题目
There are two types of soup: type A and type B. Initially we have N ml of each type of soup. There are four kinds of operations:
- Serve ;100 ml of soup A and 0 ml of soup B
- Serve ;75 ml of soup A and 25 ;ml of soup B
- Serve 50 ml of soup A and 50 ml of soup B
- Serve 25 ;ml of soup A and 75 ;ml of soup B
When we serve some soup, we give it to someone and we no longer have it. ; Each turn, ;we will choose from the four operations with equal probability 0.25. If the remaining volume of soup is not enough to complete the operation, we will serve ;as much as we can. ; We stop once we no longer have some quantity of both types of soup.
Note that we do not have the operation where all 100 ml's of soup B are used first. ; ;
Return the probability that soup A will be empty ;first, plus half the probability that A and B become empty at the same time.
;
Example:
Input: N = 50
Output: 0.625
Explanation:
If we choose the first two operations, A will become empty first. For the third operation, A and B will become empty at the same time. For the fourth operation, B will become empty first. So the total probability of A becoming empty first plus half the probability that A and B become empty at the same time, is 0.25 * (1 + 1 + 0.5 + 0) = 0.625.
Notes:
- 0 <= N <= 10^9. ;
- Answers within ;10^-6 ;of the true value will be accepted as correct.
解题思路
见程序注释