由网友(一生的笔)分享简介：假设我有一个数组 {1，2，5，4} 和 M = 3 。我需要找到：Suppose I have an array {1, 2, 5, 4} and m = 3.I need to find:1*2*5 + 1*2*4 + 1*5*4 + 2*5*4米元件从n个元素的数组的所有组合的乘法即萨姆i.e Sum...

假设我有一个数组 {1，2，5，4} 和 M = 3 。 我需要找到：

Suppose I have an array {1, 2, 5, 4} and m = 3. I need to find:

1*2*5 + 1*2*4 + 1*5*4 + 2*5*4

米元件从n个元素的数组的所有组合的乘法即萨姆

i.e Sum of multiplication of all combination of m element from an array of n elements.

其中一个可能的解决办法是找到所有组合，然后解决它，但是这将是 O（NCM）解决方案。有没有更好的算法？

One of the solution possible is to find all combination and then solving it but that would be O(nCm) solution. Is there any better algorithm?

推荐答案

这个问题等价于多项式的第M系数与给定根（计算的韦达定理）。在德尔福适应算法（O（N）内存和O（N ^ 2）时间）：

This problem is equivalent to calculation of Mth coefficient of polynom with given roots (Vieta's theorem). Adapted algorithm in Delphi (O(N) memory and O(N^2) time):

 function CalcMultiComb(const A: array of Integer; const M: Integer): Integer;
  var
    i, k, N: Integer;
    Coeff: TArray<Integer>;
  begin
    N := Length(A);
    if (N = 0) or (M > N) then
      Exit;
    SetLength(Coeff, N + 1);

    Coeff[0] := -A[0];
    Coeff[1] := 1;
    for k := 2 to N do begin
      Coeff[k] := 1;
      for i := k - 2 downto 0 do
        Coeff[i + 1] := Coeff[i] - A[k-1] * Coeff[i + 1];
      Coeff[0] := -A[k-1] * Coeff[0];
    end;

    Result := Coeff[N - M];
    if Odd(N - M) then
      Result := - Result;
  end;

调用CalcMultiComb（[1，2，3，4]，M），其中M = 1..4给予结果10，35，50，24

calls CalcMultiComb([1, 2, 3, 4], M) with M=1..4 give results 10, 35, 50, 24