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  1. PO
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    <p>This is complete conjecture, since I'm not really good with math, but I think I'd go about it like this:</p> <pre><code> int potency = 0, base = 1; while(base &lt; NumberInQuestion) { base *= 2; ++potency; } </code></pre> <p>After the loop finishes, you'll know the highest potency which still fits into 'Number'.</p> <pre><code> Number -= base/2; //Removes the base you just calculated from the number. printf("2^%d", potency); </code></pre> <p>Rinse and repeat, until Number falls to 0, which should be at 2^0 at latest.</p> <p>For your use-case, the code may look somewhat like this:</p> <pre><code>for(i=0; i &lt; n; ++i) { int Number = niz[i]; while(Number &gt; 0) { int potency = 0, base = 1; do { //Executes at least once, making a number of '1' possible. base *= 2; ++potency; } while(base &lt; Number); Number -= base/2; //Reverts the last step you made, making the 'base' smaller than 'Number'. printf("2^%d", potency); } } </code></pre> <p>There's a possible alternative, which can give you a more complete picture of things and will save you iterations. For this we use a two-step process.</p> <pre><code>for(i=0; i &lt; n; ++i) { int Number = niz[i]; int potency = 0, base = 1; do { //Executes at least once, making a number of '1' possible. base *= 2; ++potency; } while(base &lt; Number); base /= 2; //Reverses the last iteration. //At this point, we know the maximum potency, which still fits into the number. //In regards of base 2, we know the Most Significant Bit. while(base &gt; 0) { Number -= base; //Removes the MSD (Most significant digit) printf("2^%d", potency); //Prints your '1'. while(base &gt; Number) { //Executes at least once. base /= 2; //Goes back one potency. (Ends at '0' latest.) --potency; //For each potency (except for first), it's a '0'. } } } </code></pre>
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    1. POPrinting to output: integer as sum of powers of 2
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    1. USRefugnic Eternium
    1. USRefugnic Eternium
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    1. COOh ya, u gave me idea actually that i dont need to use variable sum. But how do I compare where is 1 and where is 0? I know that i will use for example > exp = ((nizb & 1)==0)?'0':'1' sec ill give it a try.. what about shifting?
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    2. CO@user2406286 There already is an answer, which explains the use of 'shifting' pretty well. My answer is for a more broad range of bases (if it doesn't have to be '2' of all things). Where there's one and where there's Zero is pretty self explanatory actually. It always looks for the 'highest multiple of base'...all those are '1'. All those who get skipped in the process are '0'. You can go reverse too. After determining the maximum multiple which fits into the original number, you decrease the base by itself and look if it's still bigger.
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