PAT A1084

最新推荐文章于 2020-06-10 13:36:40 发布

原创最新推荐文章于 2020-06-10 13:36:40 发布 · 228 阅读

0 ·

CC 4.0 BY-SA版权

PAT 专栏收录该内容

60 篇文章

订阅专栏

这篇博客解决了一个编程问题，即如何在输入字符串与实际显示字符串存在差异时找出键盘上哪些键已经磨损。通过将两个字符串进行比较，确定未出现的字符，并输出这些磨损的键。提供的解题思路包括将字符串转为大写并逐个字符比较，使用散列表记录已输出的磨损键，确保每个键只输出一次。示例输入和输出也展示了具体的应用情况。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目：
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

解题思路
1.将st1和st2中字符转化为大写字母，将st2每一个字符和st1中字符比较
2.若出现，则比较下一个字符，否则判断是否为第一次输出，借助散列表，若输出一次，则将此值设置为true，若为第一次输出，则输出，否则判断下一个字符

代码实现：

#include <iostream>
#include <string>
using namespace std;
int main()
{
    string st1, st2;
    cin >>st1 >> st2;
    int n1 = st1.length();
    int n2 = st2.length();
    bool judge[128] = {false};         //映射完的数组初始值为false
    for(int i =0; i <n1; i++){
         int j;
         char c1, c2;
         for(j = 0; j < n2; j++){
            c1 = st1[i];
            c2 = st2[j];
            if(c1 >='a' && c1 <='z')
                c1 -= 32;                 //若为小写字母，则转化为大写字母
            if(c2 >='a' && c2 <='z')
                c2 -= 32;
            if(c1 == c2) break;
         }
         if(j == n2 && judge[c1] == false)     //只输出一次
         {
             cout << c1;
             judge[c1] = true;
         }
    }
    return 0;
}