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""" 출처:프로그래머스, https://school.programmers.co.kr/learn/courses/30/lessons/181950?language=java """ import java.util.Scanner; public class Solution { public static void main(String[] args) { Scanner sc = new Scanner(System.in); String str = sc.next(); int n = sc.nextInt(); for(int i=0; i<n ; i++){ System.out.print(str); } } }
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/* 출처: 프로그래머스, https://school.programmers.co.kr/learn/courses/30/lessons/181951 */ /* import java.util.Scanner; public class Solution { public static void main(String[] args) { Scanner scanner=new Scanner(System.in); String str; str=scanner.nextLine(); System.out.println(str); } } */
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""" 출처:프로그래머스, https://school.programmers.co.kr/learn/courses/30/lessons/273709 """ #풀이 과정 """ select sum(PRICE) as TOTAL_PRICE from ITEM_INFO where RARITY = 'LEGEND' """
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""" 출처:프로그래머스, https://school.programmers.co.kr/learn/courses/30/lessons/273710 """ # 풀이 과정 """ select B.ITEM_ID,A.ITEM_NAME from ITEM_INFO as A,ITEM_TREE as B where 1=1 and A.ITEM_ID = B.ITEM_ID and B.PARENT_ITEM_ID is null order by ITEM_ID asc """
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""" 출처:프로그래머스, https://school.programmers.co.kr/learn/courses/30/lessons/273711 """ # 풀이 과정 """ -- 새로운 테이블의 형태를 조인해서 변환시킨 아이템 id의 이름을 가져오는 방향성 고민 # select A.ITEM_ID "ITEM_ID",B.ITEM_NAME,B.RARITY,B.ITEM_ID # from ITEM_TREE AS A, # (select * # from ITEM_INFO # where RARITY = "RARE") AS B # where B.ITEM_ID = A.PARENT_ITEM_ID # order by A.ITEM_ID desc select C.ITEM_ID,C.ITEM_NAME,C.RARITY from ITEM_INFO AS D, (select B.ITEM_ID,B.PARENT_ITEM_ID,A.RARITY,A.ITEM_NAME from ITEM_INFO AS A, ITEM_TREE AS B where B.ITEM_ID = A.ITEM_ID) AS C where 1=1 and D.ITEM_ID = C.PARENT_ITEM_ID and D.RARITY = "RARE" order by ITEM_ID desc """
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""" 출처:프로그래머스, https://school.programmers.co.kr/learn/courses/30/lessons/276034 """ # 풀이과정 """ -- 파이썬과 c# 스킬을 포함한 조합 구성 확보 후 테이블 조인 생각 # select ID,EMAIL,FIRST_NAME,LAST_NAME -- 이진법 구성을 보면 이진법 & 처리 생각! select distinct A.ID,A.EMAIL,A.FIRST_NAME,A.LAST_NAME from DEVELOPERS as A, (select * from SKILLCODES where NAME in ('Python','C#')) as B where A.SKILL_CODE & B.CODE = B.CODE order By A.ID asc """
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""" 출처:프로그래머스, https://school.programmers.co.kr/learn/courses/30/lessons/284527 """ select C.SCORE "SCORE", C.EMP_NO, B.EMP_NAME, B.POSITION, B.EMAIL from HR_DEPARTMENT as A,HR_EMPLOYEES as B, ( select EMP_NO,YEAR ,sum(score) "SCORE" from HR_GRADE group by EMP_NO,YEAR ) as C where B.EMP_NO = C.EMP_NO order by SCORE desc limit 1
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""" 출처:프로그래머스, https://school.programmers.co.kr/learn/courses/30/lessons/284530 """ # 풀이과정 """ select year(YM) "year",round(avg(PM_VAL1),2) "PM10",round(avg(PM_VAL2),2) "PM2.5" from AIR_POLLUTION group by year(YM),location2 having location2 = "수원" order by year asc """
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""" 출처:프로그래머스, https://school.programmers.co.kr/learn/courses/30/lessons/284531 """ """ select ROUTE, concat(cast(round( sum(D_BETWEEN_DIST),1) as char),"km") "TOTAL_DISTANCE", concat(cast(round( avg(D_BETWEEN_DIST),2) as char),"km") "AVERAGE_DISTANCE" from SUBWAY_DISTANCE group by ROUTE order by sum(D_BETWEEN_DIST) desc """
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""" 출처:프로그래머스, https://school.programmers.co.kr/learn/courses/30/lessons/293257 """ # 풀이 과정 """ select count(B.FISH_NAME) "FISH_COUNT", B.FISH_NAME from FISH_INFO as A join FISH_NAME_INFO as B on A.FISH_TYPE = B.FISH_TYPE # where A.LENGTH is not null Group by FISH_NAME order by FISH_COUNT desc """
