PAT 甲级 1112. Stucked Keyboard (20)

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

#include <iostream>#include <algorithm>#include <map>#include <string>#include <set>using namespace std;bool sureNoBroken[256];int main() {    int k, cnt = 1;    scanf("%d", &k);    string s;    cin >> s;    map<char, bool> m;    set<char> printed;    char pre = '#';    s = s + '#';    for (int i = 0; i < s.length(); i++) {        if (s[i] == pre) {            cnt++;        }        else {            if (cnt%k != 0) {                sureNoBroken[pre] = true;            }            cnt = 1;        }        if (i != s.length() - 1) m[s[i]] = (cnt%k == 0);        pre = s[i];    }    for (int i = 0; i < s.length() - 1; i++) {        if (sureNoBroken[s[i]] == true)            m[s[i]] = false;    }    for (int i = 0; i < s.length() - 1; i++) {        if (m[s[i]] && printed.find(s[i]) == printed.end()) {            printf("%c", s[i]);            printed.insert(s[i]);        }    }    printf("\n");    for (int i = 0; i < s.length()-1; i++) {        printf("%c", s[i]);        if (m[s[i]])            i = i + k - 1;    }    cin >> s;    return 0;}