zoukankan      html  css  js  c++  java
  • POJ 2039 To and Fro(模拟)

    To and Fro

    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

    题目大意:

      给出2<=N<=20和加密过的字符串(长度<=200),输出解密后的字符串。

      加密规则:例如theresnoplacelikehomeonasnowynightx,按列写出,共N列

      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

      再按照奇数行,左至右,偶数行,右到左的顺序按行写出即可

    代码如下:

     1 # include<stdio.h>
     2 # include<string.h>
     3 # define MAX 205
     4 int n,m;
     5 char s[MAX],ans[MAX][MAX];
     6 int main(){
     7     int i,j;
     8     while(scanf("%d",&n) && n){
     9         scanf("%s",s);
    10         m = strlen(s)/n;
    11         int cnt=0;
    12         for(i=1; i<=m; i++){
    13             if(i%2)
    14                 for(j=1; j<=n; j++)
    15                     ans[i][j] = s[cnt++];
    16             else
    17                 for(j=n; j>=1; j--)
    18                     ans[i][j] = s[cnt++];
    19         }
    20         for(i=1; i<=n; i++)        
    21             for(j=1; j<=m; j++)
    22                 printf("%c",ans[j][i]);
    23             puts("");
    24     }
    25     return 0;
    26 }
  • 相关阅读:
    Oracle学习笔记:oracle的表空间管理和sqlserver的文件组对比
    Oracle学习笔记:一个特殊的ORA12541错误原因
    Oracle学习笔记:通过种子数据库设置dbid为指定值
    Oracle学习笔记:使用rman duplicate {to|for} 创建数据库
    Oracle学习笔记:利用rman数据库备份,手工创建clone数据库
    使用Cufon技术实现Web自定义字体
    分享七个非常有用的Android开发工具和工具包
    60佳灵感来自大自然的网页设计作品欣赏
    20个独一无二的图片滑动效果创意欣赏
    40个幻灯片效果在网页设计中的应用案例
  • 原文地址:https://www.cnblogs.com/acm-bingzi/p/3286375.html
Copyright © 2011-2022 走看看