#25079: 最不用思考的作法


cooljamesku92@gmail.com (你要不要訂閱一塊沒有影片的餅乾owo)


#include<iostream>

using namespace std;

 

int main() {

string input, check;

int n, count_B = 0, count_A = 0;

cin >> input >> n;

for (int i = 0; i < n; i++) {

cin >> check;

for (int i = 0; i < 4; i++) {

if (check.at(i) == input.at(i))

count_A++;

else {

for (int j = 0; j < 4; j++) {

if (check.at(i) == input.at(j))

count_B++;

}

}

}

cout << count_A << "A" << count_B << "B" << endl;

count_A = 0, count_B = 0;

}

return 0;

}