1031 Hello World for U
Given any string of N (≥) 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 } 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
题意:
给出一个字符串,将字符串按照U字形的方式进行排列。
思路:
将U字形抽象成一个矩阵的三条边,且矩阵的列数一定要 >= 行数,矩阵尽可能的为方阵。然后模拟。
Code:
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 int main() { 6 string str; 7 cin >> str; 8 9 int len = str.length(); 10 int row = len / 3 + 1; 11 int col = len - 2 * row + 2; 12 if (col < row) { 13 col += 2; 14 row -= 1; 15 } 16 vector<vector<char> > martix(row, vector<char>(col, ' ')); 17 int i, j; 18 for (j = 0, i = 0; j < row; ++j, ++i) martix[j][0] = str[i]; 19 for (j = 1; j < col; ++j, ++i) martix[row - 1][j] = str[i]; 20 for (j = row - 2; j >= 0; --j, ++i) martix[j][col - 1] = str[i]; 21 for (i = 0; i < row; ++i) { 22 for (j = 0; j < col; ++j) cout << martix[i][j]; 23 cout << endl; 24 } 25 26 return 0; 27 }