Question

我正在解决此问题（https://open.kattis.com/problems/whowantstoliveforever）。如问题所述，我的程序必须根据输入0和1来确定Universe是生存还是死亡。

to determine the next value of the i-th bit, look at the current value of the bits at positions i−1 and i+1 (if they exist; otherwise assume them to be 0). If you see exactly one 1, then the next value of the i-th bit is 1, otherwise it is 0. All the bits change at once, so the new values in the next state depend only on the values in the previous state. We consider the universe dead if it contains only zeros.

我当前的解决方案适用于示例输入文件，但是在将其提交给Kattis时失败。（答案错误）

下面是我的代码。

import sys


def get_bit(bits, i):
    if 0 <= i < len(bits):
        return int(bits[i])
    else:
        return 0


def get_new_state(old_state):
    new_state = []
    for index in range(len(old_state)):
        if (get_bit(old_state, index-1) == 0 and get_bit(old_state, index+1) == 0) or (get_bit(old_state, index-1) == 1 and get_bit(old_state, index+1) == 1):
            new_state.append(0)
        elif(get_bit(old_state, index-1) == 0 and get_bit(old_state, index+1) == 1) or (get_bit(old_state, index-1) == 1 and get_bit(old_state, index+1) == 0):
            new_state.append(1)
    return new_state


def is_dead(state):
    if len(set(state)) == 1:
        return True
    else:
        return False


def foresee_fate(state):
    seen = []
    while True:
        if is_dead(state):
            return False
        if state in seen:
            return True
        seen.append(state)
        state = get_new_state(state)


def print_result(boolean):
    print("LIVES" if boolean else "DIES")


num_cases = int(sys.stdin.readline().strip())
for i in range(num_cases):
    cur_state = []
    case = sys.stdin.readline().strip()
    for char in case:
        cur_state.append(char)
    print_result(foresee_fate(cur_state))

请让我知道我可以做些什么来改进此程序。

Answer 1

根据建议，我已经在代码中添加了以下内容，并且现在可以正常工作。但是，我遇到了一个新问题。现在我得到"Time Limit Exceeded" > 3.00 s。

def is_dead(state):
    if set(state).pop() == 1:
        return False
    elif len(set(state)) == 1:
        return True
    else:
        return False

请让我知道是否有解决此问题的建议

使用列表时，Python输入错误

1 个答案: