hdu 4706:Children's Day（模拟题，模拟输出大写字母 N）

Children's Day

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 648    Accepted Submission(s): 422

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

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

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

 

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　　模拟题。

　　模拟输出大写字母‘N’，‘N’由小写字母组成，输出的尺寸大小由3增加到10，每个尺寸之间没有空格。输出示例见上。

　　做这样的模拟题就是要观察输出图形之间的数学关系。总结出关系之后，图形输出就很容易了。

　　代码：

 

 #include <iostream>

 using namespace std;

 int main()
 {
     int n;  //size
     int c = ;  //该尺寸的第一个字母的序号
     for(n=;n<=;n++){ //循环尺寸
         int i,j;  //该尺寸的第i行，和该行第j个元素
         int c1=c%,c2=(c+*n-)%;  //分别记录每一行的第一个字母和最后一个字母序号
         if(c1==) c1=;
         if(c2==) c2=;
         for(i=;i<=n;i++){  //行
             int k=n+-i;    //每一行中间元素的位置
             for(j=;j<=n;j++){   //该行每一个元素
                 if(j==)
                     cout<<char(c1+'a'-);
                 else if(j==n)
                     cout<<char(c2+'a'-);
                 else if(j==k){  //中间元素
                     int kc = (c+n+k-)%;
                     if(kc==) kc=;
                     cout<<char(kc+'a'-);
                 }
                 else
                     cout<<' ';
             }
             cout<<endl;
             c1=(c1+)%,c2=(c2+)%;
             if(c1==) c1=;
             if(c2==) c2=;
         }
         c=(c+*n-)%;
         if(c==)
             c=;
     }
     return ;
 }

Freecode : www.cnblogs.com/yym2013