UVa——400Unix ls（字典序文本处理输出iomanip）

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

Description

The computer company you work for is introducing a brand new computer line and is developing a new Unix-like operating system to be introduced along with the new computer. Your assignment is to write the formatter for thels function.

Your program will eventually read input from a pipe (although for now your program will read from the input file). Input to your program will consist of a list of (F) filenames that you will sort (ascending based on the ASCII character values) and format into (C) columns based on the length (L) of the longest filename. Filenames will be between 1 and 60 (inclusive) characters in length and will be formatted into left-justified columns. The rightmost column will be the width of the longest filename and all other columns will be the width of the longest filename plus 2. There will be as many columns as will fit in 60 characters. Your program should use as few rows (R) as possible with rows being filled to capacity from left to right.

Input

The input file will contain an indefinite number of lists of filenames. Each list will begin with a line containing a single integer ( ). There will then beN lines each containing one left-justified filename and the entire line's contents (between 1 and 60 characters) are considered to be part of the filename. Allowable characters are alphanumeric (a toz, A to Z, and 0 to 9) and from the following set{ ._- } (not including the curly braces). There will be no illegal characters in any of the filenames and no line will be completely empty.

Immediately following the last filename will be the N for the next set or the end of file. You should read and format all sets in the input file.

Output

For each set of filenames you should print a line of exactly 60 dashes (-) followed by the formatted columns of filenames. The sorted filenames 1 toR will be listed down column 1; filenames R+1 to 2R listed down column 2; etc.

Sample Input

10

tiny

2short4me

very_long_file_name

shorter

size-1

size2

size3

much_longer_name

12345678.123

mid_size_name

12

Weaser

Alfalfa

Stimey

Buckwheat

Porky

Joe

Darla

Cotton

Butch

Froggy

Mrs_Crabapple

P.D.

19

Mr._French

Jody

Buffy

Sissy

Keith

Danny

Lori

Chris

Shirley

Marsha

Jan

Cindy

Carol

Mike

Greg

Peter

Bobby

Alice

Ruben

Sample Output

------------------------------------------------------------

12345678.123         size-1

2short4me            size2

mid_size_name        size3

much_longer_name     tiny

shorter              very_long_file_name

------------------------------------------------------------

Alfalfa        Cotton         Joe            Porky

Buckwheat      Darla          Mrs_Crabapple  Stimey

Butch          Froggy         P.D.           Weaser

------------------------------------------------------------

Alice       Chris       Jan         Marsha      Ruben

Bobby       Cindy       Jody        Mike        Shirley

Buffy       Danny       Keith       Mr._French  Sissy

Carol       Greg        Lori        Peter

题意：输入N后有N段字符串，读入进行按字典序排序，然后按尽量少行数输出，每行最多60个字符，因此最大列数colum=(60-maxlen)/(maxlen+2)+1;当然maxlen最长字符串的长度，行数row=(int)(ceil( (double)t/colum) )。然后最后一列的每个字符串输出占位数为maxlen，其他为maxlen+2。题意刚开始看不懂，直接百度了。

代码：

#include<iostream>

#include<algorithm>

#include<string>

#include<cmath>

#include<cstdio>

#include<set>

#include<sstream>

#include<map>

#include<vector>

#include<iomanip>

using namespace std;

bool cmp(const string &a,const string &b)

{

	return a<b;

}

int main(void)

{

	int t,maxlen,colum,row;

	while (cin>>t)

	{

		string list[1000];

		maxlen=0;

		for (int q=0; q<t; q++)

		{

			cin>>list[q];

			maxlen=max(maxlen,(int)list[q].size());//计算字符串最大长度

		}

		sort(list,list+t,cmp);//字典序排序

		colum=(60-maxlen)/(maxlen+2)+1;//计算列数

		row=(int)(ceil((double)t/colum));//计算行数

		cout<<"------------------------------------------------------------"<<endl;

		for (int i=0; i<row; i++)//行外循环

		{

			for (int j=0; j<colum; j++)//列内循环

			{

				if(j<colum-1)

					cout<<left<<setw(maxlen)<<list[j*row+i]<<"  ";//不是最后一列，maxlen再多输俩空格

				else

					cout<<left<<setw(maxlen)<<list[j*row+i];//按maxlen输出

			}

			cout<<endl;

		}

	}

	return 0;

}