1112. Stucked Keyboard (20) 坏键盘

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本文介绍了一种算法，用于从含有重复字符的字符串中找出并修复由键盘某些按键始终被卡住导致的问题。通过分析输入字符串，算法可以识别出哪些字符可能是由于按键卡住而重复，并还原出原始输入字符串。

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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

有一种情况要特别注意：

例如

aaabbbaaabbbb

输出

abbbabbbb

这种情况，b在前面是坏键（是k的倍数），但是在后面不是坏键盘（不是k的倍数）

只要出现不是k的倍数，就是好键...

可能说的不是很清楚。就是这个例子注意一下

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
int main(){
	int k;
	cin>>k;
	getchar(); 
	char s[10000];
	int v[1000]={0}; 
	gets(s);
	set<char> se;		
	for(int i=0;i<strlen(s);i++){
		int sum=1;
		for(int j=i+1;j<strlen(s);j++){
			if(s[j]==s[i]){
				sum++;
			}
			else break;
		}
		if(sum%k==0&&v[s[i]]==0){//这种情况说明有可能是坏键盘
			se.insert(s[i]);
			i+=k-1;
		}
		else{//肯定不是坏键盘，不管前面是否为坏的。
			v[s[i]]=1;
			se.erase(s[i]);
		}
	} 
	int vv[1000]={0};
	for(int i=0;i<strlen(s);i++){
		if(v[s[i]]==0&&se.find(s[i])!=se.end()){
			cout<<s[i];
			v[s[i]]=1;
		}
	} 
	cout<<endl;
	for(int i=0;i<strlen(s);i++){
		cout<<s[i];
		if(se.find(s[i])!=se.end()){
			i+=k-1;
		}
	}
	return 0;
}