Offer_day29_60. n个骰子的点数
把n个骰子扔在地上，所有骰子朝上一面的点数之和为s。输入n，打印出s的所有可能的值出现的概率。
你需要用一个浮点数数组返回答案，其中第 i 个元素代表这 n 个骰子所能掷出的点数集合中第 i 小的那个的概率。
示例 1:
输入: 1
输出: [0.16667,0.16667,0.16667,0.16667,0.16667,0.16667]
示例 2:
输入: 2
输出: [0.02778,0.05556,0.08333,0.11111,0.13889,0.16667,0.13889,0.11111,0.08333,0.05556,0.02778]
限制：
1 <= n <= 11
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/nge-tou-zi-de-dian-shu-lcof
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
代码：
from leetcode_python.utils import *
class Solution:
def __init__(self):
pass
def dicesProbability(self, n: int) -> List[float]:
this_time = [1,1,1,1,1,1]
for time in range(n-1):
last_time = this_time
this_time=[sum(last_time[max(0,i-6):i]) for i in range(len(last_time)+6)]
res = [x/(6**n) for x in this_time if x!=0]
return res
def test(data_test):
s = Solution()
return s.dicesProbability(*data_test)
def test_obj(data_test):
result = [None]
obj = Solution(*data_test[1][0])
for fun, data in zip(data_test[0][1::], data_test[1][1::]):
if data:
res = obj.__getattribute__(fun)(*data)
else:
res = obj.__getattribute__(fun)()
result.append(res)
return result
if __name__ == '__main__':
datas = [
[1],
[2],
[5],
]
for data_test in datas:
t0 = time.time()
print('-' * 50)
print('input:', data_test)
print('output:', test(data_test))
print(f'use time:{
time.time() - t0}s')
备注：
GitHub：https://github.com/monijuan/leetcode_python
CSDN汇总：模拟卷Leetcode 题解汇总_卷子的博客-CSDN博客
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leetcode_python.utils详见汇总页说明
先刷的题，之后用脚本生成的blog，如果有错请留言，我看到了会修改的！谢谢！