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

 1 /* *******************************************
 2 Author       : kuangbin
 3 Created Time : 2013年09月08日 星期日 12时16分28秒
 4 File Name    : 1001.cpp
 5 ******************************************* */
 6 
 7 #include <stdio.h>
 8 #include <algorithm>
 9 #include <iostream>
10 #include <string.h>
11 #include <vector>
12 #include <queue>
13 #include <set>
14 #include <map>
15 #include <string>
16 #include <math.h>
17 #include <stdlib.h>
18 #include <time.h>
19 using namespace std;
20 
21 char str[100][100];
22 int main()
23 {
24     //freopen("in.txt","r",stdin);
25     //freopen("out.txt","w",stdout);
26     int cnt = 0;
27     for(int i = 3;i <= 10;i++)
28     {
29         for(int j = 0;j < i;j++)    
30         {
31             for(int k = 0;k < i;k++)
32                 str[j][k] = ' ';
33             str[j][i] = 0;
34         }
35         for(int j = 0;j < i;j++)
36         {
37             str[j][0] = 'a' + cnt;
38             cnt = (cnt+1)%26;
39         }
40         for(int j = i-2;j > 0;j--)
41         {
42             str[j][i-1-j] = 'a' + cnt;
43             cnt = (cnt+1)%26;
44         }
45         for(int j = 0;j < i;j++)
46         {
47             str[j][i-1] = 'a' + cnt;
48             cnt = (cnt+1)%26;
49         }
50         for(int j = 0;j < i;j++)
51             printf("%s\n",str[j]);
52     }
53 
54     return 0;
55 }