bzoj 3101 N皇后构造一种解数学

3101: N皇后

Time Limit: 10 Sec Memory Limit: 128 MBSec Special Judge
Submit: 70 Solved: 32
[Submit][Status]

Description

n*n的棋盘,在上面摆下n个皇后，使其两两间不能相互攻击…

Input

一个数n

Output

第i行表示在第i行第几列放置皇后

Sample Input

Sample Output

2
4
1
3

HINT

100%的数据3<n<1000000。输出任意一种合法解即可

Source

题解：http://blog.csdn.net/nike0good/article/details/41006705

 1 #include<cstring>
 2 #include<cmath>
 3 #include<iostream>
 4 #include<algorithm>
 5 #include<cstdio>
 6 
 7 using namespace std;
 8 
 9 int n;
10 
11 int main()
12 {
13     scanf("%d",&n);
14     if (n%6!=2&&n%6!=3)
15     {
16         for (int i=2;i<=n;i+=2)
17             printf("%d
",i);
18         for (int i=1;i<=n;i+=2)
19             printf("%d
",i);
20     }
21     else
22     {
23         int k=n>>1;
24         if (k&1)
25         {
26             for (int i=k;i<=n-1;i+=2)
27                 printf("%d
",i);
28             for (int i=1;i<=k-1;i+=2)
29                 printf("%d
",i);
30             for (int i=k+3;i<=n;i+=2)
31                 printf("%d
",i);
32             for (int i=2;i<=k+1;i+=2)
33                 printf("%d
",i);        
34         }
35         else
36         {
37             for (int i=k;i<=n;i+=2)
38                 printf("%d
",i);
39             for (int i=2;i<=k-1;i+=2)
40                 printf("%d
",i);
41             for (int i=k+3;i<=n-1;i+=2)
42                 printf("%d
",i);
43             for (int i=1;i<=k+1;i+=2)
44                 printf("%d
",i);            
45         }
46         if (n&1)printf("%d
",n);
47     }
48 }

bzoj 3101 N皇后构造一种解 数学