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Competition Arena > NarrowPassage2Easy
SRM 638 · 2014-08-25 · by cgy4ever · Brute Force
Class Name: NarrowPassage2Easy
Return Type: int
Method Name: count
Arg Types: (vector<int>, int)
Problem Statement

Problem Statement

There is a narrow passage. Inside the passage there are some wolves. You are given a int[] 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.

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}
3
Returns: 2

From {1, 2, 3}, you can swap 1 and 2 to get {2, 1, 3}. But you can't get other permutations.

1)
{1, 2, 3}
1000
Returns: 6

Here you can swap any two adjacent wolves. Thus, all 3! = 6 permutations are possible.

2)
{1, 2, 3}
4
Returns: 3

You can get {1, 2, 3}, {2, 1, 3} and {2, 3, 1}.

3)
{1,1,1,1,1,1}
2
Returns: 720

All of these wolves are different, even though their sizes are the same. Thus, there are 6! different permutations possible.

4)
{2,4,6,1,3,5}
8
Returns: 60

Submissions are judged against all 68 archived test cases, of which 5 are shown here. Case numbers match the judge’s.

Coding Area

Language: C++17 · define a public class NarrowPassage2Easy with a public method int count(vector<int> size, int maxSizeSum) · 68 test cases · 2 s / 256 MB per case

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