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 n​1​​ characters, then left to right along the bottom line with n​2​​ characters, and finally bottom-up along the vertical line with n​3​​ characters. And more, we would like U to be as squared as possible -- that is, it must be satisfied that n​1​​=n​3​​=max { k | k≤n​2​​ for all 3 } with n​1​​+n​2​​+n​3​​−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：

#include <bits/stdc++.h>

using namespace std;

int main() {
    string str;
    cin >> str;

    int len = str.length();
    int row = len / 3 + 1;
    int col = len - 2 * row + 2;
    if (col < row) {
        col += 2;
        row -= 1;
    }
    vector<vector<char> > martix(row, vector<char>(col, ' '));
    int i, j;
    for (j = 0, i = 0; j < row; ++j, ++i) martix[j][0] = str[i];
    for (j = 1; j < col; ++j, ++i) martix[row - 1][j] = str[i];
    for (j = row - 2; j >= 0; --j, ++i) martix[j][col - 1] = str[i];
    for (i = 0; i < row; ++i) {
        for (j = 0; j < col; ++j) cout << martix[i][j];
        cout << endl;
    }

    return 0;
}