반례:
18 48 BBBBBWBBBWWWWWBBWWBBBWBBWBWBBBWBWWBWBBBWBWWBBBWB BBWBBBWBBBWWWWWBBWBWBBBBBBBBBBBBBWWWWWBWWBWWBBWW WBWWBWBWBBBWBWBBBWWWWBWBBWWWBWWWWWWWWWWBWWWBWBBW BBWBWBBBBWWWBWWWWBBWBWWBBWWBBBBBWWBWBBBBBBWWBBBW BBBWBWWBWBBWBBWBWBWBWWBWWBBWWWBWWBBWBBBBWWBBBWBB WBWBBBWWWWBWWBBWWBWBBWBBBWBWWBBBBWBBBWBWBBWBWWBW BWWWWBBWBBBBWWWWWBBBBBBBBBBWBBWBWWWWBBBBBWBWWWWW WBWWBWBBWBBWBBWWBWWWWBWWWWWBBWWWBBWBBWWBWBBWBWWW BWBWBBWBWBBWWBBBWBWWWWBWBWWBWWBWBWWWBWBBBWBBBBBW WBWBWBWBWBWWWWWBWWBBWBBWWBBBBBBBBBBBBBWBBWWWWBBW BBBWBBWBWWBBBWWWWWBBBWBBWBWWWWWBWBWBWBBBBBBBBBWB WBWBWWWWBBBBBBWWWBBBWWBWWBWWWWWWBBWBBWWBWBBWBWBB BBWWWWWBBWBBBBWWBBBWBBWWWBBBWWWBWWWWBBWBWWWWWWBB BBWBBWBWWWBWBWBBWBBBBBWBBWWWWBWBBBWBWBBWBWBWBBWB BWBWBBWBBWBBWWWBBWBBWBBWBBBWWWBWBBBWWBBWWWWBWBBB WWWBWBWBWWWBWWBWWBBWWBBBBBBWBWWBWWWBBBBWWBWBBWWW BBBWBBWBWWBBBBBBWBBBBBWWWBWBBBBWBWWBWWWWWWBWBBBW BWBWWWWWWWWWWWBBBWWBBBWBBBWBWBBWBBBWBWBWBBWWBBWB
답: 20
출력: 21
binwon236 2년 전
코딩 시작한지 얼마 되지않아 코드가 아주 개판입니다.
알고리즘은 이렇습니다.
0. 문자열로 체스판을 입력 받는다.
1. 맨 처음의 체스 판의 색깔을 받고 key값에 저장한다.
2. check함수 호출
3. check 함수 key값이 'b' 일 경우 인덱스의 짝수 번이 'b' 이기에 b가 아니면 다음 문자열의 인덱스로 이동, count+1
4. 위 반대의 경우를 조건문으로 구현
가로 방향으로 모두 돌 경우 세로 방향으로 +1 하고 key값을 부정
대충 이렇습니다. 이것도 개판이네여 죄송합니다.
반례는 대충 다 넣어봤는데 1초만에 틀렸습니다가 뜹니다...
도와주십쇼~~~!!!!