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

 

 

Recommend

liuyiding   |   We have carefully selected several similar problems for you:   4822  4821  4820  4819  4818 

---

　　

　　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：

 

 1 #include <iostream>

 2 

 3 using namespace std;  4 

 5 int main()  6 {  7     int n;  //size

 8     int c = 1;  //该尺寸的第一个字母的序号

 9     for(n=3;n<=10;n++){ //循环尺寸

10         int i,j;  //该尺寸的第i行，和该行第j个元素

11         int c1=c%26,c2=(c+2*n-2)%26;  //分别记录每一行的第一个字母和最后一个字母序号

12         if(c1==0) c1=26; 13         if(c2==0) c2=26; 14         for(i=1;i<=n;i++){  //行

15             int k=n+1-i;    //每一行中间元素的位置

16             for(j=1;j<=n;j++){   //该行每一个元素

17                 if(j==1) 18                     cout<<char(c1+'a'-1); 19                 else if(j==n) 20                     cout<<char(c2+'a'-1); 21                 else if(j==k){  //中间元素

22                     int kc = (c+n+k-2)%26; 23                     if(kc==0) kc=26; 24                     cout<<char(kc+'a'-1); 25  } 26                 else

27                     cout<<' '; 28  } 29             cout<<endl; 30             c1=(c1+1)%26,c2=(c2+1)%26; 31             if(c1==0) c1=26; 32             if(c2==0) c2=26; 33  } 34         c=(c+3*n-2)%26; 35         if(c==0) 36             c=26; 37  } 38     return 0; 39 }

 

Freecode : www.cnblogs.com/yym2013