【CodeChef】Enormous Input Test

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The purpose of this problem is to verify whether the method you are using to read input data is sufficiently fast to handle problems branded with the enormous Input/Output warning. You are expected to be able to process at least 2.5MB of input data per second at runtime.

Input

The input begins with two positive integers n k (n, k<=107). The next n lines of input contain one positive integer ti, not greater than 109, each.

Output

Write a single integer to output, denoting how many integers ti are divisible by k.

題解：記錄這道題主要是為了記錄java中Scanner和BufferReader的區別，開始用Scanner，效率非常低，所以就TLE了。根據StackOverFlow裡面解釋：BufferReader只是從流中讀入資料，但不對資料做任何處理，Scanner按照需求解析資料並讀入，比如nextInt(),nextDouble()等。更詳細的答案還有這裡。

總結一下：

• A BufferedReader is a simple class meant to efficiently read from the underling stream.
•  BufferedReader is synchronized, so read operations on a BufferedReader can safely be done from multiple threads.
• Scanner can parse the underlying stream for primitive types and strings using regular expressions.
• A scanner however is not thread safe, it has to be externally synchronized.

對於原文中的“ A scanner can do all that a BufferedReader can do and at the same level of efficiency as well.”不太認同，因為通過OJ來看，BufferReader的效率確實比Scanner高。

BufferReader的用法就用這道題的AC代碼記錄：

 1 import java.io.BufferedReader; 2 import java.io.IOException; 3 import java.io.InputStreamReader; 4  5 public class Main { 6  7     public static void main(String[] args) { 8         // TODO Auto-generated method stub 9         BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));10         try{11             String[] line =(bf.readLine()).split(" ");12             int n = Integer.parseInt(line[0]);13             int k = Integer.parseInt(line[1]);14             int count = 0;15             while(n-- > 0){16                 int num = Integer.parseInt(bf.readLine());17                 if(num%k == 0)18                     ++count;19             }20             System.out.println(count);21         }22         catch(IOException e){23             System.out.print("input error");24         }25     }26 27 }