python 「プログラマ脳を抱える数学パズル」をPythonで解いてみた。

Posted 2021-05-09

#! /usr/bin/python
# -*- coding: utf-8 -*-


def if_even(n):
    return n / 2


def if_odd(n):
    return n * 3 + 1


def collatz(n):
    m = if_odd(n)  # 初回
    while(True):
        if m == 1:
            return False
        elif m == n:
            return True
        elif m % 2 == 0:
            m = if_even(m)
        else:
            m = if_odd(m)


def main():
    collatz_nums = []
    for n in xrange(1, 10001):
        if collatz(n):
            collatz_nums.append(n)
    print '最初の数に戻る数:', collatz_nums
    print 'その個数:', len(collatz_nums)


if __name__ == '__main__':
    main()

#! /usr/bin/python
# -*- coding: utf-8 -*-

import itertools


def exchange(money, max):
    """ money を指定枚数 max 以内で両替できるパターンを返す """
    coins = [500, 100, 50, 10]
    candidate_patterns = []
    for i in xrange(2, max+1):
        candidate_patterns.extend(list(itertools.combinations_with_replacement(coins, i)))
    answer_patterns = []
    for i, pattern in enumerate(candidate_patterns):
        if sum(pattern) == money:
            answer_patterns.append(pattern)

    print '+ 組み合わせ一覧'
    for pattern in answer_patterns:
        print pattern
    print '\n組み合わせの数 =', len(answer_patterns)


def main():
    exchange(1000, 15)


if __name__ == '__main__':
    main()

#! /usr/bin/python
# -*- coding: utf-8 -*-


class Bar:

    def __init__(self, n, done=False):
        self.len = n
        self.done = done  # その回で既に切り分けたかどうかのフラグ


def calc_min_times(n, m):
    bar_list = [Bar(n)]
    count = 0
    print 'Now state:', count, show_state(bar_list)
    while not is_finished(bar_list):
        count += 1
        reset_done(bar_list)
        for i in xrange(m):
            for j, bar in enumerate(bar_list):
                if bar.len != 1 and bar.done is False:
                    len_tmp = bar.len
                    bar_list[j].len = len_tmp / 2
                    bar_list[j].done = True
                    bar_list.append(Bar(len_tmp - len_tmp / 2, True))
                    bar_list = sorted(bar_list, key=lambda x: x.len)
                    bar_list.reverse()
                    break
        print 'Now state:', count, show_state(bar_list)
    return count


def show_state(bar_list):
    states = []
    for bar in bar_list:
        states.append(bar.len)
    return states

def is_finished(bar_list):
    for bar in bar_list:
        if bar.len != 1:
            return False
    return True


def reset_done(bar_list):
    for bar in bar_list:
        bar.done = False

if __name__ == '__main__':
    q1 = calc_min_times(20, 3)
    q2 = calc_min_times(100, 5)
    print ''
    print 'Q1 Answer:', q1
    print 'Q2 Answer:', q2

#! /usr/bin/python
# -*- coding: utf-8 -*-

import sys


if __name__ == '__main__':
    ans = []

    kigo = ['+', '-', '*', '/', '']
    for a in xrange(0,10):
            for b in xrange(0,10):
                for c in xrange(0,10):
                    for d in xrange(0,10):
                        for k1 in kigo:
                            for k2 in kigo:
                                for k3 in kigo:
                                    equation = str(a)+k1+str(b)+k2+str(c)+k3+str(d)
                                    num_rev = str(d)+str(c)+str(b)+str(a)
                                    print equation
                                    e = None
                                    try:
                                        if k1 == '' and k2 == '' and k3 == '':
                                            continue
                                        if str(eval(equation)) == num_rev:
                                            ans.append(equation)
                                    except:
                                        pass
    print ans