浙大pat 1031题解
1031. Hello World for U (20)
Given any string of N (>=5) characters, you are asked to form the characters into the shape of U. For example, "helloworld" can be printed as:
h d
e l
l r
lowo
That is, the characters must be printed in the original order, starting top-down from the left vertical line with n1 characters, then left to right along the bottom line with n2 characters, and finally bottom-up along the vertical line with n3 characters. And more, we would like U to be as squared as possible -- that is, it must be satisfied that n1 = n3 = max { k| k <= n2 for all 3 <= n2 <= N } with n1 + n2 + n3 - 2 = N.
Input Specification:
Each input file contains one test case. Each case contains one string with no less than 5 and no more than 80 characters in a line. The string contains no white space.
Output Specification:
For each test case, print the input string in the shape of U as specified in the description.
Sample Input:
helloworld!
Sample Output:
h !
e d
l l
lowor
#include"iostream"
#include "algorithm"
#include "string"
#include "vector"
using namespace std;
void OutPut(string s)
{
int n = s.size();
int n1,n2;
for(n2=3;n2<=n;n2++)
{
int temp = n+2-n2;
if(temp%2!=0)
continue;
n1 = temp/2;
if(n1<=n2)
break;
}
for(int i=0;i<n1;i++)
{
cout<<s[i];
if(i!=n1-1)
{
for(int t=0;t<n2-2;t++)
cout<<" ";
}
else
{
int k = i+1;
for(int j=0;j<n2-2;++j,++k)
cout<<s[k];
}
cout<<s[n-1-i];
cout<<endl;
}
}
int main()
{
string s;
cin >> s;
OutPut(s);
return 0;
}
浙大pat 1031题解的更多相关文章
- 浙大pat 1035题解
1035. Password (20) 时间限制 400 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue To prepare f ...
- 浙大pat 1025题解
1025. PAT Ranking (25) 时间限制 200 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Programmi ...
- 浙大pat 1011题解
With the 2010 FIFA World Cup running, football fans the world over were becoming increasingly excite ...
- 浙大PAT 7-06 题解
#include <stdio.h> #include <iostream> #include <algorithm> #include <math.h> ...
- 浙大pat 1012题解
1012. The Best Rank (25) 时间限制 400 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue To eval ...
- 浙大 pat 1003 题解
1003. Emergency (25) 时间限制 400 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue As an emerg ...
- 浙大 pat 1038 题解
1038. Recover the Smallest Number (30) 时间限制 400 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHE ...
- 浙大 pat 1047题解
1047. Student List for Course (25) 时间限制 400 ms 内存限制 64000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Y ...
- 浙大pat 1054 题解
1054. The Dominant Color (20) 时间限制 100 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard Behind the scen ...
随机推荐
- 安卓---apk反编译
转自:http://blog.csdn.net/vipzjyno1/article/details/21039349 在学习Android开发的过程你,你往往会去借鉴别人的应用是怎么开发的,那些漂亮 ...
- python reduce使用实例
通过一个简单的算法来了解reduce的巧用. 构建函数persistence(n),如果n>9,则返回0.否则继续根据n的权重来分解n,如n=999,则分解为9,9,9.那么将9*9*9=729 ...
- Python将列表中的string元素进行类型转换
例如 将 a=['1','2.0','3L'] 转换为 a=[1,2.0,3L] 只需 map(eval,['1','2.0','3L']) 即可 eval(expression[, globals[ ...
- JVM调优实战
JVM调优实战 文档修订记录 版本 日期 撰写人 审核人 批准人 变更摘要 & 修订位置 ...
- 能量项链AC了
我打算写出一个尽量看起来像是人话的解题报告. 然而这道题我还是[虽然AC但不会做] OYZ
- HUD 1171 Big Event in HDU(01背包)
Big Event in HDU Problem Description Nowadays, we all know that Computer College is the biggest depa ...
- VFL语言
VFL语言 VFL即Visual Format Language,可视化格式语言 NSDictionaryOfVariableBindings(testViewA, testViewB):此为一个宏, ...
- php笔记(四)PHP类和对象之对象继承
建立一个Truck类,扩展Car类,并覆盖speedUp方法,使速度累加50 <?php class Car { public $speed = 0; //汽车的起始速度是0 public fu ...
- linux awk命令详解2
awk是行处理器: 相比较屏幕处理的优点,在处理庞大文件时不会出现内存溢出或是处理缓慢的问题,通常用来格式化文本信息 awk处理过程: 依次对每一行进行处理,然后输出 awk命令形式: awk [-F ...
- Javascript封装弹出框控件
1.首先先定义好弹出框的HTML结构 <div class="g-dialog-contianer"> <div class="dialog-windo ...