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社区首页 >专栏 >HDOJ/HDU 1200 To and Fro(加密解密字符串)

HDOJ/HDU 1200 To and Fro(加密解密字符串)

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谙忆
发布2021-01-21 12:16:57
发布2021-01-21 12:16:57
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文章被收录于专栏:程序编程之旅程序编程之旅

Problem 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 5 toioynnkpheleaigshareconhtomesnlewx 3 ttyohhieneesiaabss 0

Sample Output theresnoplacelikehomeonasnowynightx thisistheeasyoneab

水题。 译码还原问题

题目保证输入的数据一定是一个矩形的。

代码语言:javascript
复制
import java.util.Scanner;

public class Main{

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        while(sc.hasNext()){
            int n=sc.nextInt();
            if(n==0){
                return;
            }
            String str=sc.next();
            String s[] = new String[str.length()/n];
            for(int i=0;i<str.length()/n;i++){
                s[i]="";//不要忘记初始化
                if(i%2==0){
                    for(int k=i*n;k<n+n*i;k++){
                        s[i]=s[i]+str.charAt(k);
                    }
                }else{
                    for(int k=i*n;k<n+n*i;k++){
                        s[i]=str.charAt(k)+s[i];
                    }
                }
                //System.out.println(s[i]);
            }

            for(int i=0;i<n;i++){
                for(int k=0;k<s.length;k++){
                    System.out.print(s[k].charAt(i));
                }
            }
            System.out.println();

        }

    }

}
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原始发表:2016/05/27 ,如有侵权请联系 cloudcommunity@tencent.com 删除

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