GCJ2015 problem B. Infinite House of Pancakes

I'm a weak B.

In the beginning, it was thought that for a number, half a cent would be better than the other allocation scheme, and then eventually

Half-points are not necessarily the best allocation scheme, for example, 9 may be divided into 3 3 better than half

I do not know how those Daniel think, I read the first code, he enumerated the maximum value of the final state, he is how to think about the problem?

The following I changed CAI's brute force code, plus the enumeration of each sub-method, currently can be too small data, big data timeout should

1#include <cstdio>2#include <iostream>3#include <cstring>4 5 using namespacestd;6 7 inta[1005];8 intN;9 Ten intMintime (inti) { One         intJ; A          for(j = i; J >=1; j-- ) { -                 if(A[j] >0) -                          Break; the         } -         if(J <=1) { -                 return 1= = J && A[j] >0?1:0; -         } +  -         intAns =J; +          for(intx =1; X <= (J >>1); ++x) { AA[X] + =A[j]; atA[j-x] + =A[j]; -                 intTMP = A[j] + mintime (J-1); -ans = ans > tmp?Tmp:ans; -A[X]-=A[j]; -A[j-x]-=A[j]; -         } in  -         returnans; to } +  - intMain () { the         intT; *CIN >>T; $          for(intCAS =1; CAS <= T; ++CAs) {Panax NotoginsengCIN >>N; -Memset (A,0,sizeof(a)); the                  for(inti =0; I < n; ++i) { +                         intp; ACIN >>p; the++A[p]; +                 } -                 intAns = mintime ( +); $cout <<"Case #"<< CAs <<": "<< ans <<Endl; $         } -         return 0; -}

The following is the code of the great God, I leave it to commemorate

1             intMin =10000;2              for(intLim =1; Lim <= Counts.length; lim++)3             {4                 intMoves =0;5                  for(inti =0; i < counts.length; i++)6Moves + = ((I-1)/Lim) *counts[i];//This i-1 I'm drunk, too.7                 if(Moves + Lim <min)8Min = moves +Lim;9}

GCJ2015 problem B. Infinite House of Pancakes