Stucked Keyboard （字符串，计数，恶心

求键盘怀了哪几个键，用vector存答案，map标记哪些键坏了。

一个样例：

3
aaabbbccccabc

输出应该是 aaabbbcccabc 一个键也没有坏

反正就是从左往右遍历一遍，记录连续相同的字母的出现次数，不够K次就标记为没有坏，输出时要注意这一点。

一开始想的连续K个相同的字母就pop掉，发现不对。

其实坏键可以用set存。

 

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<k≤100) 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

#include<bits/stdc++.h>
#include<set>
#include<vector>
#include<map>
#include<queue>
#include<string>
#include<stack>
using namespace std;

int main() {
	int k;
	string a;
	int cnt=1;
	vector<char>ans;
	map<char,int>mp;
	map<char,int>mp2;

	cin>>k>>a;
	ans.push_back(a[0]);

	for(int i=1; i<=a.length(); i++) {
		ans.push_back(a[i]);

		if(a[i]!=a[i-1]) {
			if(cnt<k)mp[a[i-1]]=1;
			cnt=1;
		} else {
			cnt++;
			if(cnt==k) {
				cnt=1;
				i++;
				ans.push_back(a[i]);
			}
		}
	}
	int flag=0;
	for(int i=0; i<a.length(); i++) {
		if(mp[a[i]]==0&&mp2[a[i]]==0) {
			cout<<a[i];
			mp2[a[i]]=1;
			flag=1;
		}
	}
	if(flag)cout<<endl;
	for(int i=0; i<ans.size() ; i++) {
		cout<<ans[i];
		if(mp[a[i]]==0)i+=k-1;
	}
	return 0;
}