Poj-3537-crosses and Crosses

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/* Test instructions: Draw x in 1*n's grid, when there are three consecutive, win; then when the first person draws in the I position, the sub-problem can be obtained, that is, two games in 1*i-3 and 1 * (n-i-2), and a SG template is available; */# include <stdio.h> # include <cstdlib># include <cstring># include <iostream>using namespace std;int sg[2100];inline int D FS (int x) {    if (x < 0)   return 0;    if (Sg[x] >= 0)  return sg[x];    int mex[2100] = {0};    It is worth noting that this is not written in front of the function if judgment;    for (int i = 1;i <= x;i++) {        int t = DFS (i-3) ^ dfs (x-i-2);        Mex[t] = 1;    }    for (int i = 0;;; i++) if (!mex[i]) return sg[x] = i;} int main () {    int n;    memset (sg,-1,sizeof (SG));    while (~SCANF ("%d", &n)) {        if (DFS (n))   puts ("1");        Else    puts ("2");    }    return 0;}

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Poj-3537-crosses and Crosses