pat 1084. Broken Keyboard (20)

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

解：常规思路，两层循环，为了使结果不重复，将未出现的值添加到第二个字符串中，比如说7在第二行字符串s2中没有出现，我们将7加到s2中，这样在下次遍历的时候就不会将该值添加到目标字符串中了，目标字符串存储结果。其他的都输出常规思路，还好不卡时间。

#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int main()
{
    string s1,s2;
    while(cin>>s1>>s2)
    {
        string put;
        int len1=s1.length();
        int len2=s2.length();
        int flag;
        for(int i=0;i<len1;i++)
        {
            if(s1[i]>='a'&&s1[i]<='z')
            {
                s1[i]=s1[i]-32;
            }
        }
        for(int i=0;i<len2;i++)
        {
            if(s2[i]>='a'&&s2[i]<='z')
            {
                s2[i]=s2[i]-32;
            }
        }
     //  cout<<s1<<endl;
     //   cout<<s2<<endl;
        for(int i=0;i<len1;i++)
        {
            flag=0;
            for(int j=0;j<s2.length();j++)
            {
                if(s1[i]==s2[j])
                {
                    flag=1;
                }
            }
            if(flag==0)
            {
                put.push_back(s1[i]);
                s2.push_back(s1[i]);
               // cout<<s2<<endl;
                continue;
            }
        }
        cout<<put<<endl;
    }
}

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