输入在4行中分别给出4个非空、不包含空格、且长度不超过60的字符串。
在一行中输出约会的时间，格式为“DAY HH:MM”，其中“DAY”是某星期的3字符缩写，即MON表示星期一，TUE表示星期二，WED表示星期三，THU表示星期
四，FRI表示星期五，SAT表示星期六，SUN表示星期日。题目输入保证每个测试存在唯一解。
3485djDkxh4hhGE 2984akDfkkkkggEdsb s&hgsfdk d&Hyscvnm
THU 14:04
import java.util.Scanner;
public class Main {
private final static String[] DAY = {"MON", "TUE", "WED", "THU", "FRI", "SAT", "SUN"};
public static void main(String[] args) {
StringBuilder sb = new StringBuilder();
Scanner in = new Scanner(System.in);
String l1 = in.nextLine();
String l2 = in.nextLine();
String l3 = in.nextLine();
String l4 = in.nextLine();
int len1 = Math.min(l1.length(), l2.length());
int len2 = Math.min(l3.length(), l4.length());
boolean flag1 = true;
for(int i = 0;i<len1;i++){
char c = l1.charAt(i);
if(flag1&&c>='A'&&c<='G'){
if(c==l2.charAt(i)){
flag1 = !flag1;
System.out.print(DAY[c-65]+" ");
}
}else if (!flag1&&c==l2.charAt(i)) {
if(c>='0'&&c<='9'){
System.out.print("0"+c+":");
break;
}else if (c>='A'&&c<='N') {
System.out.print((c-55)+":");
break;
}
}
}
for(int i = 0;i<len2;i++){
char c = l3.charAt(i);
if(c>='a'&&c<='z'&&c==l4.charAt(i)){
if(i<10)
System.out.print("0"+i);
else
System.out.print(i);
break;
}
}
in.close();
}
}
import java.util.Scanner; public class Main { public static void main(String[] args) { String DAY[]={"MON ","TUE ","WED ","THU ","FRI ","SAT ","SUN "}; Scanner sc = new Scanner(System.in); String a = sc.next(); String b = sc.next(); String c = sc.next(); String d = sc.next(); char m[]=xiangtongday(a,b); //星期几 System.out.print(DAY[m[0]-'A']); //小时 if(m[1]<'9'&&m[1]>='0') System.out.print("0"+m[1]+":"); else System.out.print((m[1]-'A')+10+":"); //分钟 if(count(c,d)<10&&count(c,d)>=0) System.out.print("0"+count(c,d)); else System.out.print(count(c,d)); } //返回 c[0]==星期几 c[1]=日期 public static char[] xiangtongday(String a,String b){ char x[]; char y[]; x=a.toCharArray(); y=b.toCharArray(); int length; int count1=0; int flag=0; char c[] = new char [20]; if(a.length()<b.length()) length=a.length(); else length=b.length(); for(int i=0;i<length;i++) { if(x[i]==y[i]&&x[i]>='A'&&x[i]<='G') { c[count1]=x[i]; count1++; flag=i; break; } } for(int i=flag+1;i<=length;i++) { if(x[i]==y[i]&&((x[i]>='A'&&x[i]<='G')||x[i]>='0'&&x[i]<='9')) { c[count1]=x[i]; break; } } return c; } //返回分钟 public static int count(String a,String b) { char x[] = new char[60]; char y[] = new char[60]; int count1=0; int length=0; x = a.toCharArray(); y = b.toCharArray(); if (a.length() < b.length()) length = a.length(); else length = b.length(); for (int i = 0; i < length; i++) { if (x[i] == y[i] && x[i] >= 'a' && x[i] <= 'z') { count1 = i; break; } } return count1; } }是我不够细还是这个题目太细了啊，一个=号没加愣是看了半个小时才看出来😅
import java.util.Scanner;
public class Main{
public static void main(String[] args){
Scanner input=new Scanner(System.in);
String str1=input.nextLine();
String str2=input.nextLine();
String str3=input.nextLine();
String str4=input.nextLine();
input.close();
int pos1=0,pos2=0;
StringBuilder sb1=new StringBuilder();
while(pos1<str1.length()&&pos2<str2.length()){
if(str1.charAt(pos1)!=str2.charAt(pos2)){
pos1++;
pos2++;
}else{
sb1.append(str1.charAt(pos1));
pos1++;
pos2++;
}
}
int pos3=0,pos4=0;
while(pos3<str3.length()&&pos4<str4.length()){
if(str3.charAt(pos3)!=str4.charAt(pos4)){
pos3++;
pos4++;
}else{
if(Character.isAlphabetic(str3.charAt(pos3)))
break;
else{
pos3++;
pos4++;
}
}
}
int week=0;
String Wee="";
int hour=0;
boolean flag=false;
for(int i=0;i<sb1.length();i++){
if(Character.isAlphabetic(sb1.charAt(i))&&flag==false){
week=sb1.charAt(i)-'A'+1;
flag=true;
}else if(flag==true&&(Character.isDigit(sb1.charAt(i))||Character.isAlphabetic(sb1.charAt(i)))){
if(Character.isAlphabetic(sb1.charAt(i)))
hour=9+sb1.charAt(i)-'A'+1;
else
hour=sb1.charAt(i)-'0';
break;
}
}
switch(week){
case 1:
Wee="MON";
break;
case 2:
Wee="TUE";
break;
case 3:
Wee="WED";
break;
case 4:
Wee="THU";
break;
case 5:
Wee="FRI";
break;
case 6:
Wee="SAT";
break;
case 7:
Wee="SUN";
break;
}
System.out.print(Wee+" "+(hour>9?hour:"0"+hour)+":"+(pos3>9?pos3:"0"+pos3));
}
} 
#include<iostream>
using namespace std;
int main()
{
string date[7]={"MON","TUE","WED","THU","FRI","SAT","SUN"};
string s1,s2,s3,s4;
cin>>s1>>s2>>s3>>s4;
int i,j,idx=0;
for(i=0;i<s1.length()&&i<s2.length();i++){
if(!idx){
while(s1[i]!=s2[i] || s1[i]<65 || s1[i]>71){
i++;
}
cout<<date[s1[i]-'A']<<" ";
idx++;
}else if(idx==1){
while(s1[i]!=s2[i]){
i++;
}
if(s1[i]>=48&&s1[i]<=57){
cout<<0<<(s1[i]-'0')<<":";
break;
}else if(s1[i]>=65&&s1[i]<=78){
cout<<(s1[i]-'A'+10)<<":";
break;
}
}
}
for(i=0;i<s3.length()&&i<s4.length();i++){
if(s3[i]==s4[i]&&s3[i]>=97&&s3[i]<=122){
if(i<10) cout<<0<<i;
else cout<<i;
break;
}
}
return 0;
} 题目没有说，相同的字符都出现在各个字符串同样的位置
#include<bits/stdc++.h>
using namespace std;
int main() {
string A,B,C,D;
cin>>A>>B>>C>>D;
char arrChar[2];
int arrChar2;
int k=0;
int len1,len2;
if(A.size()>B.size())
len1=B.size();
else
len1=A.size();
if(C.size()>D.size())
len2=D.size();
else
len2=C.size();
for(int i=0;i<len1;i++){
if('A'<=A[i]&&A[i]<='Z'&&A[i]==B[i]&&k==0){
arrChar[k++]=A[i];
}else if(A[i]==B[i]&&k>0){
arrChar[k++]=A[i];
break;
}
}
for(int i=0;i<len2;i++){
if(C[i]==D[i]&&(('A'<=C[i]&&C[i]<='Z')||('a'<=C[i]&&C[i]<='z'))){
arrChar2=i;
break;
}
}
//处理符号含义
switch(arrChar[0]-'A'+1){
case 1:
cout<<"MON ";break;
case 2:
cout<<"TUE ";break;
case 3:
cout<<"WED ";break;
case 4:
cout<<"THU ";break;
case 5:
cout<<"FRI ";break;
case 6:
cout<<"SAT ";break;
case 7:
cout<<"SUN ";break;
default:
break;
}
if('0'<=arrChar[1]&&arrChar[1]<='9'){
cout<<'0'<<arrChar[1]<<":";
}else if('A'<=arrChar[1]&&arrChar[1]<='N'){
cout<<arrChar[1]-'A'+10<<":";
}
if(0<=arrChar2&&arrChar2<=9)
cout<<"0"<<arrChar2<<endl;
else
cout<<arrChar2<<endl;
}
#include <stdio.h>
#include <ctype.h>
int main()
{
int i;
char strs[4][61];
const char *days[] = {"MON", "TUE", "WED", "THU", "FRI", "SAT", "SUN"};
scanf("%s%s%s%s", strs[0], strs[1], strs[2], strs[3]);
for(i = 0; strs[0][i] && strs[1][i]; ++i) {
if(strs[0][i] == strs[1][i] && strs[0][i] >= 'A' && strs[0][i] <= 'G') {
printf("%s ", days[strs[0][i] - 'A']);
break;
}
}
for(++i; strs[0][i]; ++i) {
if(strs[0][i] == strs[1][i] && ((strs[0][i] >= 'A' && strs[0][i] <= 'N') || isdigit(strs[0][i]))) {
if(isdigit(strs[0][i]))
printf("0%c:", strs[0][i]);
else
printf("%d:", strs[0][i] - 'A' + 10);
break;
}
}
for(i = 0; strs[2][i] != strs[3][i] || !isalpha(strs[2][i]); ++i)
;
printf("%02d\n", i);
return 0;
} 星期，小时和分钟都是有界限的，需要进行边界判断。星期是A-G大写字母，小时是：0-9、A-N大写字母，分钟是：0-60数字。对这些字母处理的同时需要做好边界检测。
/*
* app=PAT-Basic lang=c++
* https://pintia.cn/problem-sets/994805260223102976/problems/994805308755394560
*/
#include <cstdio>
#include <string>
#include <iostream>
using namespace std;
const string week[7] = {"MON","TUE","WED","THU","FRI","SAT","SUN"};
bool isUpperLetter(char c){
return c >= 'A' && c <= 'Z';
}
bool isWeek(char c){
return isUpperLetter(c) && (c < 'H');
}
bool isHour(char c){
return isUpperLetter(c) && (c < 'O');
}
bool isNum(char c){
return c >= '0' && c <= '9';
}
bool isSameLetter(char a,char b){
if (a != b)
return false;
return (a >= 'a'&& a <= 'z') || (a >= 'A'&& a <= 'Z');
}
int main()
{
string s1, s2, s3, s4;
cin >> s1 >> s2 >> s3 >> s4;
int minLen1 = s1.length() < s2.length() ? s1.length() : s2.length();
int minLen2 = s3.length() < s4.length() ? s3.length() : s4.length();
int i = 0;
for (; i < minLen1; i++){
if (s1[i] == s2[i] && isWeek(s1[i])){
cout << week[s1[i]-'A'];
break;
}
}
for (i = i+1; i < minLen1;i++){
if (s1[i] == s2[i]){
if (isHour(s1[i])){
printf(" %02d",10+s1[i]-'A');
break;
}
else if (isNum(s1[i])){
printf(" %02d", s1[i] - '0');
break;
}
}
}
for (int j = 0; j < 61;j++){
if (isSameLetter(s3[j], s4[j])&&j < 61){
printf(":%02d\n",j);
break;
}
}
return 0;
}
const readline = require("readline");
var Day = {
A: "MON",
B: "TUE",
C: "WED",
D: "THU",
E: "FRI",
F: "SAT",
G: "SUN"
}
var Hours = {
0: "00",
1: "01",
2: "02",
3: "03",
4: "04",
5: "05",
6: "06",
7: "07",
8: "08",
9: "09",
A: "10",
B: "11",
C: "12",
D: "13",
E: "14",
F: "15",
G: "16",
H: "17",
I: "18",
J: "19",
K: "20",
L: "21",
M: "22",
N: "23"
};
const rl = readline.createInterface({
input: process.stdin,
output: process.stdout
});
let count = 0;
let first_str = "";
let first_arr = [];
let second_str = "";
let second_arr = [];
let min_length = 0;
let Day_NUM;
let temp_arr = [];
let third_str = "";
let third_arr = "";
let forth_str = "";
let forth_arr = "";
let result = "";
rl.on("line",(line)=>{
if(count == 0){
first_str = line;
first_arr = first_str.split("");
} else if (count == 1){
second_str = line;
second_arr = second_str.split("");
if(first_arr.length <= second_arr.length){
min_length = first_arr.length;
} else if(first_arr.length > second_arr.length){
min_length = second_arr.length;
}
for(let i=0; i<min_length; i++){
if(first_arr[i] === second_arr[i]){
temp_arr.push(first_arr[i]);
}
}
for(let i=0; i<temp_arr.length; i++){
if(/^[A-G]$/.test(temp_arr[i])){
Day_NUM = i;
break;
}
}
temp_arr = temp_arr.slice(Day_NUM);
result = result + Day[temp_arr[0]] + " " + Hours[temp_arr[1]] + ":";
} else if (count == 2){
third_str = line;
third_arr = third_str.split("");
} else{
forth_str = line;
forth_arr = forth_str.split("");
if(third_arr.length <= forth_arr.length){
min_length_34 = third_arr.length;
} else if(third_arr.length > forth_arr.length){
min_length_34 = forth_arr.length;
}
for(let i=0; i<min_length_34; i++){
if(third_arr[i] === forth_arr[i] && /^[a-zA-Z]$/.test(third_arr[i])){
time_num = i;
break;
}
}
time_num<10 ? result+="0"+time_num : result+=time_num;
console.log(result);
rl.close();
}
count<4 && count++;
})
这里用JS尝试写了下，看了下别人的代码，感觉像我这么蠢地把小时对应的表一个个列出来的可能只有我吧= - =
#include <iostream>
#include <cstring>
using namespace std;
int main(){
string a,b,c,d;
cin>>a>>b>>c>>d;
int hour,min,sec,flag=0;
char tempmin;
int i=0;
for(i;a[i]!=NULL&&b[i]!=NULL;i++){
if(flag==0&&a[i]==b[i]&&((a[i]>='A'&&a[i]<='Z')||(a[i]>='a'&&a[i]<='z'))){
flag=1;
if(a[i]<='z'&&a[i]>='a'){
hour=a[i]-'a'+1;
i++;
}
else if(a[i]<='Z'&&a[i]>='A')
{
hour=a[i]-'A'+1;
i++;
}
}
if(flag==1&&a[i]==b[i]){
tempmin = a[i];
break;
}
}
for(i=0;c[i]!=NULL&&d[i]!=NULL;i++){
if(c[i]==d[i]&&((c[i]>'A'&&c[i]<'Z')||(c[i]>'a'&&c[i]<'z'))){
sec=i;
break;
}
}
//cout<<hour<<tempmin;
switch (hour){
case 1: cout<<"MON ";break;
case 2: cout<<"TUE ";break;
case 3: cout<<"WED ";break;
case 4: cout<<"THU ";break;
case 5: cout<<"FRI ";break;
case 6: cout<<"SAT ";break;
case 7: cout<<"SUN ";break;
}
if(tempmin>='0'&&tempmin<='9')
{
min=tempmin-'0';
printf("%02d:",min);
}
else{
min=tempmin-'A'+10;
cout<<min<<":";
}
printf("%02d",sec);
cout<<endl;
return 0;
}
/**
*
* @author ChopinXBP
* 大侦探福尔摩斯接到一张奇怪的字条：“我们约会吧！ 3485djDkxh4hhGE 2984akDfkkkkggEdsb s&hgsfdk d&Hyscvnm”。
* 大侦探很快就明白了，字条上奇怪的乱码实际上就是约会的时间“星期四 14:04”
* 前面两字符串中第1对相同的大写英文字母（大小写有区分）是第4个字母'D'，代表星期四；第2对相同的字符是'E'，
* 那是第5个英文字母，代表一天里的第14个钟头（于是一天的0点到23点由数字0到9、 以及大写字母A到N表示）；
* 后面两字符串第1对相同的英文字母's'出现在第4个位置（从0开始计数）上，代表第4分钟。现给定两对字符串
*
* 注意点：1.要对应位置的字符相同才算成对，2.注意序列长度不同，3.注意三个元素的判定依据
*
*/
import java.io.BufferedInputStream;
import java.io.BufferedOutputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Scanner;
public class HolmesCode { public static Scanner in = new Scanner(new BufferedInputStream(System.in)); public static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out)); public static ArrayList<String> str = new ArrayList<>(); public static String[] week = { "MON", "TUE", "WED", "THU", "FRI", "SAT", "SUN" }; public static int date = -1; public static int hour = -1; public static int min; public static void main(String[] args) { // TODO Auto-generated method stub Read(); Decode(); Print(); out.close(); in.close(); } public static void Read() { for (int i = 0; i < 4; i++) { String string = in.nextLine(); str.add(string); } } public static void Decode() { // 存放前两个序列的相同元素（注意序列长度不同） ArrayList<Character> sameword = new ArrayList<>(); for (int i = 0; i < Math.min(str.get(0).length(), str.get(1).length()); i++) { char word = str.get(0).charAt(i); char word2 = str.get(1).charAt(i); if (word == word2 && ((word >= 'A' && word <= 'Z') || (word >= '0' && word <= '9'))) { sameword.add(word); } } // 解码日期和小时 for (int i = 0; i < sameword.size(); i++) { Character word = sameword.get(i); if (date == -1) { if (word >= 'A' && word <= 'Z') { date = word - 'A'; } } else { if (word >= 'A' && word <= 'N') { hour = word - 'A' + 10; break; } else if (word >= '0' && word <= '9') { hour = Integer.parseInt(word.toString()); break; } } if (hour != -1) break; } // 解码分钟 for (int i = 0; i < Math.min(str.get(2).length(), str.get(3).length()); i++) { char word = str.get(2).charAt(i); char word2 = str.get(3).charAt(i); if (word == word2 && ((word >= 'A' && word <= 'Z') || (word >= 'a' && word <= 'z'))) { min = i; break; } } } public static void Print() { out.print(week[date] + ' '); if (hour < 10) { out.print('0'); } out.print(hour); out.print(':'); if (min < 10) { out.print('0'); } out.print(min); }
}
可以AC，这个题目最重要的就是理解题意，第一个是判断大写字母，第二个是只判断字符，可以是大写字母，当然也可以是任意的字符，只是想说这个题目出得真随便，应该是英语翻译过来的 import java.util.Scanner;
public class Main { @SuppressWarnings("resource") public static void main(String[] args) { Scanner in = new Scanner(System.in); String[] stringArray = new String[4]; for (int i = 0; i < stringArray.length; i++) { stringArray[i] = in.next(); // 这里真是坑苦我了，next()代表的是你接下来所有输入内容的第一项，nextline()代表如果你输入了一行， // 那就是这一行，没有第几项的说法 } // 现在已经拿到了四个字符串，开始进行处理 char weekday = '0'; // 建立一个字符类型的数组，存储第一个相同的字符与第二个相同的字符 char[] charArray = new char[2]; int charArrayIndex = 0;// 给字符类型的数组建立计数器 char finalFirstChar = 0; char finalSecondChar = 0; // 这里要明确的是第二个相同的字符 // 这里的相等是上下同位置相等 // 每一次循环过来这个i总会变了的，所以得到的值一直在变 int i; for (i = 0; i < stringArray[0].length(); i++) { if (stringArray[0].charAt(i) >= 'A' && stringArray[0].charAt(i) <= 'Z') { for (int j = 0; j < stringArray[1].length(); j++) { if (stringArray[0].charAt(i) == stringArray[1].charAt(j) && i == j) { charArray[0] = (char) stringArray[0].charAt(i); finalFirstChar = charArray[0]; break; } } } if (charArray[0] != '\0') { break; } } for (int i1 = i+1; i1 < stringArray[0].length(); i1++) { for (int j = 0; j < stringArray[1].length(); j++) { if (stringArray[0].charAt(i1) == stringArray[1].charAt(j) && i1 == j) { charArray[1] = (char) stringArray[0].charAt(i1); finalSecondChar = charArray[1]; break; } } if (charArray[1] != '\0') { break; } } // 做到这一步，已经拿到了星期几的，可以对这个字母做一个减去A的操作，得到的结果加一就是星期几 int weekdayNum = finalFirstChar - 'A'; String DAY = null;// 这就是星期几的前三个字母 switch (weekdayNum + 1) { case 1: DAY = "MON"; break; case 2: DAY = "TUE"; break; case 3: DAY = "WED"; break; case 4: DAY = "THU"; break; case 5: DAY = "FRI"; break; case 6: DAY = "SAT"; break; case 7: DAY = "SUN"; break; default: break; } // System.out.println(DAY); // 星期几已经可以顺利输出,接下来研究第二个相同的字符 char secondCharacter = finalSecondChar;// 这是第二个相同的字符 String hour = null; if (secondCharacter >= '0' && secondCharacter <= '9') { hour = "0" + secondCharacter; } else if (secondCharacter >= 'A' && secondCharacter <= 'N') { hour = String.valueOf(secondCharacter - 'A' + 10); } // System.out.println(hour); // 届此小时也已经打印出来了 int minute = 0; String minuteString = null; for (int i1 = 0; i1 < stringArray[2].length(); i1++) { if ((stringArray[2].charAt(i1) >= 'A' && stringArray[0].charAt(i1) <= 'Z') || (stringArray[2].charAt(i1) >= 'a' && stringArray[0].charAt(i1) <= 'z')) { for (int j = 0; j < stringArray[3].length(); j++) { if (stringArray[2].charAt(i1) - stringArray[3].charAt(j) == 0&&i1==j) { minute = i1; break; } } } if (minute != 0) { break; } } if (minute < 10) { minuteString = "0" + String.valueOf(minute); } else { minuteString = String.valueOf(minute); } System.out.println(DAY + " " + hour + ":" + minuteString); }
}
public class Question4 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
char[] chars1 = sc.next().toCharArray();
char[] chars2 = sc.next().toCharArray();
char[] chars3 = sc.next().toCharArray();
char[] chars4 = sc.next().toCharArray();
sc.close();
String[] days = { "MON", "TUE", "WED", "THU", "FRI", "SAT", "SUN" };
String day="MON";
String hour="00";
// 对前两个字符数组进行比较
int len1 = chars1.length;
int len2 = chars2.length;
int min = len1 < len2 ? len1 : len2;
int count = 0; // 遇到相同字符的次数
for (int i = 0; i < min; i++) {
if (chars1[i] == chars2[i]) {
if (count == 0 && !Character.isDigit(chars1[i]) && chars1[i] < 97) {
int d = Character.toUpperCase(chars1[i]) - 65;
day = days[d];
count++;
} else if (count == 1) {
int h;
if (!Character.isDigit(chars1[i])) {
h = Character.toUpperCase(chars1[i]) - 64 + 9;
} else {
h = chars1[i] - 48;
}
hour = h+"";
break;
}
}
}
String minute="00";
// 对后两个字符串进行比较
int len3 = chars3.length;
int len4 = chars4.length;
int min2 = len3 < len4? len3 : len4;
for (int i = 0; i < min2; i++) {
if(chars3[i]==chars4[i] && Character.isAlphabetic(chars3[i])){
minute=i+"";
break;
}
}
hour=hour.length()==1?0+hour:hour;
minute=minute.length()==1?0+minute:minute;
System.out.printf("%s %s:%s",day,hour,minute);
}
} 
#include<iostream>
#include<string>
using namespace std;
bool isSmall(char c)
{
if((c>='a'&&c<='z')||(c>='A'&&c<='Z'))
return true;
return false;
}
bool isBig(char c)
{
if(c>='A'&&c<='G')
return true;
return false;
}
bool isLegal(char c)
{
if((c>='0'&&c<='9')||(c>='A'&&c<='N'))
return true;
return false;
}
int main()
{
string s1,s2,s3,s4;
//cin>>s1>>s2>>s3>>s4;
s1="3485djDkxh4hhGE";
s2="2984akDfkkkkggEdsb";
s3="s&hgsfdk";
s4="d&Hyscvnm";
int index1=min(s1.size(),s2.size());
int index2=min(s3.size(),s4.size());
char c1,c2;
int c3;
bool frist=true;
for(int i=0;i<index1;i++)
{
if(frist)
{
if(s1[i]==s2[i]&&isBig(s1[i]))
{
c1=s1[i];
frist=false;
}
}
else
{
if(s1[i]==s2[i]&&isLegal(s1[i]))
{
c2=s1[i];
break;
}
}
}
for(int i=0;i<index2;i++)
{
if(s3[i]==s4[i]&&isSmall(s3[i]))
{
c3=i;
break;
}
}
switch(c1-'A'+1)
{
case 1:cout<<"MON"<<" ";break;
case 2:cout<<"TUE"<<" ";break;
case 3:cout<<"WED"<<" ";break;
case 4:cout<<"THU"<<" ";break;
case 5:cout<<"FRI"<<" ";break;
case 6:cout<<"SAT"<<" ";break;
case 7:cout<<"SUN"<<" ";break;
default:break;
}
if(c2>='0'&&c2<='9')
cout<<0<<c2-'0'<<":";
else
cout<<9+int(c2-'A'+1)<<":";
if(c3<9)
cout<<0<<c3<<endl;
else
cout<<c3<<endl;
return 0;
} 
#!/usr/bin/env python
#-*- coding:utf8 -*-
def findTime():
days = {'A':'MON',
'B':'TUE',
'C':'WED',
'D':'THU',
'E':'FRI',
'F':'SAT',
'G':'SUN'}
time = []
#这个循环用来找到第一个相同的大写字母，找到之后退出循环
for i in s1:
if i in s2 and i in days:
time.append(i)
break
DAY, start, end = days[time[0]], s1.index(time[0]), s2.index(time[0])
#start，end 分别用来记录第一个相同的大写字母在s1 和 s2的下标位置
#因为题目说HH的范围在0～9和A～N之间，
#找第二个相同字符就在第一个相同字母之后找符合条件的
for i in s1[start+1:]:
if i in s2[end+1:]:
if 'A' <= i <= 'N' or i in "0123456789":
time.append(i)
break
HH = time[1]
if HH in "0123456789":
HH = int(HH)
else:
HH = ord(HH) - ord('A') + 10
#找s3和s4第一个相同的字符，字符条件为A～Z和a～z 找到之后退出循环
for i in range(len(s3)):
if s3[i] == s4[i]:
if 'A'<= s3[i] <= 'Z' or 'a' <= s3[i] <= 'z':
time.append(i)
break
MM = time[2]
return DAY, HH, MM
if __name__ == '__main__':
s1 = raw_input()
s2 = raw_input()
s3 = raw_input()
s4 = raw_input()
DAY, HH, MM = findTime()
print("%s %02d:%02d" % (DAY, HH, MM))
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