poj_Crosses and Crosses_3537

Crosses and Crosses

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 2459		Accepted: 915
Case Time Limit: 2000MS

DescriptionThe game of Crosses and Crosses is played on the field of 1 × n cells. Two players make moves in turn. Each move the player selects any free cell on the field and puts a cross ‘×’ to it. If after the player’s move there are three crossesin a row, he wins.You are given n. Find out who wins if both players play optimally.InputInput file contains one integer number n (3 ≤ n ≤ 2000).OutputOutput ‘1’ if the first player wins, or ‘2’ if the second player does.Sample Input

#1
3
#2
6

Sample Output

#1
1
#2
2

考虑到如果画上一个X，就会有临近区域不能画X，也就是下一个人能画X的区域就变了，那么问题可以转换为谁不能画X谁就输了。

#include <iostream>#include <cstring>#include <cstdio>#define N 2010using namespace std;int g;int n;int sg(int num){//if(num <= 0)//   return 0;if(g[num] != -1)return g[num];bool visit[N * 2];memset(visit, 0, sizeof(visit));visit[sg(num - 3)] = 1;visit[sg(num - 4)] = 1;for (int i = 0; i <= (num - 5) / 2; i++){int t1 = sg(i);int t2 = sg(num - 5 - i);visit[t1 ^ t2] = 1;}g[num] = 0;while (visit[g[num]])g[num] ++;return g[num];}int main(){memset(g, -1, sizeof(g));g[0] = 0;g[1] = 1;g[2] = 1;g[3] = 1;g[4] = 2;g[5] = 2;while (scanf("%d", &n) != EOF){if (sg(n) == 0)printf("2\n");elseprintf("1\n");}return 0;}