PAT-Advanced-Level-Practise-1084. Broken Keyboard (20)

1084. Broken Keyboard (20)

本文来自我的Github: @jerrykcode

原题地址

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

题意：

输入两个字符串，求第一个字符串中出现过而第二个字符串中没有出现过的字符，字母不分大小写，输出的字母按

大写

思路：

先把两个字符串中所有字母字符转为大写。用map

代码：

1084. Broken Keyboard (20)_2.cpp

#include "stdafx.h"
#include <iostream>
using namespace std;
#include <string>
#include <map>

int main()
{
string s1, s2;
getline(cin, s1);
getline(cin, s2);
map<char, int> flags;
for (int i = 0; i < s1.length(); i++) {
if (isalpha(s1[i])) s1[i] = toupper(s1[i]);
if (i < s2.length() && isalpha(s2[i])) s2[i] = toupper(s2[i]);
}
for (int i = 0; i < s2.length(); i++)
flags[s2[i]] = 1;
for (int i = 0; i < s1.length(); i++)
if (flags[s1[i]] == 0) {
cout << s1[i];
flags[s1[i]] = 1;
}
return 0;
}