출처 : OEIS- A071816

의 상한이 주어진 작업 n은 방정식을 만족시키는 솔루션의 수를 찾는 것입니다.

a+b+c = x+y+z, where 0 <= a,b,c,x,y,z < n

순서는 OEIS 페이지에 설명 된대로 그리고 아래 (1)로 시작됩니다.

1, 20, 141, 580, 1751, 4332, 9331, 18152, 32661, 55252, 88913, 137292, 204763, 296492, 418503, 577744, 782153, 1040724, 1363573, 1762004, 2248575, 2837164, 3543035, 4382904, 5375005, 6539156, 7896825, 9471196, 11287235, 13371756

를 들어 n = 1, 해결책은 단 하나입니다 :(0,0,0,0,0,0)

의 경우 n = 220 가지 주문 솔루션 (a,b,c,x,y,z)이 있습니다 a+b+c = x+y+z.

(0,0,0,0,0,0), (0,0,1,0,0,1), (0,0,1,0,1,0), (0,0,1,1,0,0), (0,1,0,0,0,1),
(0,1,0,0,1,0), (0,1,0,1,0,0), (0,1,1,0,1,1), (0,1,1,1,0,1), (0,1,1,1,1,0),
(1,0,0,0,0,1), (1,0,0,0,1,0), (1,0,0,1,0,0), (1,0,1,0,1,1), (1,0,1,1,0,1),
(1,0,1,1,1,0), (1,1,0,0,1,1), (1,1,0,1,0,1), (1,1,0,1,1,0), (1,1,1,1,1,1).

난 & O를

• 입력은를 나타내는 단일 정수입니다 n.
• 출력은 단일 정수 / 문자열 인 의미하는데 f(n), f(...)상기 기능이다.
• 인덱싱은 설명 된대로 정확하게 수행되며 다른 인덱싱은 허용되지 않습니다.

이것은 code-golf , 가장 낮은 바이트 수의 승리입니다.

답변

ṗ6ḅ-ċ0

ṗ6ḅ-ċ0  Main link. Argument: n

ṗ6      Cartesian power 6; build all 6-tuples (a, x, b, y, c, z) of integers in
        [1, ..., n]. The challenge spec mentions [0, ..., n-1], but since there
        are three summands on each side, this doesn't matter.
  ḅ-    Unbase -1; convert each tuple from base -1 to integer, mapping (a, ..., z)
        to a(-1)**5 + x(-1)**4 + b(-1)**3 + y(-1)**2 + c(-1)**1 + z(-1)**0, i.e.,
        to -a + x - b + y - c + z = (x + y + z) - (a + b + c). This yields 0 if and
        only if the 6-tuple is a match.
    ċ0  Count the number of zeroes.

.55#^5+#^3/4+#/5&

Length@Solve[a+b+c==x+y+z&&And@@Table[(0<=i<#),{i,{a,b,c,x,y,z}}],Integers]&

f n=sum[1|0<-foldr1(-)<$>pure[1..n]`mapM`[1..6]]

l6:"G:gY+]X>

l        % Push 1
6:"      % Do the following 6 times
  G:g    %   Push a vector of n ones, where n is the input
  Y+     %   Convolution
]        % End
X>       % Maximum

ṗ3S€ĠL€²S

ṗ3S€ĠL€²S  Main link. Argument: n

ṗ3         Cartesian power; yield all subsets of [1, ..., n] of length 3.
  S€       Sum each.
    Ġ      Group indices by their values; for each unique sum S, list all indices whose
           values are equal to S.
     L€    Length each; for each unique sum S, yield the number of items in the original
           array that sum to S.
       ²   Square each; for each unique sum S, yield the number of pairs that both sum to S.
        S  Sum; yield the total number of equal pairs.

JsM^UQ3s/LJJ

JsM^UQ3s/LJJ
   ^UQ3         Get all triples in the range.
JsM             Save the sums as J.
        /LJJ    Count occurrences of each element of J in J.
       s        Take the sum.

lambda n:.55*n**5+n**3/4+n/5