PAT B1014 福尔摩斯的约会/ A1061 Dating (20 分) 测试点三

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 are actually referring to the coded time Thursday 14:04 -- since the first common capital English letter (case sensitive) shared by the first two strings is the 4th capital letter 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 23 in a day are represented by the numbers from 0 to 9 and the capital letters from A to N, respectively); and the English letter shared by the last two strings is s at the 4th position, representing the 4th minute. Now given two pairs of strings, you are supposed to help Sherlock decode the dating time.

以下代码：

#include 
using namespace std;

int main(){
	char* day[8] = {"0","MON","TUE","WED","THU","FRI","SAT","SUN"};
	int d, h, m;
	string str1, str2, str3, str4;
	char s1[100], s2[100], s3[100], s4[100];
	cin >> str1 >> str2 >> str3 >> str4;
	int cnt = 0;
	for(int i = 0; i= 'A' && s1[i] <= 'G' && cnt == 0)){
			d = s1[i] - 'A' + 1;
			cout << day[d];
			cnt++;
		}else if(s1[i] == s2[i] && s1[i] >= 'A' && s1[i] <= 'N' && cnt == 1){
			h = s1[i] - 'A' + 10;
			printf(" %02d:", h);
			cnt++;
		}else if(s1[i] == s2[i] && s1[i] >= '0' && s1[i] <= '9' && cnt == 1){
			h = s1[i] - '0';
			printf(" %02d:", h);
			cnt++;
		}
	}
	for(int i = 0; i < str3.length() && i < str4.length(); i++){
		s3[i] = str3[i];
		s4[i] = str4[i];
		if(s3[i] == s4[i] && ((s3[i] >= 'a' && s3[i] <= 'z') || (s3[i] >= 'A' && s3[i] <= 'Z'))){
			printf("%02d", i);
			break;
		}
	}
	return 0;
}