1084 Broken Keyboard (20 分)（字符串处理）

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
hssaes

Sample Output:

7TI

题目大意：

旧键盘上坏了几个键，于是在敲一段文字的时候，对应的字符就不会出现。现在给出应该输入的一段文字、以及实际被输入的文字，请你列出肯定坏掉的那些键～

分析：

用string的find函数～遍历字符串s1，当当前字符s1[i]不在s2中，它的大写也不在ans中时，将当前字符的大写放入ans中，最后输出ans字符串即可～

ps：感谢github上的@xiaorong61给我发的pull request中strchr()函数带来的灵感～让我想到了曾经用过的string的find函数～

原文链接：https://blog.csdn.net/liuchuo/article/details/51994896

错误思路

用set呢，确实可以去重，但是他的顺序是字母序的；
而unordered_set是随机顺序。
因此，两者都不可。还是乖乖用hash。

#include <bits/stdc++.h>

using namespace std;

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif
    string s1,s2;
    unordered_set<char> st;
    cin>>s1>>s2;
    for(int i=0;i<s1.size();i++){
        if(s2.find(s1[i])==string::npos){
            st.insert(toupper(s1[i]));
        }
    }
    for(auto it=st.begin();it!=st.end();it++)
        cout<<*it;
    return 0;
}

题解

柳神的思路总是这么简洁明了~

#include <bits/stdc++.h>

using namespace std;

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif
    string s1,s2,ans;
    cin>>s1>>s2;
    for(int i=0;i<s1.size();i++){
        if(s2.find(s1[i])==string::npos&&ans.find(toupper(s1[i]))==string::npos){
            ans+=toupper(s1[i]);
        }
    }
    cout<<ans;
    return 0;
}