George has recently entered the BSUCP (Berland State University for Cool Programmers). George has a friend Alex who has also entered the university. Now they are moving into a dormitory.
George and Alex want to live in the same room. The dormitory has n rooms in total. At the moment the i-th room has pi people living in it and the room can accommodate qi people in total (pi ≤ qi). Your task is to count how many rooms has free place for both George and Alex.
Input
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of rooms.
The i-th of the next n lines contains two integers pi and qi (0 ≤ pi ≤ qi ≤ 100) — the number of people who already live in the i-th room and the room’s capacity.
Output
Print a single integer — the number of rooms where George and Alex can move in.
Eamples
Input
3 1 1 2 2 3 3
Output
0
Input
3 1 10 0 10 10 10
Output
2
Solution
In our solution, we need to iterate through all rooms and check if each room has at least two available spots for George and Alex to move in.
C# Solution
Solution1
int n = int.Parse(Console.ReadLine()); //total count of rooms
int res = 0;
for(int i = 0; i<n ; i++)
{
string[] arr = Console.ReadLine().Split(' '); //split by space
int alreadyUsed = int.Parse(arr[0]);
int capacity = int.Parse(arr[1]);
if(capacity-alreadyUsed >= 2){
res+=1;
}
}
Console.WriteLine(res);
Java Solution
Solution1
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = Integer.parseInt(scanner.nextLine()); //total count of rooms
int res = 0;
for (int i = 0; i < n; i++)
{
String[] arr = scanner.nextLine().split(" ");
int alreadyUsed = Integer.parseInt(arr[0]);
int capacity = Integer.parseInt(arr[1]);
if(capacity-alreadyUsed >= 2){
res+=1;
}
}
System.out.println(res);
}
}
Conclusion
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