[已解决]Python：每日一题 261

Unicorn#

阴阳神万物主 发表于 2019-10-23 16:10
其实吧，我的思路，用小数来说明是非常困难的，所以我就用分数来说明了。（在草稿上本来就是用分数在表 ...

用频数表示频率当然理解
困惑的是怎么实现从n-1到n的推进的

danteer

Unicorn# 发表于 2019-10-23 14:17
Dynamic Programing 动态规划

那。。。我写的程序也算是DP吧。。。大概？

Unicorn#

danteer 发表于 2019-10-23 17:06
那。。。我写的程序也算是DP吧。。。大概？

应该不是
DP是用过去的状态来描述下一阶段的状态

你的方法很有意思：逐步推进，分配概率而不是频数
我还没能验证它的可行性，无论如何，很有创意的点子

阴阳神万物主

Unicorn# 发表于 2019-10-23 17:05
用频数表示频率当然理解
困惑的是怎么实现从n-1到n的推进的

哦，这个啊，其实就是模拟了掷骰子的过程：在 n-1 时已经求出了(n-1)*6-(n-1)个点数，在此基础上再增加一个骰子的变数。假如说：扔三次，得到六点，所有的可能性如下：
1->1->4:p=(1/6)*(1/6)*(1/6) = p(n=2,x=2) * (1/6)
1->2->3与2->1->3:p=((1/6)*(1/6)+(1/6)*(1/6))*(1/6) = p(n=2,x=3) * (1/6)
1->3->2与2->2->2与3->1->2:p=((1/6)*(1/6)+(1/6)*(1/6)+(1/6)*(1/6))*(1/6) = p(n=2,x=4) * (1/6)
1->4->1与2->3->1与3->2->1与4->1->1:p=((1/6)*(1/6)+(1/6)*(1/6)+(1/6)*(1/6)+(1/6)*(1/6))*(1/6) = p(n=2,x=5) * (1/6)
此时p全部加起来，就是扔三次得到六点的总概率了。
所涉及的知识点：古典概型、条件概率

阴阳神万物主

Unicorn# 发表于 2019-10-23 17:05
用频数表示频率当然理解
困惑的是怎么实现从n-1到n的推进的

补个传送门：古典概型、条件概率

Unicorn#

danteer 发表于 2019-10-23 17:06
那。。。我写的程序也算是DP吧。。。大概？

很厉害的思路，学到了学到了
直接对概率进行操作。太厉害了
代码有些地方可以优化，帮你改了下，可读性变强

1. def func(n):
   
2.     if n == 0:
   
3.         return []
   
4.     else:
   
5.         p = 0
   
6.         result = {}
   
7.         for i in range(6*n+1): #初始化
   
8.             result[i] = 0
   
9. 
   
10.         while p < n:  #用p模拟掷骰子过程，p = 当前骰子数
    
11.             for i in range(p,6*p+1):
    
12.                 for j in range(1,7):
    
13.                     e = 1/(6**(p+1)) #单位概率
    
14.                     result[i+j] += e
    
15.                     result[i] -= e  #原有的某种情况中分出另一种情况的概率
    
16.             p += 1
    
17.         
    
18.         for i in range(1,n):
    
19.             result[i] = 0
    
20.         list1 = []
    
21.         for each in result:
    
22.             if result[each] > 0:
    
23.                 list1.append([each,result[each]])
    
24.         return list1

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Unicorn#

阴阳神万物主 发表于 2019-10-23 17:54
哦，这个啊，其实就是模拟了掷骰子的过程：在 n-1 时已经求出了(n-1)*6-(n-1)个点数，在此基础上再增加一 ...

思路都差不多嘛
从状态n-1推出状态n，这就是DP哦。只是我们的程序实现方式不同啦

Unicorn#

阴阳神万物主 发表于 2019-10-23 17:59
补个传送门：古典概型、条件概率

这个当然知道的啦
<等待下一期中........>

danteer

Unicorn# 发表于 2019-10-23 18:02
很厉害的思路，学到了学到了
直接对概率进行操作。太厉害了
代码有些地方可以优化，帮你改了下，可读性 ...

多谢多谢

阴阳神万物主

Unicorn# 发表于 2019-10-23 18:05
思路都差不多嘛
从状态n-1推出状态n，这就是DP哦。只是我们的程序实现方式不同啦

哦，原来是这样啊，对于算法，我了解得还不够呢。

zltzlt

Unicorn# 发表于 2019-10-22 21:26
最终版 dp

输入：1
输出：[[1,0.17],[1,0.17],[1,0.17],[1,0.17],[1,0.17],[1,0.17]]
预期结果：[[1,0.17],[2,0.17],[3,0.17],[4,0.17],[5,0.17],[6,0.17]]

动心成神

1. def points_count(n):
   
2.     '''
   
3.         求S集合
   
4.     '''
   
5.     num = [1,2,3,4,5,6]
   
6.     if n==1:
   
7.         return num
   
8.     else:
   
9.         points = points_count(n-1)
   
10.         points_num = []
    
11.         for i in num:
    
12.             points_num.extend(map(lambda x:x+i,points))
    
13.         return points_num
    
14. 
    
15. 
    
16. def prob_count(points_num):
    
17.     '''
    
18.         计算相应的概率
    
19.     '''
    
20.     num = len(points_num)
    
21.     points_set_list = list(set(points_num))
    
22.     prob_list = map(lambda x:[x,points_num.count(x)/num],points_set_list)
    
23.     return prob_list
    
24. 
    
25. 
    
26. 
    
27. print(list(prob_count(list(points_count(2)))))

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zltzlt

danteer 发表于 2019-10-22 22:10
这是纯粹拿浮点数算的，可能后面几位数字就不大对了。不过今天没空改上去了。
超时警报

输入：3
输出：[[3,0.03],[4,0.06],[5,0.10],[6,0.13],[7,0.16],[8,0.14],[9,0.11],[10,0.08],[11,0.06],[12,0.03],[13,0.03],[14,0.02],[15,0.02],[16,0.01],[17,0.01],[18,0.00]]
预期结果：[[3,0.00],[4,0.01],[5,0.03],[6,0.05],[7,0.07],[8,0.10],[9,0.12],[10,0.13],[11,0.13],[12,0.12],[13,0.10],[14,0.07],[15,0.05],[16,0.03],[17,0.01],[18,0.00]]

zltzlt

panheng 发表于 2019-10-22 22:18

输入：3
输出：[[3,0.00],[4,0.01],[5,0.03],[6,0.05],[7,0.07],[8,0.10],[9,0.12],[10,0.12],[11,0.12],[12,0.12],[13,0.10],[14,0.07],[15,0.05],[16,0.03],[17,0.01],[18,0.00]]
预期结果：[[3,0.00],[4,0.01],[5,0.03],[6,0.05],[7,0.07],[8,0.10],[9,0.12],[10,0.13],[11,0.13],[12,0.12],[13,0.10],[14,0.07],[15,0.05],[16,0.03],[17,0.01],[18,0.00]]

zltzlt

wisky 发表于 2019-10-22 22:19

输入 2 出错

zltzlt

战场原 发表于 2019-10-22 22:25
def solution(n: int) -> list:
    import collections
    count = 6**n

输入：3
输出：[[3,0.00],[4,0.01],[5,0.03],[6,0.05],[7,0.07],[8,0.10],[9,0.12],[10,0.12],[11,0.12],[12,0.12],[13,0.10],[14,0.07],[15,0.05],[16,0.03],[17,0.01],[18,0.00]]
预期结果：[[3,0.00],[4,0.01],[5,0.03],[6,0.05],[7,0.07],[8,0.10],[9,0.12],[10,0.13],[11,0.13],[12,0.12],[13,0.10],[14,0.07],[15,0.05],[16,0.03],[17,0.01],[18,0.00]]

zltzlt

阴阳神万物主 发表于 2019-10-22 23:27
数学是个很好的工具，这里我使用概率学的古典概型
毕竟，满足古典概型的特性：
1.所有可能出现的基本事件 ...

恭喜通过！

执行用时：988 ms

zltzlt

__mrsq__ 发表于 2019-10-23 10:57
我这个好像蛮快的，但是不知道结果是否正确

输入：3
输出：[[3,0.00],[4,0.01],[5,0.03],[6,0.05],[7,0.07],[8,0.10],[9,0.12],[10,0.12],[11,0.12],[12,0.12],[13,0.10],[14,0.07],[15,0.05],[16,0.03],[17,0.01],[18,0.00]]
预期结果：[[3,0.00],[4,0.01],[5,0.03],[6,0.05],[7,0.07],[8,0.10],[9,0.12],[10,0.13],[11,0.13],[12,0.12],[13,0.10],[14,0.07],[15,0.05],[16,0.03],[17,0.01],[18,0.00]]

zltzlt

塔利班 发表于 2019-10-23 12:57
理清了顺序重写了

这才是真正的大佬

执行用时：1063 ms

__mrsq__

zltzlt 发表于 2019-10-23 20:20
输入：3
输出：[[3,0.00],[4,0.01],[5,0.03],[6,0.05],[7,0.07],[8,0.10],[9,0.12],[10,0.12],[11,0.12] ...

这个好像是round函数输出0.125没进位，那我不用round你看看可以吗

1. def dice_outcome(dice_num):
   
2.     pro = [1] * 6
   
3.     n = dice_num
   
4.     d = 6**dice_num
   
5.     res = []
   
6.     while dice_num > 1:
   
7.         media_list = [0] * (len(pro) + 5)
   
8.         length = len(media_list)
   
9.         for index in range(int(length / 2 + 0.5)):
   
10.             if index > 5:
    
11.                 start = index - 5
    
12.             else:
    
13.                 start = 0
    
14.             for ele in pro[start:index + 1]:
    
15.                 media_list[index] += ele
    
16.             media_list[length - 1 - index] = media_list[index]
    
17.         pro = media_list
    
18.         dice_num -= 1
    
19.     for num in pro:
    
20.         res.append([n,num/d)])
    
21.         n += 1
    
22.     return res
    
23. 
    
24. print(dice_outcome(100))

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