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back to boardWhat idea in solution? I thought that we must do with heads (n) next: 2*2*2*...*(n div 2) ---> for odd(n)=false and 2*2*2*...*3*((n-3) div 2)---> for odd(n)=true. What why answer for n=: 5-->2*3=6 6-->2*2*2=8 7-->2*2*3=12 8-->2*2*2*2=16 9-->2*2*2*3=24 but when I write solution, a saw that ans for 6=3*3=9(not an 8), for 9=3*3*3=27(not a 24). COULD YOU HELP ME WITH RIGHT IDEA? Edited by author 29.05.2006 18:25 Re: What idea in solution? 6--> 3*3=9, isn't it? Re: What idea in solution? The last number must be 3,and others must be 2.So you have to judge whether n is an odd number or an even number first.This step is very important.And the n must be: n=2+2+2...+3(if n was an even number,there is no 3.) Re: What idea in solution? if n mod 3=0 then ansver is 3 in pover (n div 3); in other cases answer is 2*2*2*...*3 Re: What idea in solution? Posted by denton 17 Feb 2008 19:45 if n mod 3=0 then ansver is 3 in pover (n div 3); in other cases answer is 2*2*2*...*3 3*3 > 2*2*2, so your assumption is wrong. The answer is 2*3*3*...*3, when n is even and 3*3*...*3 otherwise. Re: What idea in solution? for 8-->3*3*2=18(not a 16) Re: What idea in solution? Edited by author 03.08.2010 17:40 No DP, O(1) Posted by void 3 Aug 2010 03:30 Answer is a simple multiplication: 3*3*...*3. If n % 3 == 2, then it ends with ...*2*2. It's somehow linked with E, but how? Re: No DP, O(1) If n could be divided on rational numbers, then the max product will be always (n/k)^k, where k is integer, such that abs(n/k - E) is minimal possible value. It is a well-known theorem. |
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