Fix indices
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@ -411,10 +411,10 @@ The following code implements this algorithm:
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\begin{lstlisting}
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\begin{lstlisting}
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int best = 0;
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int best = 0;
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for (int first = 0; first < n; first++) {
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for (int a = 0; a < n; a++) {
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for (int last = first; last < n; last++) {
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for (int b = a; b < n; b++) {
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int sum = 0;
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int sum = 0;
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for (int k = first; k <= last; k++) {
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for (int k = a; k <= b; k++) {
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sum += array[k];
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sum += array[k];
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}
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}
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best = max(best,sum);
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best = max(best,sum);
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@ -423,9 +423,9 @@ for (int first = 0; first < n; first++) {
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cout << best << "\n";
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cout << best << "\n";
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\end{lstlisting}
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\end{lstlisting}
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The variables \texttt{first} and \texttt{last} determine the range
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The variables \texttt{a} and \texttt{b} fix the first and
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of the subarray,
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last index of the subarray,
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and the sum of the numbers is calculated to the variable \texttt{sum}.
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and the sum of values is calculated to the variable \texttt{sum}.
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The variable \texttt{best} contains the maximum sum found during the search.
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The variable \texttt{best} contains the maximum sum found during the search.
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The time complexity of the algorithm is $O(n^3)$,
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The time complexity of the algorithm is $O(n^3)$,
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@ -442,10 +442,10 @@ The result is the following code:
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\begin{lstlisting}
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\begin{lstlisting}
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int best = 0;
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int best = 0;
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for (int first = 0; first < n; first++) {
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for (int a = 0; a < n; a++) {
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int sum = 0;
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int sum = 0;
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for (int last = first; last < n; last++) {
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for (int b = a; b < n; b++) {
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sum += array[last];
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sum += array[b];
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best = max(best,sum);
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best = max(best,sum);
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}
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}
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}
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}
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