and the Chinese letter gkfx by the last of the strings is ' s ' at the 4th position, representing the 4th minute. Now given and pairs of strings, you is supposed to help Sherlock decode the dating time.

Input Specification:

Each input file contains the one test case. Each case gives 4 non-empty strings of no more than, characters without white space in 4 lines.

Output Specification:

For each test case, print the decoded time on one line, in the format ' Day hh:mm ', where ' Day ' is a 3-character abbreviati On for the days in a week--that's, ' MON ' for Monday, ' TUE ' for Tuesday, ' WED ' for Wednesday, ' THU ' for Thursday, ' FRI ' For Friday, ' SAT ' for Saturday, and ' SUN ' for Sunday. It is guaranteed, the result is a unique for each case.

Really torture people, the line has been dragged all the time, before said 8th, now said 11th, can not imagine what kind of mood to finish this question

#include <stdio.h> #include <stdlib.h> #include <string.h> #define N 65bool Check1 (char A,char b); bool Check2 (char A,char b); bool Check3 (char A,char b); void Day (char a); void Hour (char a); int main () {char First[n],seco    Nd[n],third[n],forth[n];    Gets (first);    Gets (second);    Gets (third);    Gets (forth);    int Len1=strlen (first), Len2=strlen (second), Len3=strlen (third), Len4=strlen (forth);    int i,j;         for (i=0;i<len1&&i<len2;i++) {if (Check1 (First[i],second[i])) break;     } day (First[i]);    i++;         for (; i<len1&&i<len2;i++) {if (Check2 (First[i],second[i])) break;    } Hour (First[i]);         for (i=0;i<len3&&i<len4;i++) {if (Check3 (Third[i],forth[i])) break;    } printf (":%.2d\n", i);    System ("pause");    return 0;         }void Hour (char a) {if (A-' 0 ' >=0&&a-' 9 ' <=0) {printf ("%.2d", A-' 0 '); } else printf ("%.2d", A-' a ' +10);            }void Day (char a) {switch (a) {case ' a ':p rintf ("MON"); return;            Case ' B ':p rintf ("TUE"); return;            Case ' C ':p rintf ("WED"); return;            Case ' D ':p rintf ("THU"); return;            Case ' E ':p rintf ("FRI"); return;            Case ' F ':p rintf ("SAT"); return;            Case ' G ':p rintf ("SUN"); return;         }}bool Check1 (char A,char b) {if (a-b==0) {if (A-' a ' >=0 && a ' G ' <=0) return true;         else return false;     } else return false; } bool Check2 (char A,char b) {if (a-b==0) {if ((A-' 0 ' >=0 &&a-' 9 ' <=0) | | (A-' a ' >=0&&a-' N ' <=0))         return true;         else return false;     } else return false;         } bool Check3 (char A,char b) {if (a-b==0) {if (A-' a ' >=0&&a-' z ' <=0) return true;         else return false;     } return false;  }

1061. Dating