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[100][100];
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int cnt = 0;
    for(int i = 3;i <= 10;i++)
    {
        for(int j = 0;j < i;j++)    
        {
            for(int k = 0;k < i;k++)
                str[j][k] = ' ';
            str[j][i] = 0;
        }
        for(int j = 0;j < i;j++)
        {
            str[j][0] = 'a' + cnt;
            cnt = (cnt+1)%26;
        }
        for(int j = i-2;j > 0;j--)
        {
            str[j][i-1-j] = 'a' + cnt;
            cnt = (cnt+1)%26;
        }
        for(int j = 0;j < i;j++)
        {
            str[j][i-1] = 'a' + cnt;
            cnt = (cnt+1)%26;
        }
        for(int j = 0;j < i;j++)
            printf("%s\n",str[j]);
    }

    return 0;
}