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15 changes: 15 additions & 0 deletions 지민/week5/1204.py
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# import sys
#
# sys.stdin = open("input.txt", "r")

T = int(input())
# 여러개의 테스트 케이스가 주어지므로, 각각을 처리합니다.
for test_case in range(1, T + 1):
rounds = int(input())
scores = list(map(int, input().split()))
scores_set = list(set(scores))
frequency = []
for i in range(len(scores_set)):
frequency.append(scores.count(scores_set[i]))
max_index = frequency.index(max(frequency))
print(f"#{rounds} {scores_set[max_index]}")
17 changes: 17 additions & 0 deletions 지민/week5/12916.py
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# Level 1 - 문자열 내 p와 y의 개수
def solution(s):
answer = True

lower_s = s.lower()
count_p = lower_s.count('p')
count_y = lower_s.count('y')
if (count_p == count_y == 0):
return True
elif (count_p == count_y):
return True
else:
return False

# 다른 풀이 - 비교해서 값 바로 반환
def numPY(s):
return s.lower().count('p') == s.lower().count('y')
7 changes: 7 additions & 0 deletions 지민/week5/12928.py
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# Level 1 - 약수의 합
def solution(n):
answer = 0
for i in range(1, n+1):
if (n % i == 0):
answer += i
return answer
7 changes: 7 additions & 0 deletions 지민/week5/12937.py
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# Level 1 - 짝수 홀수 문제
def solution(num):
if (num % 2 == 0):
return "Even"

else:
return "Odd"
7 changes: 7 additions & 0 deletions 지민/week5/12944.py
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# Level 1 - 평균 구하기
def solution(arr):
answer = 0
for i in range(0, len(arr)):
answer += arr[i]
answer /= len(arr)
return answer
6 changes: 6 additions & 0 deletions 지민/week5/12954.py
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# Level 1 - x만큼 간격이 있는 n개의 숫자
def solution(x, n):
answer = []
for i in range(1, n+1):
answer.append(x*i)
return answer
15 changes: 15 additions & 0 deletions 지민/week5/138477.js
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//Level 1 - 명예의 전당 문제 풀이
function solution(k, score) {
let scores = [];
let answers = [];
for (let i = 0; i < score.length; i++) {
scores.push(score[i]);
scores.sort((a, b) => b - a);
if (scores.length <= k) {
answers.push(scores[scores.length - 1]);
} else {
answers.push(scores[k - 1]);
}
}
return answers;
}
17 changes: 17 additions & 0 deletions 지민/week5/1545.py
Original file line number Diff line number Diff line change
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import sys
sys.stdin = open("input.txt", "r")

input_num = int(input())
while (input_num >= 0):
print(input_num, end=' ')
input_num -= 1

# 다른 풀이 - 왜인지 이 풀이가 더 실행시간이 빠르다!
n = int(input())
answer = []
for i in range(n + 1):
answer.append(n)
n -= 1

for j in range(len(answer)):
print(answer[j], end=' ')
19 changes: 19 additions & 0 deletions 지민/week5/1859.py
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# import sys
#
# sys.stdin = open("input.txt", "r")

# 배열 끝을 최댓값으로 설정, 앞으로 오면서 최댓값과의 차이를 비교(최댓값 갱신일 떄는 계산 안 함)

T = int(input())
# 여러개의 테스트 케이스가 주어지므로, 각각을 처리합니다.
for test_case in range(1, T + 1):
days = int(input())
answer = 0
costs = list(map(int, input().split()))
max_cost = costs[-1]
for i in range(days-2, -1, -1):
if max_cost <= costs[i]:
max_cost = costs[i]
else:
answer += max_cost - costs[i]
print(f"#{test_case} {answer}")
51 changes: 51 additions & 0 deletions 지민/week5/1926.py
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# import sys
#
# sys.stdin = open("input.txt", "r")
#
# rounds = int(input())
# for i in range(rounds):
# counts = 0
# check = i + 1
# answers = []
# while check > 0:
# if check % 10 == 3 or check % 10 == 6 or check % 10 == 9:
# if answers and answers[-1] == "-":
# answers.append("-")
# else:
# answers.append(" -")
# elif check % 10 > 0:
# if answers and answers[-1] == '-':
# answers.append(f" {str(check % 10)} ")
# else:
# answers.append(f" {str(check % 10)}")
# check //= 10
# answers.reverse()
# print(''.join(answers), end='')

# 풀이 1 - count로 완전 간단히
# i라는 숫자에 3 6 9 가 들어간 수 만큼 -출력, 띄어쓰기 없이
# 그 외에는 일반 숫자 출력

T = int(input())
for i in range(1, T+1): # 1 ~ 100

i = str(i)
clap = i.count('3') + i.count('6') + i.count('9')

if clap == 0:
print(i, end=' ')
else:
print("-" * clap, end=' ')

# 풀이 2 - list 사용
N = int(input())
clap = ['3', '6', '9']

for i in range(1, N+1):
count = 0
for j in str(i):
if j in clap:
count += 1
if count > 0:
i = '-' * count
print(i, end=' ')
1 change: 1 addition & 0 deletions 지민/week5/1928.py
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ord
21 changes: 21 additions & 0 deletions 지민/week5/1933.py
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import sys
sys.stdin = open("input.txt", "r")

input_num = int(input())
answers = []

i = 1
while i <= input_num:
if input_num % i == 0:
answers.append(i)
i += 1
for nums in answers:
print(nums, end=' ')

# 다른 풀이
n = int(input())
answer = []
for i in range(1,n+1) :
if n % i == 0 :
answer.append(i)
print(*answer)
27 changes: 27 additions & 0 deletions 지민/week5/1936.py
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a, b = map(int, input().split())
if (a == 1 and b == 2) or (a == 2 and b == 3) or (a==3 and b == 1):
print('B')
else:
print('A')

# 다른 풀이 - 리스트 사용
A, B = map(int, input().split())
# 가위, 바위, 보
# 1 2 3
win = [0, 3, 1, 2]

if A == B:
print('비김')
elif win[A] == B:
print('A')
else:
print('B')

#다른 풀이 2 - 뺼셈 값으로 정답 반환
A , B = map(int,input().split())
answer = ''
if A-B == 1 or B-A == 2:
answer = 'A'
elif B-A == 1 or A-B == 2:
answer = 'B'
print(answer)
19 changes: 19 additions & 0 deletions 지민/week5/1938.py
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import sys
sys.stdin = open("input.txt", "r")

a, b = map(int, input().split())
sum_result = a + b
minus_result = a - b
multiply_result = a * b
division_result = a // b
print(sum_result)
print(minus_result)
print(multiply_result)
print(division_result)

# 다른 풀이
a, b = map(int, input().split())
print(a+b)
print(a-b)
print(a*b)
print(a//b)
7 changes: 7 additions & 0 deletions 지민/week5/1954.py
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import sys
sys.stdin = open("input.txt", "r")

T = int(input())
# 여러개의 테스트 케이스가 주어지므로, 각각을 처리합니다.
for test_case in range(1, T + 1):
3
36 changes: 36 additions & 0 deletions 지민/week5/1974.py
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# import sys
# sys.stdin = open("input.txt", "r")

T = int(input())
# 여러개의 테스트 케이스가 주어지므로, 각각을 처리합니다.
for test_case in range(1, T + 1):
answer = 0
board = [list(map(int, input().split())) for _ in range(9)]

for i in range(9):
# 가로 검사
row_check = [0] * 10
# 세로 검사
col_check = [0] * 10
for j in range(9):
row_check[board[i][j]] += 1
col_check[board[j][i]] += 1
# 중복 체크
for k in range(1, 10):
if row_check[k] != 1:
answer = 0
break

if col_check[k] != 1:
answer = 0
break

# 3*3 검사
for i in range(3):
for j in range(3):
square_check = [0]*10
for k in range(3):
for l in range(3):
square_check[board[3i + k]]

# print(f"#{test_case) {answer}"
17 changes: 17 additions & 0 deletions 지민/week5/1984.py
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# import sys
#
# sys.stdin = open("input.txt", "r")

T = int(input())
# 여러개의 테스트 케이스가 주어지므로, 각각을 처리합니다.
for test_case in range(1, T + 1):
nums = list(map(int, input().split()))
min_num = min(nums)
max_num = max(nums)
for number in range(len(nums)):
if min_num in nums:
nums.remove(min_num)
if max_num in nums:
nums.remove(max_num)
answer = round(sum(nums) / len(nums))
print(f"#{test_case} {answer}")
19 changes: 19 additions & 0 deletions 지민/week5/2001.py
Original file line number Diff line number Diff line change
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import sys

sys.stdin = open("input.txt", "r")

T = int(input())
# 여러개의 테스트 케이스가 주어지므로, 각각을 처리합니다.
for test_case in range(1, T + 1):
n, m = map(int, input().split())
flies = [list(map(int, input().split())) for _ in range(n)]
max_sum = 0

for i in range(n - m + 1):
for j in range(n - m + 1):
current_sum = sum(
sum(flies[x][y] for y in range(j, j + m))
for x in range(i, i + m)
)
max_sum = max(max_sum, current_sum)
print(f"#{test_case} {max_sum}")
19 changes: 19 additions & 0 deletions 지민/week5/2019.py
Original file line number Diff line number Diff line change
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import sys
sys.stdin = open("input.txt", "r")


total = int(input())
init = 1
print(1, end=' ')
for _ in range(1, total+1):
print(2*init, end=' ')
init = 2*init

# 다른 풀이 - 더 깔끔하다
N = int(input())

result = 1

for i in range(N + 1):
print(result, end=' ')
result *= 2
10 changes: 10 additions & 0 deletions 지민/week5/2025.py
Original file line number Diff line number Diff line change
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rounds = int(input())
sum = 0
for i in range(1, rounds+1):
sum += i
print(sum)

# 다른 풀이 - list 활용
n = int(input())
n1 = list(range(1, n+1))
print(sum(n1))
2 changes: 2 additions & 0 deletions 지민/week5/2027.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,2 @@
for i in range(0, 5):
print('+'*i+'#'+'+'*(4 - i))
21 changes: 21 additions & 0 deletions 지민/week5/2029.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,21 @@
# import sys
#
# sys.stdin = open("input.txt", "r")

T = int(input())
# 여러개의 테스트 케이스가 주어지므로, 각각을 처리합니다.
for test_case in range(1, T + 1):
a, b = map(int, input().split())
div, mod = divmod(a, b)
print(f'#{test_case} {div} {mod}')

# 다른 풀이 - 내장함수 없는 버전, 다른 출력 형태
T = int(input())
# 여러개의 테스트 케이스가 주어지므로, 각각을 처리합니다.
for test_case in range(1, T + 1):
a, b = map(int, input().split())

quotient = a // b
remainder = a % b

print('#{0} {1} {2}'.format(test_case, quotient, remainder))
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