[Java] Baekjun No. 15552 - Fast A+B
We have compiled a solution using BufferedReader, StringTokenizer, and BufferedWriter for fast input/output in Java.
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[Java] Baekjun No. 15552 - Fast A+B
Solution
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import java.io.*;
import java.util.StringTokenizer;
public class Main {
public static void main(String args[]) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
int n = Integer.parseInt(br.readLine());
for(int i=0; i<n; i++) {
StringTokenizer st = new StringTokenizer(br.readLine());
int a = Integer.parseInt(st.nextToken());
int b = Integer.parseInt(st.nextToken());
bw.write((a+b) + "\n");
}
bw.flush();
bw.close();
}
}
많은 수의 입력과 출력이 있을 때 in Java
- In Java, it is better than the previously used
ScannerandSystem.out.print. - Better to use
BufferedReader + StringTokenizer + BufferedWriterBufferedReader (input)
- Read one line quickly
- Use:
import java.io.*; - Default:
BufferedReder br = new BufferedReader(new InputStreamReader(System.in)); - One line input:
String s = br.readLine();-> string - Convert to number:
int n = Integer.parseInt(br.readLine());StringTokenizer (Split)
- Faster than
split(), whitespace-based separator - Use:
import java.util.StringTokenizer; - Default:
StringTokenizer st = new StringTokenizer(br.readLine()); - Split:
int a = Integer.parseInt(st.nextToken());BufferedWriter (output)
- Print quickly
- Use:
import java.io.*; - Default:
BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out)); - Output:
bw.write("hello\n");-> Only strings are allowed inwrite() - Finalization (required):
bw.flush(); bw.close();-> Use only one last timeWhy can a+b, not a string, be entered in write() in the above code?
- In Java,
"문자열" + 숫자is treated as문자열, so it can be used because it is converted to a string through"\n"
import java.io.*
- Import input/output related classes at once ex) BufferedReader, BufferedWriter, InputStreamReader, OutputStreamWriter