高效的算法来计算所有k乘积的和 [英] Efficient algorithm to calculate the sum of all k-products

问题描述

假设您得到一个由n个数字组成的列表L和一个整数k<n.是否有一种有效的方法来计算L中k个不同数字的所有乘积之和?

Suppose you are given a list L of n numbers and an integer k<n. Is there an efficient way to calculate the sum of all products of k distinct numbers in L?

例如，以L=[1,3,4,6]和k=2为例.然后我要寻找的号码是

As an example, take L=[1,3,4,6] and k=2. Then the number I am looking for is

1*3 + 1*4 + 1*6 + 3*4 + 3*6 + 4*6.

您能想到一种避免生成所有大小为k的子集的方法吗?

Can you think of a way of doing it which avoids generating all the subsets of size k?

推荐答案

让F(X，k，n)为数组X的前n个元素的k乘积之和.

Let F(X,k,n) be the k-product sum of first n elements of array X.

F(X，k，n)= F(X，k，n-1)+ F(X，k-1，n-1)* X [n]

F(X,k,n) = F(X,k,n-1)+F(X,k-1,n-1)*X[n]

您可以使用动态编程解决的问题.复杂度= O(kn).

which you can solve using dynamic programming. Complexity = O(kn).

F(X，k，n)的结束条件:当n = k F(X，k，k)= X [1] * X [2] * ... * X [n]

End conditions for F(X,k,n): When n=k F(X,k,k) = X[1]* X[2]*...*X[n]

更多详细信息:

F(X,1,1) = X[1]
F(X,1,i) = F(X,1,i-1)+X[i] for i=2...n 

For j=2..n:
    For i = 1..k:
        if i<j:
            F(X,i,j) = F(X,i,j-1)+F(X,i-1,j-1)*X[j]
        else if i==j:
            F(X,i,j) = F(X,i-1,j-1)*X[j]
        else:
            pass

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