SG[i]表示一个数二进制下有i个1的SG值
SG[0]=0
打表:
#include<bits/stdc++.h> using namespace std; int sg[100]; int vis[30000000]; int maxnnow = 0, x; int main() { sg[0] = 0; for (int i = 1; i <= 64; i++) { for (int j = 0; j <= maxnnow; j++) { vis[j] = 0; } maxnnow = 0; for (int j1 = 0; j1 < i; j1++) for (int j2 = j1; j2 < i; j2++) for (int j3 = j2; j3 < i; j3++) for (int j4 = j3; j4 < i; j4++) for (int j5 = j4; j5 < i; j5++) for (int j6 = j5; j6 < i; j6++) for (int j7 = j6; j7 < i; j7++) { x = sg[j1] ^ sg[j2] ^ sg[j3] ^ sg[j4] ^ sg[j5] ^ sg[j6] ^ sg[j7]; maxnnow = max(maxnnow, x); vis[x] = 1; } for (int j = 0; j <= maxnnow + 1; j++) { if (vis[j] == 0) { sg[i] = j; break; } } } for(int i=1;i<=64;i++) cout<<sg[i]<<endl; }
表打出来后面就很简单了