第3回オフラインリアルタイムどう書くの参考問題 pythonで解く
BallCount.py
# -*- coding:utf-8 -*-
# ボールカウントディクショナリの定義と初期化
ball_count_dict = {'o':0, 's':0, 'b':0}
def dict_initialize():
ball_count_dict['o'] = 0
ball_count_dict['s'] = 0
ball_count_dict['b'] = 0
# アウトのカウントアップ
def count_up_o():
if 2 == ball_count_dict.get('o'):
ball_count_dict['o'] = 0
ball_count_dict['s'] = 0
ball_count_dict['b'] = 0
elif 2 > ball_count_dict.get('o'):
ball_count_dict['o'] += 1
ball_count_dict['s'] = 0
ball_count_dict['b'] = 0
# ストライクのカウントアップ
def count_up_s(pitch_type):
if 2 == ball_count_dict['s'] and 's' == pitch_type:
count_up_o()
elif 2 > ball_count_dict['s']:
ball_count_dict['s'] += 1
# ボールのカウントアップ
def count_up_b():
if 3 > ball_count_dict.get('b'):
ball_count_dict['b'] += 1
else:
ball_count_dict['s'] = 0
ball_count_dict['b'] = 0
# 実行メソッド
def execute(pitch_types):
count_list = []
dict_initialize()
for pitch_type in pitch_types:
if 's' == pitch_type:
count_up_s(pitch_type)
elif 'f' == pitch_type:
count_up_s(pitch_type)
elif 'b' == pitch_type:
count_up_b()
elif 'p' == pitch_type:
count_up_o()
elif 'h' == pitch_type:
ball_count_dict['s'] = 0
ball_count_dict['b'] = 0
count_list.append(str(ball_count_dict.get('o')) + str(ball_count_dict.get('s')) + str(ball_count_dict.get('b')))
return count_list
# テスト実行メソッド
def test(pitch_types, exp):
result = execute(pitch_types)
if exp == ','.join(result):
print pitch_types + ' => ' + exp + " : OK"
if __name__ == "__main__":
test('s' , '010')
test('sss' , '010,020,100')
test('bbbb' , '001,002,003,000')
test('ssbbbb' , '010,020,021,022,023,000')
test('hsbhfhbh' , '000,010,011,000,010,000,001,000')
test('psbpfpbp' , '100,110,111,200,210,000,001,100')
test('ppp' , '100,200,000')
test('ffffs' , '010,020,020,020,100')
test('ssspfffs' , '010,020,100,200,210,220,220,000')
test('bbbsfbppp' , '001,002,003,013,023,000,100,200,000')
test('sssbbbbsbhsbppp' , '010,020,100,101,102,103,100,110,111,100,110,111,200,000,100')
test('ssffpffssp' , '010,020,020,020,100,110,120,200,210,000')
Author And Source
この問題について(第3回オフラインリアルタイムどう書くの参考問題 pythonで解く), 我々は、より多くの情報をここで見つけました https://qiita.com/mtdtysk/items/b8e32d2d3818534aac7d著者帰属:元の著者の情報は、元のURLに含まれています。著作権は原作者に属する。
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