WERTYU
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 8371   Accepted: 4007

Description


A common typing error is to place the hands on the keyboard one row to the right of the correct position. So "Q" is typed as "W" and "J" is typed as "K" and so on. You are to decode a message typed in this manner.

Input

Input consists of several lines of text. Each line may contain digits, spaces, upper case letters (except Q, A, Z), or punctuation shown above [except back-quote (`)]. Keys labelled with words [Tab, BackSp, Control, etc.] are not represented in the input.

Output

You are to replace each letter or punctuation symbol by the one immediately to its left on the QWERTY keyboard shown above. Spaces in the input should be echoed in the output.

Sample Input

O S, GOMR YPFSU/

Sample Output

I AM FINE TODAY.

算法分析:下面的算法,写起来实在是过于麻烦而且容易出错,现提供常量数组的做法。初始化定义一个字符数组s[]={",,,,,,,,,,,,"}
将全部的键盘输进去,输出时,先遍历一遍找到对应的字符,然后输出该字符的钱一个字符就可以了,此写法不容易出错!
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std; char s[1000000]; int main()
{
char t[10000];
int i, j;
int len;
//定义区
s['1']='~'; s['2']='1'; s['3']='2'; s['4']='3'; s['5']='4'; s['6']='5'; s['7']='6'; s['8']='7';
s['9']='8'; s['0']='9'; s['-']='0'; s['=']='-';
s['W']='Q'; s['E']='W'; s['R']='E'; s['T']='R'; s['Y']='T'; s['U']='Y';
s['I']='U'; s['O']='I'; s['P']='O'; s['[']='P'; s[']']='[';
s['\\']=']';
s['\'']=';';
s['S']='A'; s['D']='S'; s['F']='D'; s['G']='F'; s['H']='G'; s['J']='H'; s['K']='J'; s['L']='K'; s[';']='L'; s['X']='Z'; s['C']='X'; s['V']='C'; s['B']='V'; s['N']='B'; s['M']='N'; s[',']='M'; s['.']=','; s['/']='.'; while(gets(t)!=NULL)
{
len = strlen(t);
for(i=0; i<len; i++)
{
if(t[i]==' ')
{
printf(" ");
}
else
{
printf("%c", s[t[i]] );
}
}
printf("\n");
}
return 0;
}

POJ 之 WERTYU的更多相关文章

  1. POJ 2538 WERTYU水的问题

    [题目简述]:题意非常easy,没有trick. [分析]:事实上这题还是挺有趣的,在 算法竞赛入门经典中也有这一题. 详见代码: // 120K 0Ms /* 边学边做 -- */ // 字符串:W ...

  2. POJ 3370. Halloween treats 抽屉原理 / 鸽巢原理

    Halloween treats Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 7644   Accepted: 2798 ...

  3. POJ 2356. Find a multiple 抽屉原理 / 鸽巢原理

    Find a multiple Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7192   Accepted: 3138   ...

  4. POJ 2965. The Pilots Brothers' refrigerator 枚举or爆搜or分治

    The Pilots Brothers' refrigerator Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 22286 ...

  5. POJ 1753. Flip Game 枚举or爆搜+位压缩,或者高斯消元法

    Flip Game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 37427   Accepted: 16288 Descr ...

  6. POJ 3254. Corn Fields 状态压缩DP (入门级)

    Corn Fields Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 9806   Accepted: 5185 Descr ...

  7. POJ 2739. Sum of Consecutive Prime Numbers

    Sum of Consecutive Prime Numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 20050 ...

  8. POJ 2255. Tree Recovery

    Tree Recovery Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11939   Accepted: 7493 De ...

  9. POJ 2752 Seek the Name, Seek the Fame [kmp]

    Seek the Name, Seek the Fame Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 17898   Ac ...

随机推荐

  1. Synchronized与ReentrantLock的区别

    1.ReentrantLock 拥有Synchronized相同的并发性和内存语义,此外还多了 锁投票,定时锁等候和中断锁等候 线程A和B都要获取对象O的锁定,假设A获取了对象O锁,B将等待A释放对O ...

  2. springnodejs

    作者 : solq 最新文档请看 http://www.springnodejs.com 本文不再更新 blog : http://www.cnblogs.com/solq/p/3574640.htm ...

  3. C#控件之ListView

    1.添加一行 listView1.Items.Add(new ListViewItem(new string[] {str1, str2, str3 })); 2.鼠标悬停在listview某项时弹出 ...

  4. hdu1595 find the longest of the shortest(Dijkstra)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1595 find the longest of the shortest Time Limit: 100 ...

  5. 每天5道面试题(二)java基础

    说出Servlet的生命周期,并说出Servlet和CGI的差别 Servlet被server实例化后,容器执行其init方法,请求到达时执行其service方法,service方法自己主动派遣执行与 ...

  6. Shell脚本之:变量

    与编译型语言不同,shell脚本是一种解释型语言. 执行这类程序时,解释器(interpreter)需要读取我们编写的源代码(source code),并将其转换成目标代码(object code), ...

  7. ruby 作为嵌入脚本时使用的注意事项

    近期一直在採坑... 假设是作为嵌入式脚本使用ruby的话... 一定会遇到这2个问题... gem安装的,无法在嵌入时使用..为啥.? 由于你没require 'ruby gem' 出现 找不到 E ...

  8. [javase学习笔记]-8.2 成员变量与静态变量的差别

    这一节我们看一看成员变量与静态变量的差别所在. 什么是静态变量呢?我们上节用statickeyword时就提到了静态变量.也就是说用statickeyword修饰的变量就是静态变量. 我们在6.4节学 ...

  9. 转:HDMI介绍与流程

    HDMI介绍与流程   HDMI,全称为(High Definition Multimedia Interface)高清多媒体接口,主要用于传输高清音视频信号. HDMI引脚: HDMI有A,B,C, ...

  10. mysql中的类型转换和精确位数

    select round(123.5); 四舍五入 select floor(123.5);取整数部分 select ceil(123.5);四舍五入