HDU 4706:Children's Day

Children's Day

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1733    Accepted Submission(s): 1121

Problem Description

Today is Children's Day. Some children ask you to output a big letter 'N'. 'N' is constituted by two vertical linesand one diagonal. Each pixel of this letter is a character orderly. No tail blank is allowed.
For example, this is a big 'N' start with 'a' and it's size is 3.

a e
bdf
c g

Your task is to write different 'N' from size 3 to size 10. The pixel character used is from 'a' to 'z' continuously and periodic('a' is reused after 'z').

 

Input

This problem has no input.

 

Output

Output different 'N' from size 3 to size 10. There is no blank line among output.

 

Sample Output

  
  
  
  
  

   
   
   
   
   [pre]
a e
bdf
c g
h  n
i mo
jl p
k  q
.........
r        j
[/pre]


   
   
   
   
   
 
    
 
     Hint 
    
 Not all the resultsare listed in the sample. There are just some lines. The ellipsis expresseswhat you should write. 
   
   
   
   
   
    
  
  
  
  
  

 

迷失在幽谷中的鳥兒，獨自飛翔在這偌大的天地間，殊不知本身該飛往何方……

#include <iostream>
#include <cmath>
#include <stdio.h>
#include <string>
#include <cstring>
#include <map>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <iomanip>
#include <algorithm>
using namespace std;
char mp[20][20];
char nextc(char c)
{
    if(c=='z')return 'a';
    else return ++c;
}
int main()
{
    char cur='a';
    int i,j,k;
    for(i=3; i<=10; i++)
    {
        memset(mp,' ',sizeof(mp));
        for(j=0; j<i; j++)
        {
            mp[j][0]=cur;
            cur=nextc(cur);
        }
        for(j=i-2,k=1; j>0; j--,k++)
        {
            mp[j][k]=cur;
            cur=nextc(cur);
        }
        for(j=0; j<i; j++)
        {
            mp[j][i-1]=cur;
            cur=nextc(cur);
        }
        for(j=0; j<i; j++)
        {
            for(k=0; k<i; k++)
                printf("%c",mp[j][k]);
            printf("\n");
        }
    }
    return 0;
}