PAT甲级 1084 Broken Keyboard (20point(s))

最新推荐文章于 2020-07-07 23:40:04 发布

原创最新推荐文章于 2020-07-07 23:40:04 发布 · 309 阅读

CC 4.0 BY-SA版权

本文介绍了一种用于检测键盘上磨损键位的算法。通过对比预期输入字符串与实际输入字符串，该算法能有效找出并列出所有磨损的键位，适用于英语字母、数字及空格键的检测。示例代码使用C++实现，展示了如何通过散列表记录键位状态，从而快速定位故障键。

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
经典散列问题
AC代码：

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
bool Hash[128]={false};
int main(){
    char str1[81],str2[81];
    cin.get(str1,81);
    getchar();
    cin.get(str2,81);
    int len1=strlen(str1), len2=strlen(str2);
    for(int i=0;i<len1;i++)
    if(str1[i]>='a'&&str1[i]<='z') str1[i]-='a'-'A';
for(int i=0;i<len2;i++){
    if(str2[i]>='a'&&str2[i]<='z')
str2[i]-='a'-'A';
Hash[str2[i]]=true;
}
for(int i=0;i<len1;i++)
if(Hash[str1[i]]==false){
    printf("%c",str1[i]);
    Hash[str1[i]]=true;
}
system("pause");
return 0;
}