5-1 代码对齐 UVA1593
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1、问题描述:
You are working in a team that writes Incredibly Customizable Programming Codewriter (ICPC) which is basically a text editor with bells and whistles. You are working on a module that takes a piece of code containing some definitions or other tabular information and aligns each column on a fixed vertical position, while keeping the resulting code as short as possible, making sure that only whitespaces that are absolutely required stay in the code. So, that the first words on each line are printed at position p1 = 1; the second words on each line are printed at the minimal possible position p2, such that all first words end at or before position p2 - 2; the third words on each line are printed at the minimal possible position p3, such that all second words end at or before position p3 - 2, etc.
For the purpose of this problem, the code consists of multiple lines. Each line consists of one or more words separated by spaces. Each word can contain uppercase and lowercase Latin letters, all ASCII punctuation marks, separators, and other non-whitespace ASCII characters (ASCII codes 33 to 126 inclusive). Whitespace consists of space characters (ASCII code 32).
Input
The input file contains one or more lines of the code up to the end of file. All lines (including the last one) are terminated by a standard end-of-line sequence in the file. Each line contains at least one word, each word is 1 to 80 characters long (inclusive). Words are separated by one or more spaces. Lines of the code can have both leading and trailing spaces. Each line in the input file is at most 180 characters long. There are at most 1000 lines in the input file.
Output
Write to the output file the reformatted, aligned code that consists of the same number of lines, with the same words in the same order, without trailing and leading spaces, separated by one or more spaces such that i-th word on each line starts at the same position pi.
Note for the Sample:
The `
Sample Input
start: integer; // begins here
stop: integer; // ends here
s: string;
c: char; // temp
Sample Output
start: integer; // begins here
stop: integer; // ends here
s: string;
c: char; // temp
2、本题采用向量来储存数据,在向向量中推送数据的同时对每部分进行计数,以便在输出是可以对齐
#include <iostream>#include<vector>#include<cstring>#include<string>#include<queue>#include<algorithm>#include<iomanip>using namespace std;#define maxn 1000+1vector<string> s[maxn];int maxl[4];/* run this program using the console pauser or add your own getch, system("pause") or input loop */void print (vector<string>s){ for (int i=0;i<s.size();i++){ cout<<setisofalgs(ios::left)<<setw(maxl[i])<<s[i]; } cout<<endl;}int main(int argc, char** argv) { int n; cin>>n; int m=0; string s; while(n--){ string l; memset(maxl,0,sizeof(maxl)); l=""; int t=0; int count=0; cin.getline(s,sizeof(s)); for(int i=0;i<strlen(s);i++){ if(s[i]==' ')continue; l+=s[i]; if(s[i]==':'){ s[m].push_back(l); l=""; maxl[t]=(count>maxl[t])?count:maxl[t]; t++; count=0; continue; } if(s[i]==';'){ s[m].push_back(l); l=""; maxl[t]=(count>maxl[t])?count:maxl[t]; t++; count=0; continue; } if(s[i]=='/'&&s[i+1]=='/')continue; if(s[i]=='/'&&s[i-1]=='/'){ s[m].push_back(l); l=""; maxl[t]=(count>maxl[t])?count:maxl[t]; t++; count=0; continue; } count++; } } for_each(s.begin(),s.end(),print); return 0;}
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