PAT_A1112#Stucked Keyboard

Source:

PAT A1112 Stucked Keyboard (20 分)

Description:

On a broken keyboard, some of the keys are always stucked. So when you type some sentences, the characters corresponding to those keys will appear repeatedly on screen for k times.

Now given a resulting string on screen, you are supposed to list all the possible stucked keys, and the original string.

Notice that there might be some characters that are typed repeatedly. The stucked key will always repeat output for a fixed k times whenever it is pressed. For example, when k=3, from the string thiiis iiisss a teeeeeest we know that the keys i and e might be stucked, but s is not even though it appears repeatedly sometimes. The original string could be this isss a teest.

Input Specification:

Each input file contains one test case. For each case, the 1st line gives a positive integer k (1) which is the output repeating times of a stucked key. The 2nd line contains the resulting string on screen, which consists of no more than 1000 characters from {a-z}, {0-9} and _. It is guaranteed that the string is non-empty.

Output Specification:

For each test case, print in one line the possible stucked keys, in the order of being detected. Make sure that each key is printed once only. Then in the next line print the original string. It is guaranteed that there is at least one stucked key.

Sample Input:

3
caseee1__thiiis_iiisss_a_teeeeeest

Sample Output:

ei
case1__this_isss_a_teest

Keys:

• 哈希映射

Code:

 1 /*
 2 Data: 2019-07-19 18:45:30
 3 Problem: PAT_A1112#Stucked Keyboard
 4 AC: 29:33
 5
 6 题目大意：
 7 键盘上坏掉的键会重复打印K次，给出结果字符串，找出坏掉的键和原始字符串
 8 输入：
 9 第一行给出，重复次数k
10 第二行给出，结果字符串
11 输出：
12 坏掉的字符
13 原始字符串
14 */
15
16 #include<cstdio>
17 #include<string>
18 #include<iostream>
19 #include<algorithm>
20 using namespace std;
21 char mp[128]={0};
22
23 int main()
24 {
25 #ifdef    ONLINE_JUDGE
26 #else
27     freopen("Test.txt", "r", stdin);
28 #endif
29
30     int k;
31     scanf("%d", &k);
32     string s,brk="";
33     cin >> s;
34     for(int i=0; i<s.size(); i++)
35     {
36         int cnt=0;
37         while(i+1<s.size() && s[i]==s[i+1]){
38             i++;cnt++;
39         }
40         if((cnt+1)%k==0 && mp[s[i]]==0)
41             mp[s[i]]=1;
42         else if((cnt+1)%k!=0)
43             mp[s[i]]=2;
44     }
45     for(int i=0; i<s.size(); i++){
46         if(mp[s[i]]==1 || mp[s[i]]==3)
47         {
48             if(mp[s[i]]==1){
49                 brk += s.substr(i,1);
50                 mp[s[i]]=3;
51             }
52             s.erase(i,k-1);
53         }
54     }
55     cout << brk << endl;
56     cout << s << endl;
57
58     return 0;
59 }

原文地址：https://www.cnblogs.com/blue-lin/p/11215320.html