http://acm.hdu.edu.cn/showproblem.php?pid=6708
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Problem Description
In recent years, CCPC has developed rapidly and gained a large number of competitors .One contestant designed a design called CCPC Windows .The 1-st order CCPC window is shown in the figure:
And the 2-nd order CCPC window is shown in the figure:
We can easily find that the window of CCPC of order k is generated by taking the window of CCPC of order k−1 as C of order k,and the result of inverting C/P in the window of CCPC of order k−1 as P of order k.
And now I have an order k ,please output k-order CCPC Windows , The CCPC window of order k is a 2k∗2k matrix.
Input
The input file contains T test samples.(1<=T<=10)
The first line of input file is an integer T.
Then the T lines contains a positive integers k , (1≤k≤10)
Output
For each test case,you should output the answer .
Sample Input
3 1 2 3
Sample Output
CC PC CCCC PCPC PPCC CPPC CCCCCCCC PCPCPCPC PPCCPPCC CPPCCPPC PPPPCCCC CPCPPCPC CCPPPPCC PCCPCPPC
Source
这题比较麻烦的就是换行符的处理了，感觉可以用打表和递归做
递归
解题思路：
最开始是4个字符左下角那个和其余3个不一样，
用最初的可以拼成第2个，把第2个分成4部分，左下角和第一个相反，也就是P变为C，C变为P，其余相同。
一共要输出2^n行，那么可以一行一行的输出，假设我要输出总行为8行，现在要输出第1行，
那么其实是输出总行为4行的第1行输出两遍，
当输出左下角的部分时，这是总行为4行的相应行相反输出1遍，在输出1遍相同的。
1 #include <stdio.h>
2 #include <string.h>
3 #include <iostream>
4 #include <string>
5 #include <math.h>
6 #include <algorithm>
7 #include <vector>
8 #include <queue>
9 #include <set>
10 #include <map>
11 #include <math.h>
12 const int INF=0x3f3f3f3f;
13 typedef long long LL;
14 const int mod=1e9+7;
15 //const double PI=acos(-1);
16 const int maxn=1e5+10;
17 using namespace std;
18 //ios::sync_with_stdio(false);
19 // cin.tie(NULL);
20
21 //n表示图案一共有n层，row表示目前的层数 ，f表示是正常输出还是反着输出
22 void solve(int n,int row,int f)
23 {
24 if(n==2)
25 {
26 if(f==1)
27 {
28 if(row==1)
29 printf("CC");
30 else
31 printf("PC");
32 }
33 else
34 {
35 if(row==1)
36 printf("PP");
37 else
38 printf("CP");
39 }
40 return ;
41 }
42 int t=row%(n/2);//t表示该图案row行是上一阶的多少行
43 if(t==0)
44 t=n/2;
45 if(f==1)
46 {
47 if(row>n*1.0/2)
48 solve(n/2,t,0);
49 else
50 solve(n/2,t,1);
51 solve(n/2,t,1);
52 }
53 else if(f==0)
54 {
55 if(row>n*1.0/2)
56 solve(n/2,t,1);
57 else
58 solve(n/2,t,0);
59 solve(n/2,t,0);
60 }
61 }
62
63 int main()
64 {
65 int n,T;
66 scanf("%d",&T);
67 while(T--)
68 {
69 scanf("%d",&n);
70 n=1<<n;
71 for(int i=1;i<=n;i++)
72 {
73 solve(n,i,1);
74 printf("\n");
75 }
76 }
77 return 0;
78 }
STL打表
1 #include <stdio.h>
2 #include <string.h>
3 #include <iostream>
4 #include <string>
5 #include <math.h>
6 #include <algorithm>
7 #include <vector>
8 #include <queue>
9 #include <set>
10 #include <map>
11 #include <math.h>
12 const int INF=0x3f3f3f3f;
13 typedef long long LL;
14 const int mod=1e9+7;
15 //const double PI=acos(-1);
16 const int maxn=1e5+10;
17 using namespace std;
18 //ios::sync_with_stdio(false);
19 // cin.tie(NULL);
20
21 vector<string> vt[15];
22
23 int main()
24 {
25 ios::sync_with_stdio(false);
26 cin.tie(NULL);
27
28 vt[1].push_back("CC");
29 vt[1].push_back("PC");
30 for (int i = 2; i <= 10; i++)
31 {
32 for (vector<string>::iterator it1 = vt[i - 1].begin(); it1 != vt[i - 1].end(); it1++)
33 {
34 vt[i].push_back(*it1 + *it1);
35 }
36 for (vector<string>::iterator it1 = vt[i - 1].begin(); it1 != vt[i - 1].end(); it1++)
37 {
38 string s1 = *it1;
39 string s2 = "";
40 for (string::iterator it2 = s1.begin(); it2 != s1.end(); it2++) {
41 if (*it2 == ‘C‘)
42 s2 += ‘P‘;
43 else
44 s2 += ‘C‘;
45
46 }
47 vt[i].push_back(s2 + *it1);
48 }
49 }
50 int T;
51 cin >> T;
52 while (T--)
53 {
54 int i;
55 cin >> i;
56 for (vector<string>::iterator it1 = vt[i].begin(); it1 != vt[i].end(); it1++)
57 {
58 cout << *it1 << endl;
59 }
60 }
61 return 0;
62 }
还可以根据行数找规律
提前打表G[2^10+1][2^10+1]，输出时两个for 1->2^k就行了，有空可以尝试一下
原文地址：https://www.cnblogs.com/jiamian/p/11403395.html