HDU 4559 Coloring Game (SG)

Stain on a 2*n grid every time a lattice or 2*2 lattice that can dye 1*1 is the most full of all the squares. M is the number of stained squares Alice initiator

The 1*1 point SG value of 1 SG "I" means the SG value of continuous 2*i SG "3" is a continuous 2*3 blank lattice

If n = 3
The case of a 1*1 lattice, then the subsequent state may be
State 1 Continuous 0-column lattice + continuous 2-column grid + remaining that column coated with 1 lattices
State 2 continuous 1-column lattice + continuous 1-column grid + remaining that column coated with 1 lattices

The case of a 2*3 lattice, then the subsequent state may be
State 1 Continuous 0-column lattice + continuous 1-column lattice + remaining 2*2 lattice

Finally, the whole checkerboard is segmented continuously blank 2*i lattice and a column of the color-coated lattice

Sample Input
2
2 0//N m
2 2//n m
1 1//The coordinates of the dyed lattice
2 2

Sample Output
Case 1:alice
Case 2:bob

1# include <iostream>2# include <cstdio>3# include <cstring>4# include <algorithm>5# include <string>6# include <cmath>7# include <queue>8# include <list>9# define LLLong LongTen using namespacestd; One Const intmaxn= the; A intSG[MAXN]; - BOOLVIS[MAXN]; - BOOLg[2][MAXN]; the intMexintx) - { -     if(sg[x]!=-1)returnSg[x]; -memset (Vis,false,sizeof(Vis)); +      for(intI=0; i<=x-1-i;i++)//Dye 1*1 -     { +         intTmp=mex (i) ^mex (X-1-i) ^1; Avis[tmp]=true; at     } -      for(intI=0; i<=x-2-i;i++)//Dye 2*2 -     { -         intTmp=mex (i) ^mex (X-2-i); -vis[tmp]=true; -     } in      for(intI=0;; i++) -         if(!Vis[i]) to         { +sg[x]=i; -              Break; the         } *     returnSg[x]; $ }Panax Notoginseng  - intMain () the { +  Amemset (sg,-1,sizeof(SG)); thesg[0]=0; +      for(intI=1;i< the; i++) -sg[i]=Mex (i); $     intT; $scanf"%d",&T); -     intIcase=0; -      while(t--) the     { -         intn,m;Wuyiscanf"%d%d",&n,&m); theMemset (G,false,sizeof(g)); -         intu,v; Wu          while(m--) -         { Aboutscanf"%d%d",&u,&v); $u--;v--; -g[u][v]=true; -         } -         intlen=0; A         intans=0; +          for(intI=0; i<n;i++) the         { -             if(g[0][i]| | g[1][i]) $             { theans^=Sg[len]; thelen=0; the                 if(g[0][i]&&g[1][i])Continue; theans^=1; -             } in             Elselen++; the         } theans^=Sg[len]; Abouticase++; the         if(ANS) printf ("Case %d:alice\n", icase); the         Elseprintf"Case %d:bob\n", icase); the     } +     return 0; -}

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HDU 4559 Coloring Game (SG)