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Dont think so, hello Green for me. |
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Nice.. i missed this |
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I got into similar issue during contest. I have used Math.sqrt()[java], but not sure why it was failing in 2nd testcase. |
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I guess I knew in the back of my mind about the "Two ways of explaining thing" bit. But it sure is a good bit of advice and that's my take from this blog. |
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I was not sure about the upper bound of i, where 1 <= i <= p. Submitted with bound to n, but then realised it should rather be 2 * n, as the formula (x + (I*(I+1))/2) has a division of 2, so the bound will be multiplied by 2 instead. |
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same here. Just finished writing a working code with example pretest. Waiting for the judge to finish to submit it to see if my idea is correct. |
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What is the meaning of this line in solution F if (brr[i] >> c & 1 ^ 1) |
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in the -1& case, i guess it just trying to find the mask where only the position of letter c is unset. So in my view, -1& is completely not required. |
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When will the editorial be out? |
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Could you please elaborate on your F solution? How do you calculate this Therefore we can calculate the contribution of every mask[i]=(1<<26)-1-(1<<i) for i in range [0, 25] using a map. in O(n) complexity? |
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Thanks for the help :) |
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Why does the following implementation for problem D is giving TLE in TC 67 (I have followed the idea given in editorial) . Here is my implementation [CODE] |
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After reading your post I am totally surprised how easily you have explained the solution. Thanks though. |
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Thanks for replying. I have looked at the problem and I can think that GSS1 can be solved using for states in a node in the segment tree maxsum, prefixsum, suffixsum and sum. But I am not sure how to apply the reverse operation in the problem given in my post. |
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can you please elaborate why binary search works here. i didn't get the last part. |
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Please someone help me with using Test Class for creating custom input for my Java code. |
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