1084. Broken Keyboard (20)

最新推荐文章于 2022-09-01 23:04:18 发布

原创最新推荐文章于 2022-09-01 23:04:18 发布 · 197 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#PAT甲

PAT 甲级专栏收录该内容

97 篇文章

订阅专栏

本文介绍了一个算法，用于检测输入字符串中哪些字符可能因为键盘损坏而未被正确输入。该算法通过对比预期输入与实际输入的字符串来找出损坏的键。

1084. Broken Keyboard (20)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

On a broken keyboard, some of the keys are worn out. So when you type some sentences, the characters corresponding to those keys will not appear on screen.

Now given a string that you are supposed to type, and the string that you actually type out, please list those keys which are for sure worn out.

Input Specification:

Each input file contains one test case. For each case, the 1st line contains the original string, and the 2nd line contains the typed-out string. Each string contains no more than 80 characters which are either English letters [A-Z] (case insensitive), digital numbers [0-9], or "_" (representing the space). It is guaranteed that both strings are non-empty.

Output Specification:

For each test case, print in one line the keys that are worn out, in the order of being detected. The English letters must be capitalized. Each worn out key must be printed once only. It is guaranteed that there is at least one worn out key.

Sample Input:

7_This_is_a_test
_hs_s_a_es

Sample Output:

7TI

提交代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <queue>
#include <stack>

using namespace std;

int main()
{
    freopen("in.txt","r",stdin);

    int i;
    char a[100];
    char b[100];
    scanf("%s",a);
    scanf("%s",b);
    int len1=strlen(a);
    int len2=strlen(b);
    int book[200]={0};
    queue<char> q1;
    queue<char> q2;
    for(i=0;i<len1;i++)
    {
        int t;
        t=a[i];
        if(t>=97&&t<=122)
            a[i]-=32;
        q1.push(a[i]);
    }
    for(i=0;i<len2;i++)
    {
        int t2;
        t2=b[i];
        if(t2>=97&&t2<=122)
            b[i]-=32;
        q2.push(b[i]);
    }
    while(!q1.empty())
    {
        int t3=q1.front();
        if(  (q1.front()!=q2.front())&&book[t3]==0  )
        {
            book[t3]=1;
            cout<<q1.front();
            q1.pop();
        }
        else if( (q1.front()!=q2.front())&&book[t3]==1    )
        {
            q1.pop();
        }
        else if(  q1.front()==q2.front() )
        {
            q1.pop();
            q2.pop();
        }

    }


    return 0;
}