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2nd optimisation |
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j = j + calcval;
}</pre></code>
This can then be further optimized, leading to less overall executed code for larger values of maxval and/or smaller values of calcval.
<code><pre>
int calcval = (4+array[k])*pi+5;
int j = j + integer_part(maximum - 1) / calcval) * calcval;
</pre></code>
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