UVA 11534-say Goodbye to Tic-Tac-Toe (NIM and)

Topic Link: UVA 11534-say Goodbye to Tic-Tac-Toe

Topic: Given a 1*n of the sub, each operation can be selected an unfinished sub-picture x or O, if the operation formed XX or OO, then the operator as a failure, man-made, for a given state (note the current state is counted in the steps), asked whether it can beat the computer.

Problem-solving ideas: For fixed length, the two ends may be empty, x,o, the combination of a total of 9 kinds, although some cases equivalent, but for the convenience of treatment, divided into 9 cases. The SG values of 9 cases in each length are pre-processed. Then for the given state, the enumeration position is X and O, and the Nim and the new state are judged.

#include <cstdio>#include <cstring>#include <algorithm>using namespace STD;Const intMAXN = -;ints[maxn+5][9];intSG (intLintXintY) {intvis[maxn+5];memset(Vis,0,sizeof(VIS)); for(inti =1; I <= l; i++) {if((I! =1|| X! =1) && (i! = L | | Y! =1)) {intX = s[i-1][x*3+1] ^ s[l-i][1*3+y]; VIS[X] =1; }if((I! =1|| X! =2) && (i! = L | | Y! =2)) {intO = s[i-1][x*3+2] ^ s[l-i][2*3+y]; Vis[o] =1; }    }intMV =-1; while(VIS[++MV]);returnMV;}voidInit (intN) {memset(s[0],0,sizeof(s[0])); for(inti =0; I <9; i++) s[1][i] =1; s[1][1*3+2] = s[1][2*3+1] =0; for(inti =2; I <= N; i++) { for(intx =0; X <3; X + +) { for(inty =0; Y <3; y++) s[i][x*3+y] = SG (i, x, y); }    }}Charstr[maxn+5];BOOLJudge () {intLen =strlen(str+1);intL =0, r =0, cnt =0, ans =0, k =0; for(inti =1; I <= Len; i++) {if(Str[i] = ='. ') k++;Else{r = (Str[i] = =' X '?1:2); Ans ^= s[k][l*3+R];            cnt++;            L = r; K =0; }} ans ^= s[k][l*3];if(cnt&1) ans = (ans = =0?1:0);returnAns;}intMain () {init (MAXN);intCAsscanf("%d", &cas); while(cas--) {scanf('%s ', str+1);printf("%s\n", Judge ()?"Possible.":"impossible."); }return 0;}