水题，就是一个暴力。大力出奇迹。
Problem Statement
There is a narrow passage. Inside the passage there are some wolves. You are given a vector size that contains the sizes of those wolves, from left to right.
The passage is so narrow that some pairs of wolves cannot pass by each other. More precisely, two adjacent wolves may swap places if and only if the sum of their sizes is maxSizeSum or less. Assuming that no wolves leave the passage, what is the number of different permutations of wolves in the passage? Note that two wolves are considered different even if they have the same size.
Compute and return the number of permutations of wolves that can be obtained from their initial order by swapping a pair of wolves zero or more times.
Definition
Class: NarrowPassage2Easy Method: count Parameters: vector , int Returns: int Method signature: int count(vector size, int maxSizeSum) (be sure your method is public)
Limits
Time limit (s): 2.000 Memory limit (MB): 256
Constraints
– size will contain between 1 and 6 elements, inclusive. – Each element in size will be between 1 and 1,000, inclusive. – maxSizeSum will be between 1 and 1,000, inclusive.
Examples
0) @@######@@
{1, 2, 3}
登录后复制登录后复制登录后复制 From {1, 2, 3}, you can swap 1 and 2 to get {2, 1, 3}. But you can’t get other permutations. 1) @@######@@ @@######@@
Returns: 2
登录后复制 Here you can swap any two adjacent wolves. Thus, all 3! = 6 permutations are possible. 2) @@######@@
{1, 2, 3}
登录后复制登录后复制登录后复制 You can get {1, 2, 3}, {2, 1, 3} and {2, 3, 1}. 3) @@######@@
1000
登录后复制登录后复制 All of these wolves are different, even though their sizes are the same. Thus, there are 6! different permutations possible. 4) @@######@@
Returns: 6
登录后复制 5) @@######@@ @@######@@
{1, 2, 3}
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Returns: 3
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{1,1,1,1,1,1}
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Returns: 720
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{2,4,6,1,3,5}
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Returns: 60
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{1000}
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1000
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Returns: 1
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#include #include #include #include #include #include #include #include #include #include #include
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