sicily 1007 to and from 密文解碼 數組與下標

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1007. To and FroConstraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

Mo and Larry have devised a way of encrypting messages. They first decide secretly on the number of columns and write the message (letters only) down the columns, padding with extra random letters so as to make a rectangular array of letters. For example, if the message is "There‘s no place like home on a snowy night" and there are five columns, Mo would write down t o i o y h p k n n e l e a i r a h s g e c o n h s e m o t n l e w x Note that Mo includes only letters and writes them all in lower case. In this example, Mo used the character `x‘ to pad the message out to make a rectangle, although he could have used any letter. Mo then sends the message to Larry by writing the letters in each row, alternating left-to-right and right-to-left. So, the above would be encrypted as toioynnkpheleaigshareconhtomesnlewx Your job is to recover for Larry the original message (along with any extra padding letters) from the encrypted one.

Input

There will be multiple input sets. Input for each set will consist of two lines. The first line will contain an integer in the range 2 . ..20 indicating the number of columns used. The next line is a string of up to 200 lower case letters. The last input set is followed by a line containing a single 0, indicating end of input.

Output

Each input set should generate one line of output, giving the original plaintext message, with no spaces.

Sample Input

5toioynnkpheleaigshareconhtomesnlewx3ttyohhieneesiaabss0

Sample Output

theresnoplacelikehomeonasnowynightxthisistheeasyoneab

#include <iostream>using namespace std;//產生密文用，此處無用 void filter(string &str) {string temp;for (int i = 0; i < str.length(); i++) {if (str[i] >= 97 && str[i] <= 122) {temp += str[i];} else if (str[i] >= 65 && str[i] <= 90) {temp += (str[i] + 32);}}str = temp;} int main() {int n;string str;int col_capacity; while (cin >> n && n) {string sub_str[n]; //存放密文分解出來的字串，一共n列 cin >> str;//確定每列的字元數 if (str.length() % n != 0) {col_capacity = str.length() / n + 1; } else {col_capacity = str.length() / n;}int k = 0; //用於遍曆密文 for (int i = 0; i < col_capacity; i++) { //依次填充每列從上到下的字元，構成一行  if (i % 2 == 0) { // 偶數行則從左至右填充 for (int j = 0; j < n; j++)sub_str[j].push_back(str[k++]);} else { //奇數行則從右至左填充 for (int j = n-1; j >= 0; j--) {sub_str[j].push_back(str[k++]);}}}//依次輸出各列即得到明文 for (int i = 0; i < n; i++) {cout << sub_str[i];}cout << endl;}return 0;} 

　　

sicily 1007 to and from 密文解碼 數組與下標