[ACM] hdu 4706 Children's Day

Children's Day

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 10   Accepted Submission(s) : 5

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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

解题思路：

样例中怎么给，我们就跟着怎么做，模拟输出图形。该题属于比较简单的。字母从a开始一直到z，再从a开始，所以我们只要用一个全局变量 char s='a'。对于每个'N'图形，先最左边竖着在数组中填字母，然后从左上角向右上角填字母，最后从最右边竖着填字母。记得判断s的值，如果大于'z'，要重新赋值为’a' .

代码：

#include <iostream>
#include <string.h>
using namespace std;
char tu[11][11];
char s='a';

void output(int n)
{
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        tu[i][j]=' ';//初始化数组
    for(int i=1;i<=n;i++)//最左边竖着填字母
        {
            tu[i][1]=s++;
            if(s>'z')
                s='a';
        }
    for(int i=n-1;i>=2;i--)//斜着从左向右填字母
        {
            tu[i][n-i+1]=s++;
            if(s>'z')
                s='a';
        }
    for(int i=1;i<=n;i++)//最右边竖着填字母
        {
            tu[i][n]=s++;
            if(s>'z')
                s='a';
        }
    for(int i=1;i<=n;i++)//输出“N'图形
    {
        for(int j=1;j<=n;j++)
            cout<<tu[i][j];
        cout<<endl;
    }
}

int main()
{
    for(int i=3;i<=10;i++)
        output(i);
    return 0;
}

运行截图：