poj 2538(簡單題)

題意是一個錯誤的鍵盤，輸入的一個字母或字元時會變成這個字母或字元在鍵盤上位置右邊的一個。 然後已知輸出的字元，求你原本輸入的是什麼。

有一種很土的辦法就是一個字元一個字元的裝換，但是這樣不知道要用多少個if。

可以把這這些字元存在一起 g[1000] ={"1234567890-=QWERTYUIOP[]\ASDFGHJKL;'ZXCVBNM,./"} ,這樣只要找到了一個字元在這個g串中的位置i，然後g[i-1]就是我們要的哪個字元。 這樣編程量就少了很多

 

WERTYU Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7513   Accepted: 3538 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. Source Waterloo local 2001.01.27

 

#include <stdio.h>#include <string.h>#include <iostream>using namespace std;char str[1100];char g[1100]={"1234567890-=QWERTYUIOP[]\ASDFGHJKL;'ZXCVBNM,./"};int main(){    while(gets(str)!=0)    {                int len=strlen(str);        for(int i=0;i<len;i++)        {            char tmp=str[i];            int j;            for(j=0;j<1100;j++)            {                if(g[j]==tmp)                    break;            }            if(j==1100) printf("%c",tmp);            else            printf("%c",g[j-1]);        }        printf("\n");    }    return 0;}