审题:(有坑)
1. 判断是星期几,是根据第一对相同的大写字母(A–G之间),并不是第一对相同的大写字母(可能超出范围).
2. 判断小时是判断字符而不是字母,包括0–9,A–N.
以下是我完成的代码:
#include <iostream>
#include <iomanip>
#include <string>
using namespace std;
int main()
{
string str1, str2, str3, str4;
cin >> str1 >> str2 >> str3 >> str4;
int l = 0;
//循环使用较短的字符串长度即可,本代码有一定的冗余
if (str1.length() > str2.length()){
l = str2.length();
}
else
l = str1.length();
char ss;
int pos; //记录第一次查找到星期的位置,+1为查找hour的起始位置
for (int i = 0; i < l; i++){
if (str1[i] == str2[i] && isupper(str1[i]) && str1[i] <='G'){
ss = str1[i];
pos = i;
break;
}
}
switch (ss){
case 'A': cout << "MON ";
break;
case 'B': cout << "TUE ";
break;
case 'C': cout << "WED ";
break;
case 'D': cout << "THU ";
break;
case 'E': cout << "FRI ";
break;
case 'F': cout << "SAT ";
break;
case 'G': cout << "SUN ";
break;
}
char s1;
for (int i = pos+1; i < l; i++){
if (str1[i] == str2[i] && ((isupper(str1[i]) && str1[i] <= 'N') || (str1[i] >= '0'&& str1[i] <= '9'))){
s1 = str1[i];
break;
}
}
if (isupper(s1)){
cout << ((s1 - 'A') + 10) << ":";
}
else
cout << "0" << s1 - '0' << ":";
int m = 0;
if (str3.length() > str4.length()){
m = str4.length();
}
else
m = str3.length();
for (int i = 0; i < m; i++){
if (str3[i] == str4[i] && (islower(str3[i])|| isupper(str3[i]))){
if (i < 10)
cout << "0" << i << endl;
else
cout << i << endl;
break;
}
}
return 0;
}