PARTITION (number theory) ALSO explosive number in Codewars

The problem is codewars, in short: finding and combining different integers for an integer. Https://en.wikipedia.org/wiki/Partition_ (Number_theory)

For example: The combination of integers with 6 has the following 11 kinds,

6

5 + 1

4 + 2

4 + 1 + 1

3 + 3

3 + 2 + 1

3 + 1 + 1 + 1

2 + 2 + 2

2 + 2 + 1 + 1

2 + 1 + 1 + 1 + 1 + 1

1 + 1 + 1 + 1 + 1 + 1

Solution One:

To avoid repetition, we can choose to gradually decrement the first addend until it is not less than half of N (keep the number of decomposition), then apply this strategy to the first addend, or take 6 as an example:

6

5 + 1

4 + 2

3 + 3

Then apply this strategy to the first Addend 5,

4 + 1

3 + 2

Then apply this strategy to decomposition 4 and 3 to get the following:

6

5 + 1

4 + 1 + 1

3 + 1 + 1 + 1

2 + 1 + 1 + 1 + 1

1 + 1 + 1 + 1 + 1 + 1

2 + 2 + 1 + 1

3 + 2 + 1

4 + 2

2 + 2 + 2

3 + 3

Because we already have the 3+2+1, for 4 is not decomposed into 3+1+2, here the limitations of the decomposition 4 can not produce than 2 (4 after the 2) small number, so can only be the three, the same, 3 will no longer decompose, because the decomposition is not more than 3 (the latter) is larger than the number.

1 defexp_sum (n):2     ifN <0:3         return04     ifn = =0:5         return16     returnHelp (N, 1)7 8 defHelp (n, mini):9TMP = 1Ten     ifN <= 1: One         return1 A      forIinchRange (1, n/2+1):##保证n-I not less than I -         ifI >=Mini: # # Ensure that the number of decomposition is not less than the smaller number behind -TMP = tmp + Help (N-I, i) the     returntmp - PrintExp_sum (6)

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Of course, this recursive speed is very slow, t (n) = t (n-1) + t (n-2) +...+ t (ceil (N/2)), the time complexity is exponential level.

Solution Two:

You can open a list of size n to store the results that have been computed:

1 defexp_sum (n):2     ifN <0:3         return04     ifn = = 0orn = = 1:5         return16solution = [1] + [0]*N7 8      forIinchRange (1, N):9          forJinchRange (I, n+1):TenSOLUTION[J] + = solution[j-i] One     returnSolution[n]+1

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Solution Three:

According to generation function on Wikipedia:P(k) = p(k − 1) + p(k − 2) − p (k −5) − p(k −7) + p(k −12) + p(k −15) − P (k −22) − ... + ( -1) ^ (k+1) *P(k − (3k^2-k)/2) + ( -1) ^ (k+1) *P(k − (3k^2 + k)/2)

1 defexp_sum (n):2Solutions = [1]* (n + 1)3 4      forIinchXrange (1, n + 1):5J, K, s = 1, 1, 06          whileJ >0:7j = i-(3 * k * k + k)/28                 ifJ >=0:9s + = ( -1) * * (k+1) *Solutions[j]Tenj = i-(3 * k * k-k)/2 One                 ifJ >=0: As + = ( -1) * * (k+1) *Solutions[j] -K + = 1 -Solutions[i] =s the  -     returnSolutions[n]

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Reference Links:

Https://en.wikipedia.org/wiki/Partition_ (Number_theory)

Http://stackoverflow.com/questions/438540/projecteuler-sum-combinations

2015-07-15 17:03:25

PARTITION (number theory) ALSO explosive number in Codewars