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[백준 Java]15552 : 빠른 A+B

by 종이빨대 2023. 9. 1.
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목차

    1. 문제

    2. 맞춘코드

    // [백준 Java]15552 : 빠른 A+B
    import java.io.BufferedReader;
    import java.io.InputStreamReader;
    import java.io.IOException;
    import java.util.StringTokenizer;
    
    
    public class bj15552{
    	public static void main(String[] args) throws IOException {
    		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    		StringTokenizer st;
    		StringBuilder sb = new StringBuilder();
    
    		int num=Integer.parseInt(br.readLine());
    
    		for(int i=0; i<num; i++){
    			st = new StringTokenizer(br.readLine(), " ");
    			sb.append(Integer.parseInt(st.nextToken()) + Integer.parseInt(st.nextToken())).append("\n");
    		}
    		br.close();
    		System.out.println(sb);
    	}
    }

    3. 틀린코드

    3.1. 시간초과: 테스트 케이스마다 System.out.println() 을 호출

    // [백준 Java]15552 : 빠른 A+B
    import java.io.BufferedReader;
    import java.io.InputStreamReader;
    import java.io.IOException;
    import java.util.StringTokenizer;
    
    
    public class bj15552{
    	public static void main(String[] args) throws IOException {
    		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    		StringTokenizer st;
    
    		int num=Integer.parseInt(br.readLine());
    
    		for(int i=0; i<=num; i++){
    			st = new StringTokenizer(br.readLine(), " ");
    			int a=Integer.parseInt(st.nextToken());
    			int b=Integer.parseInt(st.nextToken());
    			
    			System.out.println(a+b);
    
    		}
    	}
    }

    4. 참고사이트

    https://st-lab.tistory.com/30