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    <p>You do 64 sums, one for each bit, for each of the sums you calculate sum mod n, this calculation return m for each bit that should to be set in the result, and 0 for each bit that should not be set.</p> <p><strong>Example:</strong><br> n = 3, m = 2. list = [5 11 5 2 11 5 2 11]</p> <pre><code> 5 11 5 2 11 5 2 11 sum of bit 0: 1 + 1 + 1 + 0 + 1 + 1 + 0 + 1 = 6 6 % 3 = 0 sum of bit 1: 0 + 1 + 0 + 1 + 1 + 0 + 1 + 1 = 5 5 % 3 = 2 sum of bit 2: 1 + 0 + 1 + 0 + 0 + 1 + 0 + 0 = 3 3 % 3 = 0 sum of bit 3: 0 + 1 + 0 + 0 + 1 + 0 + 0 + 1 = 3 3 % 3 = 0 </code></pre> <p>So only bit 1 is set, which means the result is 2.</p> <p><strong>Optimizing implementation:</strong><br> <em>(Tricks and considerations that are also useful for real problems)</em><br> It is worth noting that when iterating over an array, execution speed will to some extend be limited by memory access, if one need to perform multiple operations with each element it is usually fastest to perform them all on one element at a time, thus the processor only need to load each element from memory once. <a href="http://igoro.com/archive/gallery-of-processor-cache-effects/">Interesting blog post on memory and cache.</a></p> <p>It is possible to sum multiple bits in a single integer, rather than applying 64 different bitmasks to get each bit on it's own one could for instance use only 4 bitmasks that each extract 16 bits with 3 bits of space between each, as long as no overflow occur a normal addition operation will work exactly as if dealing with 16 4-bit integers, thus this method will work for 15 numbers. After 15 numbers have been processed this way the results must be added to a storage capable of holding bigger integers (could be 8 64-bit integers each holding 8 8-bit integers, they must of course in turn be emptied into bigger integers etc.).<br> The result is that rather than for each value doing 64 bitmasks, 63 bitshifts and 64 additions one need only do 4 bitmasks, 3 bitshifts and 4 additions, plus for every 15 values 8 bitmasks, 4 bitshifts and 8 additions, plus for every 255 values 16 bitmasks, 8 bitshifts and 16 additions etc.</p> <p><strong>Visualization:</strong><br> <em>(Summing 4x4-bit integers using 16-bit integers)</em></p> <pre><code>1000 1000 1000 1000 + 1000 0000 0000 1000 + 0000 0000 0000 1000 + 1000 1000 0000 0000 + 1000 0000 1000 0000 + 0000 0000 1000 1000 = 0010 0100 1100 0010 </code></pre> <p>The result is the same whether you consider this to be 4 columns of 4-bit integers or 1 column of 16-bit integers, this is only true as long as long the 4-bit integers do not overflow.</p>
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    1. COWow, it is the **most easy, clear** and **elegant** solution! And it [works](http://ideone.com/9H2q2) perfectly.
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    2. COElegant. This can also be thought of as generalising xor, just with modulo m instead of modulo 2.
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    3. COAnd _that's_ why they told him the exact size of integers. Great solution.
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