HDU-6708 Windows Of CCPC (play table, recursion)

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

Input

The input file contains

Output

For each test case,you should output the answer .

Sample Input

Sample Output

CC
PC
CCCC
PCPC
PPCC
CPPC
CCCCCCCC
PCPCPCPC
PPCCPPCC
CPPCCPPC
PPPPCCCC
CPCPPCPC
CCPPPPCC
PCCPCPPC

Source

2019 Chinese Collegiate Programming Contest (CCPC) - Trials Network

This problem is more trouble handling line breaks, and feel you can do with a play table and recursive

Recursion

Problem-solving ideas:

Starting with the four characters in the lower left corner and the remaining three are not the same,

May be used initially makes up the second, the second is divided into four parts, a lower left corner and the opposite, i.e. P becomes C, C becomes P, remaining the same.

To export a total of 2 ^ n rows, it can output line by line, suppose I want to total output behavior 8 lines, now the output line 1,

In fact, it is twice the total output behavior output line 14 line,

When the output of the lower left corner portion, which is the overall behavior of the respective row 4 opposite to the output line 1 times, 1 times the output of the same.

 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= 1E9 + MOD . 7 ;
 15  // const Double the PI = ACOS (-1); 
16  const  int MAXN 1E5 + = 10 ;
 . 17  the using  namespace STD;
 18 is  // iOS :: sync_with_stdio (to false);
 . 19  //     cin.tie (NULL );
 20 is  
21 is  // n represents a total of the n-layer pattern, row represents the current number of layers, f denotes the opposite of normal output or output 
22 is  void Solve ( int n, int Row, int F)
 23 is  {
 24      IF (n = = 2 )
 25      {
 26 is          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 is      }
 42 is      int T% = row (n-/ 2 ); // T represents the line pattern row is the last row number order 
43 is      IF (T == 0 )
 44 is          T = n-/ 2 ;
 45      IF (F == . 1 )
 46 is      {
 47          IF (Row> * n- 1.0 / 2 )
 48              Solve (n-/ 2 , T, 0 );
 49          the 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 play table

 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 }

You can also find the law according to the number of lines

Advance the playing table G [2 ^ 10 + 1] [2 ^ 10 + 1], the output of two for 1-> 2 ^ k on the line, the time can try