N皇后问题的构造方法(2007-06-03 19:48)

 

貌似网上搜的都一样的版本(有错...貌似还不少).先把代码弄上来,如果POJ的SPJ没错的话,这代码就应该没错了.POJ 3239

有时间好好整理一下再把规则写一下.

1. #include "stdio.h"
3. int main()
4. {
5.           int n,k;
6.           int i;
7.           while(scanf("%d",&n)){
8.                   if(n==0){
9.                           break;
10.                   }
11.                   if(n%6!= 2){
12.                           if(n&1){
13.                                   for(i=2;i< =n-1;i+=2){
14.                                           printf("%d ",i);
15.                                   }
16.                                   for(i=1;i< =n;i+=2){
17.                                           printf("%d ",i);
18.                                   }
19.                           }
20.                           else{
21.                                   for(i=2;i< =n;i+=2){
22.                                           printf("%d ",i);
23.                                   }
24.                                   for(i=1;i< =n-1;i+=2){
25.                                           printf("%d ",i);
26.                                   }
27.                           }
28.                   }
29.                   else{
30.                           k=n/2;
31.                           if(k&1){
32.                                   for(i=k;i< =n-1;i+=2){
33.                                           printf("%d ",i);
34.                                   }
35.                                   for(i=1;i< =k-1;i+=2){
36.                                           printf("%d ",i);
37.                                   }
38.                                   for(i=k+3;i< =n;i+=2){
39.                                           printf("%d ",i);
40.                                   }
41.                                   for(i=2;i< =k+1;i+=2){
42.                                           printf("%d ",i);
43.                                   }
44.                           }
45.                           else{
46.                                   for(i=k;i< =n;i+=2){
47.                                           printf("%d ",i);
48.                                   }
49.                                   for(i=2;i< =k-2;i+=2){
50.                                           printf("%d ",i);
51.                                   }
52.                                   for(i=k+3;i< =n-1;i+=2){
53.                                           printf("%d ",i);
54.                                   }
55.                                   for(i=1;i< =k+1;i+=2){
56.                                           printf("%d ",i);
57.                                   }
58.                           }
60.                   }
61.                   printf("/n");
62.           }
63.           return 0;
64. }