Uva 10082 POJ2538 WERTU(简单字符操作)

最新推荐文章于 2020-01-29 21:06:25 发布

AC_Dreameng

最新推荐文章于 2020-01-29 21:06:25 发布

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分类专栏：算法竞赛入门经典训练指南 UVa POJ ACM_HDU刷题录文章标签： Uva10082 POJ2538 WERTU

本文链接：https://blog.youkuaiyun.com/hurmishine/article/details/50898325

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这篇博客介绍了如何解决Uva在线判题系统中的10082题和POJ判题系统中的2538题——WERTU。题目相对简单，但需要注意输入处理，不能使用scanf()，应采用gets()或getchar()，以防因空格导致的输入问题。博主分享了通过此题的AC（Accepted）代码。

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WERTYU

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8971		Accepted: 4288

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.

原题链接:https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1023

POJ链接:http://poj.org/problem?id=2538

Virtual OJ链接:http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=19338

水题,但是要注意输入不能用scanf(),因为输入中含有空格,用gets(),或者getchar()

AC代码:

#include <stdio.h>
char *s = "`1234567890-=QWERTYUIOP{}\\ASDFGHJKL:'ZXCVBNM,./";
int main()
{
    int i,c;
    while((c = getchar()) != EOF)
    {
        for (i = 1; s[i] && s[i]!=c; i++);
        if(s[i]) putchar(s[i-1]);
        else putchar(c);
    }
    return 0;
}

上面那个代码是<<算法竞赛入门经典>>上面的,下面那个代码是我的...

#include <stdio.h>
int main()
{
    char x[100]="`1234567890-=QWERTYUIOP[]\\ASDFGHJKL;'ZXCVBNM,./";
    int i,j;
    char a[1000];
    while(gets(a))
    {
        for (i=0;a[i]!='\0';i++)
        {
            if (a[i]==' ')
            {
                printf(" ");
                continue ;
            }
            for (j=0;x[j]!='\0';j++)
            {
                if (a[i]==x[j])
                {
                    printf("%c",x[j-1]);
                }
            }
        }
        printf("\n");
    }
    return 0;
}