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\r\n\r\n | \r\n \n \n \n  \n \n | \r\n \r\n \r\n \r\n \r\n \r\n \r\n \r\n In problem C we can consider only _prime_ divisors of n (at most 6 :), as number of vertices. But there is one tricky case that i've missed, unfortunately. \r\n | \r\n
| \r\n \n | \r\n\r\n \r\n \r\n \r\n \r\n \r\n \r\n In problem E, can you explain what is a mask and its submasks ? \r\n | \r\n
| \r\n \n | \r\n\r\n \r\n \r\n 13 years ago,\r\n #\r\n |\r\n \r\n\r\n \r\n \r\n \r\n ←\r\n Rev. 2\r\n \r\n \r\n \r\n\r\n\r\n \r\n \r\n0\r\n\r\n\r\n \r\n \r\n \r\n \r\n \r\n edit: I'm sorry that I didn't notice Lepetrandr's post. For (C) it suffices to check polygons with a prime number of sides (except for 2, then check 4), that gives O(log n) divisors to check. | \r\n
| \r\n \n | \r\n\r\n \r\n \r\n 13 years ago,\r\n #\r\n ^\r\n |\r\n \r\n\r\n \r\n \r\n \r\n\r\n\r\n \r\n \r\n0\r\n\r\n\r\n \r\n \r\n \r\n \r\n \r\n Yes , because if the - polygon with length equal to n - is regular , we can create another regular polygons which have the length of any prime divisors of n..but this number is not equal to log(n). This is equal to prime divisors of n. \r\n | \r\n
| \r\n \n | \r\n\r\n \r\n \r\n \r\n \r\n \r\n \r\n In the first problem explanation there's a type error: \r\n If L > 10, output word without any changes, otherwise output the first letter, next L - 2 and finally the last letter. Should be "If L ≤ 10" | \r\n
| \r\n \n | \r\n\r\n \r\n \r\n \r\n \r\n \r\n \r\n In author's solution of problem E, \r\n does dp1 calculation really need dynamic programming? I think, this part can be done by simple 2 loops. dp1[mask]=0; for(int i=0;i<n;i++){ if(mask&(1<<i)) dp1[mask]+=(weight of i-th atom); } } | \r\n
| \r\n \n | \r\n\r\n \r\n \r\n 13 years ago,\r\n #\r\n ^\r\n |\r\n \r\n\r\n \r\n \r\n \r\n\r\n\r\n \r\n \r\n0\r\n\r\n\r\n \r\n \r\n \r\n \r\n \r\n You are right. It can be calculated by your way. Also it is calculated in this way in the author's solution. \r\n | \r\n
| \r\n \n | \r\n\r\n \r\n \r\n \r\n \r\n \r\n \r\n can you please explain O(nlog(n)) solution for question C ? \r\n | \r\n
| \r\n \n | \r\n\r\n \r\n \r\n 13 years ago,\r\n #\r\n ^\r\n |\r\n \r\n\r\n \r\n \r\n \r\n ←\r\n Rev. 3\r\n \r\n \r\n \r\n\r\n\r\n \r\n \r\n0\r\n\r\n\r\n \r\n \r\n \r\n \r\n \r\n assume n= p1q1 * p2q2 * ... * pmqm, where all the pi is prime. then for each i, (1<=i<=m), check if exist good polygyn with exactly pi vertex. if at least one of this question has positive answer then output "YES", otherwise output "NO". because if we have a good polygyn with A*B vertex, then we have a good polygyn with A vertex. sorry for my poor english! | \r\n
| \r\n \n | \r\n\r\n \r\n \r\n 13 years ago,\r\n #\r\n ^\r\n |\r\n \r\n\r\n \r\n \r\n \r\n\r\n\r\n \r\n \r\n0\r\n\r\n\r\n \r\n \r\n \r\n \r\n \r\n sorry, couldn't understand, can you explain with one example ? \r\n | \r\n
| \r\n \n | \r\n\r\n \r\n \r\n 13 years ago,\r\n #\r\n ^\r\n |\r\n \r\n\r\n \r\n \r\n \r\n ←\r\n Rev. 2\r\n \r\n \r\n \r\n\r\n\r\n \r\n \r\n0\r\n\r\n\r\n \r\n \r\n \r\n \r\n \r\n assume this good polygyn with A*B vertex: S1, S2, ... , SA*B now we have subpolygyn which have only A vertex, like this: S1*B, S2*B, ... , SA*B. now it's obviously that if the circle has a good polygyn then it has a polygyn with prime vertex. it's algo is O(nlgn) because for each i, pi >=2 so m<=lgn, and check function is O(n). therefore main algo is O(nlgn). | \r\n
| \r\n \n | \r\n\r\n \r\n \r\n 12 years ago,\r\n #\r\n |\r\n \r\n\r\n \r\n \r\n \r\n ←\r\n Rev. 3\r\n \r\n \r\n \r\n\r\n\r\n \r\n \r\n0\r\n\r\n\r\n \r\n \r\n \r\n \r\n \r\n
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