SOJ 1007. To and Fro

题目大意：还原（解码）字符串。该字符串str首先被平均分成n列，每列r行，其中n * r  ≤ len(str) ，多余位置用'x'添补。字符串按列填充，然后将偶数列翻转，然后按行获得输入串。

解题思路：逆向思维。先将输入串按行填充 r * n的矩阵，然后偶数列翻转，最后按列输出输出串。

代码如下：

#include <iostream>
#include <cstring>
using namespace std;

const int ROWS = 105;
const int COLS = 25;

char m[ROWS][COLS];
char str[ROWS * 2];
char ans[ROWS * 2];
int n;

int main() {
    while (cin >> n, n) {
        cin >> str;
        int len = strlen(str);

        int r = (len + n - 1) / n;

        for (int i = 0; i < r; i++) {
            for (int j = 0; j < n; j++) {
                m[i][j] = str[i*n + j];
            }
        }

        for (int i = 1; i < r; i += 2) {
            for (int j = 0; j * 2 < n; j++) {
                char temp = m[i][j];
                m[i][j] = m[i][n - j - 1];
                m[i][n - j - 1] = temp;
            }
        }

        int k = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < r; j++) {
                ans[k++] = m[j][i];
            }
        }
        ans[k++] = '';

        cout << ans << endl;
    }
    return 0;
}