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 ‘⊔’ character in the example below denotes a space character in the actual files (ASCII code 32).
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
题目类型：字符处理题
算法分析：边输入边处理，使用一个cnt数组记录每一列字符串的最大长度，使用num数组记录每行有多少字符串。输出时对于第一个字符串直接输出；对于之后的字符串要输出相应的前导空格，前导空格的数目为前一列字符串的最大长度减去前一个字符串的长度再加一：num_space = cnt[j-1] - ans[k][j-1].length () + 1;，注意每个字符串的最后不要输出多余的空格，下面第一个代码是更好的实现
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/************************************************** filename :a.cpp author :maksyuki created time :2018/1/22 17:04:25 last modified :2018/1/22 17:49:23 file location :C:\Users\abcd\Desktop\TheEternalPoet ***************************************************/ #pragma comment(linker, "/STACK:102400000,102400000") #include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("in", "r", stdin) #define CFO freopen ("out", "w", stdout) #define CPPFF ifstream cin ("in") #define CPPFO ofstream cout ("out") #define DB(ccc) cout << #ccc << " = " << ccc << endl #define DBT printf("time used: %.2lfs\n", (double) clock() / CLOCKS_PER_SEC) #define PB push_back #define MP(A, B) make_pair(A, B) typedef long long LL; typedef unsigned long long ULL; typedef double DB; typedef pair <int, int> PII; typedef pair <int, bool> PIB; const int INF = 0x7F7F7F7F; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 1e3 + 6666; vector<string> dict[maxn]; int cnt[maxn], col[maxn], acol, len; void OutSpace(int n) { for(int i = 1; i <= n; i++) cout << " "; } int main() { #ifdef LOCAL CFF; CFO; #endif string s; stringstream ss; len = acol = 0; while(getline(cin, s)) { ss.clear(); ss << s; while(ss >> s) { dict[len].emplace_back(s); } col[len] = (int) dict[len].size(); acol = max(acol, col[len]); len++; } for(int i = 0; i < acol; i++) { cnt[i] = -INF; for(int j = 0; j < len; j++) if(col[j] >= i + 1) cnt[i] = max(cnt[i], (int) dict[j][i].size()); } for(int i = 0; i < len; i++) { for(int j = 0; j < col[i]; j++) { if(j) cout << " "; cout << dict[i][j]; if(j < col[i] - 1) OutSpace(cnt[j] - (int) dict[i][j].size()); } cout << endl; } return 0; } |
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#include <iostream> #include <fstream> #include <sstream> #include <algorithm> #include <cstring> #include <cstdio> #include <iomanip> #include <cmath> #include <map> #include <string> #include <vector> #include <stack> #include <queue> #include <set> using namespace std; #define EPS 1e-8; string ans[1000+66][200+66]; int num[1000+66], cnt[200+66]; int main() { // ifstream cin ("aaa.txt"); // freopen ("aaa.txt", "r", stdin); int i = 0, j = 0; string input; while (getline (cin, input)) { stringstream ss (input); j = 0; while (ss >> ans[i][j]) { if (cnt[j] < ans[i][j].length ()) cnt[j] = ans[i][j].length (); j++; } num[i] = j; i++; } int k; for (k = 0; k < i; k++) { for (j = 0; j < num[k]; j++) { if (j >= 1) { int q = 1, len_output = cnt[j-1] - ans[k][j-1].length () + 1; while (q <= len_output) { cout << " "; q++; } cout << ans[k][j]; } else cout << ans[k][j]; } cout << endl; } return 0; } |
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