フルナビゲーション

模擬試験

def solution(answers):
    tester1 = [1, 2, 3, 4, 5]
    tester2 = [2, 1, 2, 3, 2, 4, 2, 5]
    tester3 = [3, 3, 1, 1, 2, 2, 4, 4, 5, 5]
    rights = [0,0,0]
    count = 0
    for answer in answers:
        if answer == tester1[count%len(tester1)]:
            rights[0] += 1
        if answer == tester2[count%len(tester2)]:
            rights[1] += 1
        if answer == tester3[count%len(tester3)]:
            rights[2] += 1 
        count += 1
        
    max = rights[0]
    result = [1]
    if rights[1] > max:
        max = rights[1]
        result = [2]
    elif rights[1] == max:
        result.append(2)
    if rights[2] > max:
        max = rights[2]
        result = [3]
    elif rights[2] == max:
        result.append(3)

小数点を検索

#coding:utf8
# 소수 찾기
# 완전 탐색

from itertools import permutations

def solution(numbers):
    answer = 0

    # 중복 조합 생성
    perList = []
    for i in range(len(numbers)):
        pers = list(permutations(list(numbers),i+1))
        for per in pers:
            per = int("".join(per))
            if per not in perList:
                perList.append(per)
    # 소수인지 체크
    for num in perList:
        primeBool = True
        if num == 1 or num == 0:
            primeBool = False
        else:
            for j in range(2,num):
                if num%j == 0:
                    primeBool = False
                    break
        if primeBool:
            answer += 1

    return answer

print(solution("011"))

じゅうたん

#coding:utf8
# 카펫
# 완전탐색

def solution(brown, yellow):
    answer = []
    divisors = [] 

    # yellow의 약수구하기
    for i in range(1,yellow):
        if yellow % i == 0:
            if yellow/i < i :
                break
            divisors.append(yellow/i)
            divisors.append(i)
    if yellow == 1:
        divisors = [1,1]

    # 가로세로 길이 구하기
    for j in range(0, len(divisors), 2):
        if brown == divisors[j]*2 + divisors[j+1]*2 + 4:
            answer = [divisors[j]+2,divisors[j+1]+2]
            break
    
    return answer

print(solution(10, 2))