B. Center Alignment

题目连接:

http://www.codeforces.com/contest/5/problem/B

Description

Almost every text editor has a built-in function of center text alignment. The developers of the popular in Berland text editor «Textpad» decided to introduce this functionality into the fourth release of the product.

You are to implement the alignment in the shortest possible time. Good luck!

Input

The input file consists of one or more lines, each of the lines contains Latin letters, digits and/or spaces. The lines cannot start or end with a space. It is guaranteed that at least one of the lines has positive length. The length of each line and the total amount of the lines do not exceed 1000.

Output

Format the given text, aligning it center. Frame the whole text with characters «*» of the minimum size. If a line cannot be aligned perfectly (for example, the line has even length, while the width of the block is uneven), you should place such lines rounding down the distance to the left or to the right edge and bringing them closer left or right alternatively (you should start with bringing left). Study the sample tests carefully to understand the output format better.

Sample Input

This is

Codeforces

Beta

Round

5

Sample Output


  • This is *
  •      *

Codeforces

  • Beta *
  • Round *
  • 5    *

Hint

题意

给你一堆字符串,让你居中

如果恰好居中的话,那就居中

如果不是的话,先偏左边,然后再偏右边去

题解:

模拟题,按照题意模拟就好了。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2005;
string s[maxn],s1;
int mx,tot;
int main()
{
while(getline(cin,s1))
{
s[tot++]=s1;
mx = max((int)s1.size(),mx);
}
for(int i=0;i<mx+2;i++)
cout<<"*";
cout<<endl;
int d = 0;
for(int i=0;i<tot;i++)
{
cout<<"*";
int len = (mx-s[i].size());
if(len%2==1)d++;
int l = len/2;
int r = len/2;
if(len%2==1&&d%2==1)r++;
else if(len%2==1)l++;
for(int j=0;j<l;j++)cout<<" ";
cout<<s[i];
for(int j=0;j<r;j++)cout<<" ";
cout<<"*"<<endl;
}
for(int i=0;i<mx+2;i++)
cout<<"*";
cout<<endl;
}

Codeforces Beta Round #5 B. Center Alignment 模拟题的更多相关文章

  1. Codeforces Beta Round #7 B. Memory Manager 模拟题

    B. Memory Manager 题目连接: http://www.codeforces.com/contest/7/problem/B Description There is little ti ...

  2. Codeforces Beta Round #51 A. Flea travel 水题

    A. Flea travel Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/55/problem ...

  3. Codeforces Beta Round #77 (Div. 2 Only)

    Codeforces Beta Round #77 (Div. 2 Only) http://codeforces.com/contest/96 A #include<bits/stdc++.h ...

  4. Codeforces Beta Round #75 (Div. 2 Only)

    Codeforces Beta Round #75 (Div. 2 Only) http://codeforces.com/contest/92 A #include<iostream> ...

  5. Codeforces Beta Round #74 (Div. 2 Only)

    Codeforces Beta Round #74 (Div. 2 Only) http://codeforces.com/contest/90 A #include<iostream> ...

  6. Codeforces Beta Round #73 (Div. 2 Only)

    Codeforces Beta Round #73 (Div. 2 Only) http://codeforces.com/contest/88 A 模拟 #include<bits/stdc+ ...

  7. Codeforces Beta Round #72 (Div. 2 Only)

    Codeforces Beta Round #72 (Div. 2 Only) http://codeforces.com/contest/84 A #include<bits/stdc++.h ...

  8. Codeforces Beta Round #67 (Div. 2)

    Codeforces Beta Round #67 (Div. 2) http://codeforces.com/contest/75 A #include<bits/stdc++.h> ...

  9. Codeforces Beta Round #65 (Div. 2)

    Codeforces Beta Round #65 (Div. 2) http://codeforces.com/contest/71 A #include<bits/stdc++.h> ...

随机推荐

  1. Coursera在线学习---第六节.构建机器学习系统

    备: High bias(高偏差) 模型会欠拟合    High variance(高方差) 模型会过拟合 正则化参数λ过大造成高偏差,λ过小造成高方差 一.利用训练好的模型做数据预测时,如果效果不好 ...

  2. ubuntu之一些安装配置的坑

    前言 本博客记录自己使用ubuntu的一些错误和坑. ubuntu不支持yum下载安装机制 命令 sudo apt install yum 是可以安装yum的,但安装好后执行: $ yum insta ...

  3. flask基础之jijia2模板语言进阶(三)

    前言 前面学习了jijia2模板语言的一些基础知识,接下来继续深挖jijia2语言的用法. 系列文章 flask基础之安装和使用入门(一) flask基础之jijia2模板使用基础(二) 控制语句 和 ...

  4. NEERC2014

    NEERC2014 A - Alter Board 题目描述:给出一个\(n \times m\)的国际象棋棋盘,每次选定一个矩形,使得矩形中的每个格子的颜色翻转,求出最少次数的方案使得最终棋盘只有一 ...

  5. Python 根据地址获取经纬度及求距离

    方法一: 使用Geopy包 : https://github.com/geopy/geopy   (仅能精确到城镇,具体街道无结果返回) from geopy.geocoders import Nom ...

  6. windows 下安装 nginx + php

    1. 下载软件 需要的软件有nginx,php,mysql(如不需要可不安装) nginx.org,www.php.net上面有得下 全部解压出来 php基本不用做任何修改(下载直接解压版本的) 用c ...

  7. 【PAT】1008. 数组元素循环右移问题 (20)

    1008. 数组元素循环右移问题 (20) 一个数组A中存有N(N>0)个整数,在不允许使用另外数组的前提下,将每个整数循环向右移M(M>=0)个位置,即将A中的数据由(A0A1……AN- ...

  8. Linux的权限对于文件与目录的意义

    权限对文件: r:可读取此文件的实际内容. w:可以编辑.新增或者是修改该文件的内容(但不含删除该文件),如果没有r权限,无法w. x :该文件具有被系统执行的权限.可以删除. 权限对目录: r:re ...

  9. MVC Partial页面的使用

    先建立Action: public PartialViewResult CurrentCount() { ViewBag.Count = CurrentUserCount; return Partia ...

  10. Unity 2D游戏开发教程之使用脚本实现游戏逻辑

    Unity 2D游戏开发教程之使用脚本实现游戏逻辑 使用脚本实现游戏逻辑 通过上一节的操作,我们不仅创建了精灵的动画,还设置了动画的过渡条件,最终使得精灵得以按照我们的意愿,进入我们所指定的动画状态. ...