HDU 4706 Children's Day （水题）

Children's Day

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 158    Accepted Submission(s): 72

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.

 

Source

2013 ACM/ICPC Asia Regional Online —— Warmup

 

Recommend

liuyiding

 

今天比赛的水题。

记录一发

 /* *******************************************
 Author       : kuangbin
 Created Time : 2013年09月08日 星期日 12时16分28秒
 File Name    : 1001.cpp
 ******************************************* */

 #include <stdio.h>
 #include <algorithm>
 #include <iostream>
 #include <string.h>
 #include <vector>
 #include <queue>
 #include <set>
 #include <map>
 #include <string>
 #include <math.h>
 #include <stdlib.h>
 #include <time.h>
 using namespace std;

 char str[][];
 int main()
 {
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     int cnt = ;
     for(int i = ;i <= ;i++)
     {
         for(int j = ;j < i;j++)
         {
             for(int k = ;k < i;k++)
                 str[j][k] = ' ';
             str[j][i] = ;
         }
         for(int j = ;j < i;j++)
         {
             str[j][] = 'a' + cnt;
             cnt = (cnt+)%;
         }
         for(int j = i-;j > ;j--)
         {
             str[j][i--j] = 'a' + cnt;
             cnt = (cnt+)%;
         }
         for(int j = ;j < i;j++)
         {
             str[j][i-] = 'a' + cnt;
             cnt = (cnt+)%;
         }
         for(int j = ;j < i;j++)
             printf("%s\n",str[j]);
     }

     return ;
 }