"POJ" "3537" crosses and crosses

Game theory

Equivalent to put X position, left and right 4 grid can no longer put X, who has nowhere to put lose.

n<=2000

The direct enumeration of subsequent States, the force of the SG function can be.

Example: 0000000->x. 0000/. x. 000/.. X.. 00/0..x. 0/00..x.

Memory Search write hung up ... or sequential DP-based = = (similar to S-nim, brute force for SG function)

1 Source Code2Problem:3537user:sdfzyhy3 memory:692k time:141ms4language:g++result:accepted5 6 Source Code7 8     //POJ 35379#include <cmath>Ten#include <queue> One#include <vector> A#include <string> -#include <cstdio> -#include <cstring> the#include <cstdlib> -#include <iostream> -#include <algorithm> -     #defineRep (i,n) for (int i=0;i<n;++i) +     #defineF (i,j,n) for (int i=j;i<=n;++i) -     #defineD (i,j,n) for (int i=j;i>=n;--i) +     using namespacestd; A  at     intGetint () { -         intR=0, c=1;CharCh=GetChar (); -          for(;! IsDigit (CH); Ch=getchar ())if(ch=='-') c=-1; -          for(; isdigit (ch); Ch=getchar ()) r=r*Ten+ch-'0'; -         returnr*C; -     } in     Const intn= ., inf=~0u>>2; -     Const Doubleeps=1e-9; to     /******************template***********************/ +     intDp[n],n; -     BOOLVis[n],mark[n]; the  *     intMain () { $n=getint ();Panax Notoginsengdp[0]=0; dp[1]=dp[2]=dp[3]=1; -F (I,4, N) { thememset (Mark,0,sizeofmark); +mark[dp[i-3]]=mark[dp[i-4]]=mark[dp[i-5]]=1; A              for(intj=1; j<=i-5-j;j++) themark[dp[j]^dp[i-5-j]]=1; +F (J,0, N)if(!mark[j]) {dp[i]=j; Break;} -         } $printf"%d\n", Dp[n]?1:2); $         return 0; -}

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"POJ" "3537" crosses and crosses