A1112. Stucked Keyboard

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<cstdio>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
int hashTB[128];
int main(){
    char str[1001], prt[1001] = {0};
    int N, K;
    fill(hashTB, hashTB + 128, 1);
    scanf("%d", &K);
    scanf("%s", str);
    N = strlen(str);
    for(int i = 0; i < N; i++){
        int tag = 1;
        if(i + K > N){
            hashTB[str[i]] = 0;
            continue;
        }
        for(int j = i; j < i + K; j++){
            if(str[j] != str[i]){
                hashTB[str[i]] = 0;
                tag = 0;
                break;
            }
        }
        if(tag == 1){
            i += (K - 1);
        }
    }
    int flag = 0;
    for(int i = 0; i < N; i++){
        if(hashTB[str[i]] == 1 && prt[str[i]] == 0){
            printf("%c", str[i]);
            prt[str[i]] = 1;
            flag = 1;
        }
    }
    if(flag == 1)
        printf("
");
    for(int i = 0; i < N; i++){
        if(hashTB[str[i]] == 0){
            printf("%c", str[i]);
        }else{
            printf("%c", str[i]);
            i += K - 1;
        }
    }
    cin >> N;
    return 0;
}

总结：

1、题意：给出一个字符串和粘连次数K，检查这个键是否是坏键。

2、使用哈希表。初始时假设所有键都是坏的，然后遍历，发现a没有粘连，则说明a是好的。 注意不能假设所有键是好的，然后发现坏件再设置为坏。因为aaaba，则会错误将a识别为坏件。

3、注意最后K个字符的判断。