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.
中文翻譯
George 最近進入了 BSUCP ( Berland 州立酷工程師大學 )。George 有一個朋友 Alex 也進入了這間大學,現在他們要搬進宿舍。
George 跟 Alex 希望可以住在同一個房間。整個宿舍共有 n 個房間,而目前第 i 個房間共有 pi 個人住在裡面,而且整個房間最多可以容納 qi 個人 (pi ≤ qi)。你的任務是要去確認有多少間房間有足夠的空間讓 George 跟 Alex 一起搬進去。
輸入
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.
第一行包含了一個整數 n (1 ≤ n ≤ 100) — 房間的數量
接下來 n 列中,第 i 列包含了兩個整數 pi 跟 qi (0 ≤ pi ≤ qi ≤ 100) — 分別代表了已經入住的人數以及房間最多可以容納幾個人。
輸出
Print a single integer — the number of rooms where George and Alex can move in.
印出一個整數 — 有多少個房間是 George 跟 Alex 可以入住的。
範例
輸入
3 1 1 2 2 3 3
輸出
0
每一個房間都住滿人了
輸入
3 1 10 0 10 10 10
輸出
2
第一個房間還有9個位置,第二個房間沒人入住,還有10個位置
第三個房間已經住滿了,所以答案是 「2」間
解題思路
在我們的解答中,就是要迴圈拜訪每一行輸入並判斷剩餘空位是否大於等於2個 ( George 跟 Alex )。
C#解決方案
方案1
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解決方案
方案1
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);
}
}
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