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Thieves

TIME LIMIT = 1 SEC.

Every evening, hundreds of students gather up at canteen and buy various snacks. Many of these students prefer to pay through E-Wallets and UPI methods such as PayTM, PhonePe and Google Pay etc.
Due to the rush, the owner can't keep track of all payments and verify if a student has actually made the payment or not. But the canteen owner has visual memory i.e. he can remember a few digits of phone numbers after looking at it just once. The digits that he can't remember, he marks them with the symbol '%'.
Ex. For a phone number 1234567890, he may remember it as 1234%67%9%.
Now, it is your job to find out whether a given set of students have made the payment or not.

Input	Output
First line will contain N, number of students. Then the N phone numbers will follow. Each of the next N lines consists of two ten digit numbers: the number that the canteen owner remembers and the corresponding actual number.	Print two space-separated integers - the number of students who made the payment and the number of students who did not.

Constraints

1 ≤ T ≤ 100
Maximum number of digits per phone number in each test case = 10

Example Test Case

Input	Output
3 1234%67%9% 1234567890 349%%%%810 3497831233 56%%172273 5612172273	2 1

Solution

#include <iostream> #include <string> using namespace std; int main() { int n, i, j, c = 0; //c denotes the number of mismatched pairs of phone numbers. cin >> n; //take in number of numbers. string ip[n][2];//there are n pairs of phone numbers. for (i = 0; i < n; i++) ///take in all the pairs of phone numbers cin >> ip[i][0] >> ip[i][1]; for (i = 0; i < n; i++) //checking each pair. { for (j = 0; j < 10; j++) //checking corresponding characters in the phone numbers. { if (ip[i][0][j] != ip[i][1][j] && ip[i][0][j] != '%') { //if the digit is known and it is not equal then the numbers do not match. c++; break; } } } cout << n - c << " " << c; // n-c gives the matching pairs of phone numbers return 0; }