Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10397 Accepted Submission(s): 4682
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; int tot=0; int n; int c[100]; int vis[3][100]; int ans[12]; void solve(int cur) { int i; if(cur==n) tot++; else for(i=0; i<n; i++) { if(!vis[0][i]&&!vis[1][cur+i]&&!vis[2][cur-i+n]) { c[cur]=i; vis[0][i]=vis[1][cur+i]=vis[2][cur-i+n]=1;//修改全局变量 solve(cur+1); //及时恢复被修改的值 vis[0][i]=vis[1][cur+i]=vis[2][cur-i+n]=0; } } } int main() { //之前一直TLE,是忘了打表了 int m,k=1; for(int i=1;i<=10;i++) {memset(vis,0,sizeof(vis));tot=0;n=i;solve(0);ans[k++]=tot;} while(scanf("%d",&m)&&m) { printf("%d\n",ans[m]); } }
原文:http://www.cnblogs.com/orchidzjl/p/4351809.html