poj 3239(n皇后问题)

（网上找的。。。）构造法公式（序列）：
一、当n mod 6 != 2 && n mod 6 != 3时：

[2,4,6,8,...,n],[1,3,5,7,...,n-1]        (n为偶数)

[2,4,6,8,...,n-1],[1,3,5,7,...,n ]       (n为奇数)

二、当n mod 6 == 2 或 n mod 6 == 3时

(当n为偶数,k=n/2；当n为奇数,k=(n-1)/2)

[k,k+2,k+4,...,n],[2,4,...,k-2],[k+3,k+5,...,n-1],[1,3,5,...,k+1]         (k为偶数,n为偶数)

[k,k+2,k+4,...,n-1],[2,4,...,k-2],[k+3,k+5,...,n-2],[1,3,5,...,k+1],[n]     (k为偶数,n为奇数)

[k,k+2,k+4,...,n-1],[1,3,5,...,k-2],[k+3,...,n],[2,4,...,k+1]              (k为奇数,n为偶数)

[k,k+2,k+4,...,n-2],[1,3,5,...,k-2],[k+3,...,n-1],[2,4,...,k+1],[n ]      (k为奇数,n为奇数)

 

(上面有六条序列。一行一个序列，中括号是我额外加上的，方便大家辨认子序列，子序列与子序列之间是连续关系，无视中括号就可以了。第i个数为ai，表示在第i行ai列放一个皇后；... 省略的序列中，相邻两数以2递增。)

 

 1 #include <iostream>

 2 #include <cstring>

 3 #include <cstdio>

 4 #include <algorithm>

 5 #include <cmath>

 6 

 7 using namespace std;

 8 

 9 

10 void queens_puzzle(int n)//n>=8

11 {

12     if(n%6!=2 && n%6!=3)

13     {

14         printf("2");

15         for(int i=4;i<=n;i+=2)

16             printf(" %d",i);

17         for(int i=1;i<=n;i+=2)

18             printf(" %d",i);

19         printf("\n");

20     }

21     else

22     {

23         int k=n/2;

24         if(n%2==0 && k%2==0)

25         {

26             printf("%d",k);

27             for(int i=k+2;i<=n;i+=2)

28                 printf(" %d",i);

29             for(int i=2;i<=k-2;i+=2)

30                 printf(" %d",i);

31             for(int i=k+3;i<=n-1;i+=2)

32                 printf(" %d",i);

33             for(int i=1;i<=k+1;i+=2)

34                 printf(" %d",i);

35         }

36         else if(n%2==1 && k%2==0)

37         {

38             printf("%d",k);

39             for(int i=k+2;i<=n-1;i+=2)

40                 printf(" %d",i);

41             for(int i=2;i<=k-2;i+=2)

42                 printf(" %d",i);

43             for(int i=k+3;i<=n-2;i+=2)

44                 printf(" %d",i);

45             for(int i=1;i<=k+1;i+=2)

46                 printf(" %d",i);

47             printf(" %d",n);

48         }

49         else if(n%2==0 && k%2==1)

50         {

51             printf("%d",k);

52             for(int i=k+2;i<=n-1;i+=2)

53                 printf(" %d",i);

54             for(int i=1;i<=k-2;i+=2)

55                 printf(" %d",i);

56             for(int i=k+3;i<=n;i+=2)

57                 printf(" %d",i);

58             for(int i=2;i<=k+1;i+=2)

59                 printf(" %d",i);

60         }

61         else

62         {

63             printf("%d",k);

64             for(int i=k+2;i<=n-2;i+=2)

65                 printf(" %d",i);

66             for(int i=1;i<=k-2;i+=2)

67                 printf(" %d",i);

68             for(int i=k+3;i<=n-1;i+=2)

69                 printf(" %d",i);

70             for(int i=2;i<=k+1;i+=2)

71                 printf(" %d",i);

72             printf(" %d",n);

73         }

74         printf("\n");

75     }

76 }

77 

78 int main()

79 {

80     int n;

81     freopen("input.txt","r",stdin);

82     freopen("out1.txt","w",stdout);

83     while(~scanf("%d",&n))

84     {

85         if(!n) break;

86         queens_puzzle(n);

87     }

88     return 0;

89 }