hduoj 4706 Children's Day 2013 ACM/ICPC Asia Regional Online —— Warmup
http://acm.hdu.edu.cn/showproblem.php?pid=4706
Children's Day
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 32768/32768 K (Java/Others)
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').
Not all the resultsare listed in the sample. There are just some lines. The ellipsis expresseswhat you should write.
分析:
输出由字符组成的‘N’ , 三到十的大小。(模拟一下就可以啦)
AC代码:
#include<cstdio>
#include<cstring>
using namespace std;
char map[][];
int main() {
int k = ;
for(int n = ;n <= ;n++) { //memset(map,'\n',sizeof(map));
for(int i = ;i <= n;i++) {
for(int j = ;j < n;j++) {
if(i == || i == n - || i + j == n - ) {
map[j][i] = k % + 'a';
k++;
} else if(i == n) {
map[j][i] = '\0';
} else map[j][i] = ' ';
}
} for(int i = ;i < n;i++)
printf("%s\n",map[i]);
}
return ;
}
hduoj 4706 Children's Day 2013 ACM/ICPC Asia Regional Online —— Warmup的更多相关文章
- hduoj 4706 Herding 2013 ACM/ICPC Asia Regional Online —— Warmup
hduoj 4706 Children's Day 2013 ACM/ICPC Asia Regional Online —— Warmup Herding Time Limit: 2000/1000 ...
- hduoj 4712 Hamming Distance 2013 ACM/ICPC Asia Regional Online —— Warmup
http://acm.hdu.edu.cn/showproblem.php?pid=4712 Hamming Distance Time Limit: 6000/3000 MS (Java/Other ...
- hduoj 4710 Balls Rearrangement 2013 ACM/ICPC Asia Regional Online —— Warmup
http://acm.hdu.edu.cn/showproblem.php?pid=4710 Balls Rearrangement Time Limit: 6000/3000 MS (Java/Ot ...
- hduoj 4708 Rotation Lock Puzzle 2013 ACM/ICPC Asia Regional Online —— Warmup
http://acm.hdu.edu.cn/showproblem.php?pid=4708 Rotation Lock Puzzle Time Limit: 2000/1000 MS (Java/O ...
- hduoj 4715 Difference Between Primes 2013 ACM/ICPC Asia Regional Online —— Warmup
http://acm.hdu.edu.cn/showproblem.php?pid=4715 Difference Between Primes Time Limit: 2000/1000 MS (J ...
- hduoj 4707 Pet 2013 ACM/ICPC Asia Regional Online —— Warmup
http://acm.hdu.edu.cn/showproblem.php?pid=4707 Pet Time Limit: 4000/2000 MS (Java/Others) Memory ...
- 2013 ACM/ICPC Asia Regional Online —— Warmup
1003 Rotation Lock Puzzle 找出每一圈中的最大值即可 代码如下: #include<iostream> #include<stdio.h> #inclu ...
- HDU 4714 Tree2cycle(树状DP)(2013 ACM/ICPC Asia Regional Online ―― Warmup)
Description A tree with N nodes and N-1 edges is given. To connect or disconnect one edge, we need 1 ...
- HDU 4749 Parade Show 2013 ACM/ICPC Asia Regional Nanjing Online
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4749 题目大意:给一个原序列N,再给出一个序列M,问从N中一共可以找出多少个长度为m的序列,序列中的数 ...
随机推荐
- Python For Data Analysis -- Pandas
首先pandas的作者就是这本书的作者 对于Numpy,我们处理的对象是矩阵 pandas是基于numpy进行封装的,pandas的处理对象是二维表(tabular, spreadsheet-like ...
- poj1012约瑟夫
#include<stdio.h>int a[14];int f(int k,int m){ int n,i,s; n=2*k;s=0; for(i=0;i<k;i ...
- 用c++builder读取一个一行有多行变量的文件
文件内容如下: C DXDY.INP FILE, IN FREE FORMAT ACROSS COLUMNS for 83658 Active CellsC 2013-5-25 上午 10:43 ...
- (转)JAVA 调用matlab
本文仅用于学习. 原文地址链接:http://blog.csdn.net/wannshan/article/details/5907877 前段时间摸索了java调用matlab东西,不说学的有多深, ...
- Dom4j
Dom4j http://baike.baidu.com/link?url=2XOnr06saKUd-9By1GyPxIolXMQhf_C-CnMFll_yhtR4m00i27zphbkI5-dGpw ...
- pro5
1.本次课学到的知识点 (1)循环结构的概念 在我们需要重复进行某个步骤是就需要运用到循环结构. (2)三种循环语句 for,while,do-while是三种常用的循环语句,其中while的适用范围 ...
- aspx后台页面添加服务器控件
System.Text.StringBuilder stringBuilder = new System.Text.StringBuilder(); System.IO.StringWriter st ...
- eclipse有时候会报错:Cannot change version of project facet Dynamic Web Module to 2.5。这个错误不会影响程序的运行,不过看着总是不舒服。这个问题现在可以解决啦。
把项目WEB-INF底下的web.xml文件头部的: <?xml version="1.0" encoding="UTF-8"?> < ...
- JavaScipt选取文档元素的方法
摘自JavaScript权威指南(jQuery根据样式选择器查找元素的终极方式是 先用getElementsByTagName(*)获取所有DOM元素,然后根据样式选择器对所有DOM元素进行筛选) 选 ...
- RTSP协议详解
RTSP(Real Time Streaming Protocol)是由Real Network和Netscape共同提出的如何有效地在IP网络上传输流媒体数据的应用层协议.RTSP对流媒体提 ...