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    <p>Yes, there is a way. Consider the polynomial</p> <pre><code>(X + a[0]) * (X + a[1]) * ... * (X + a[n-1]) </code></pre> <p>Its coefficients are just the sums of the <code>k</code>-products, its degree is <code>n</code>, so you can calculate the sum of all <code>k</code>-products for all <code>k</code> simultaneously in O(n^2) steps.</p> <p>After <code>s</code> steps, the coefficient of X<sup>s-k</sup> is the sum of the <code>k</code>-products of the first <code>s</code> array elements. The <code>k</code>-products of the first <code>s+1</code> elements fall into two classes, those involving the <code>(s+1)</code><sup>st</sup> element - these have the form <code>a[s]*((k-1)</code>-product of the first <code>s</code> elements) - and those not involving it - these are the <code>k</code>-products of the first <code>s</code> elements.</p> <p>Code such that <code>result[i]</code> is the coefficient of X<sup>i</sup> (the sum of the <code>(n-i)</code>-products):</p> <pre><code>int *k_products_1(int *a, int n){ int *new, *old = calloc((n+1)*sizeof(int)); int d, i; old[0] = 1; for(d = 1; d &lt;= n; ++d){ new = calloc((n+1)*sizeof(int)); new[0] = a[d-1]*old[0]; for(i = 1; i &lt;= d; ++i){ new[i] = old[i-1] + a[d-1]*old[i]; } free(old); old = new; } return old; } </code></pre> <p>If you only want the sum of the <code>k</code>-products for one <code>k</code>, you can stop the calculation at index <code>n-k</code>, giving an O(n*(n-k)) algorithm - that's good if <code>k &gt;= n/2</code>. To get an O(n*k) algorithm for <code>k &lt;= n/2</code>, you have to organise the coefficient array the other way round, so that <code>result[k]</code> is the coefficient of X<sup>n-k</sup> (and stop the calculation at index <code>k</code> if you want only one sum):</p> <pre><code>int *k_products_2(int *a, int n){ int *new, *old = calloc((n+1)*sizeof(int)); int d, i; old[0] = 1; for(d = 1; d &lt;= n; ++d){ new = calloc((n+1)*sizeof(int)); new[0] = 1; for(i = 1; i &lt;= d; ++i){ new[i] = old[i] + a[d-1]*old[i-1]; } free(old); old = new; } return old; } </code></pre>
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    1. POEfficient algorithm to calculate the sum of all k-products
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    1. USDaniel Fischer
    1. USDaniel Fischer
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    1. COThis answer is referring to the fact that you want to evaluate the [elementary symmetric polynomial](http://en.wikipedia.org/wiki/Elementary_symmetric_polynomial) `e_k` at the values given by `L`.
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      1. USPengOne
 

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