Problem:
Score : 500 points

Problem Statement

Takahashi has decided to work on K days of his choice from the N days starting with tomorrow.

You are given an integer C and a string S. Takahashi will choose his workdays as follows:

After working for a day, he will refrain from working on the subsequent C days.

If the i-th character of S is x, he will not work on Day i, where Day 1 is tomorrow, Day 2 is the day after tomorrow, and so on.

Find all days on which Takahashi is bound to work.

Constraints

1 \leq N \leq 2 \times 10^5

1 \leq K \leq N

0 \leq C \leq N

The length of S is N.

Each character of S is o or x.

Takahashi can choose his workdays so that the conditions in Problem Statement are satisfied.

Input

Input is given from Standard Input in the following format:

N K C
S

Output

Print all days on which Takahashi is bound to work in ascending order, one per line.

Sample Input 1

11 3 2
ooxxxoxxxoo

Sample Output 1

6

Takahashi is going to work on 3 days out of the 11 days. After working for a day, he will refrain from working on the subsequent 2 days.

There are four possible choices for his workdays: Day 1,6,10, Day 1,6,11, Day 2,6,10, and Day 2,6,11.

Thus, he is bound to work on Day 6.

Sample Input 2

5 2 3
ooxoo

Sample Output 2

1
5

There is only one possible choice for his workdays: Day 1,5.

Sample Input 3

5 1 0
ooooo

Sample Output 3

There may be no days on which he is bound to work.

Sample Input 4

16 4 3
ooxxoxoxxxoxoxxo

Sample Output 4

11
16

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <malloc.h>
#include <string.h>
#include <stdlib.h>

int main(void) {
  int n, k, c, i, tmp, cou=0, j;
  char ox;
  scanf(" %d %d %d", &n, &k, &c);
  int s[n], p[n], pt[n], p2[n]; //ox, 労働確定日か, pの仮定, どこから労働確定日でないと分かっているか
  for(i=0; i < n; i++){
    scanf(" %c", &ox);
    if(ox == 'o') s[i]=0;
    else s[i]=1;
  }

  /*******************************************************/
  //規定回数より多く働けるなら確定日無し
  for(i=0; i < n; i++){
    if(s[i] == 1){
      continue;
    } //労働不可の日はスキップ
    tmp=0; //tmp日目
    for(j=i; j < n; j++){
      if(s[j] == 1) continue;
      tmp=j;
      cou++;
      break;
    }

    while(tmp < n){
      tmp+=c + 1;
      while(tmp < n && s[tmp] == 1)
        tmp++;
      if(tmp < n) cou++;
      if(cou > k) return 0; //規定回数より多く働けるなら確定日無し
    }
    break; //そうでなければ通常通り
  }
  /*******************************************************/

  ////前から予定を詰めていく////
  memcpy(p, s, sizeof(s));
  memcpy(pt, p, sizeof(p));

  for(i=0; i < n; i++){
    if(p[i] == 1){
      //  printf("XXcontinueXX i=%d\n", i);
      continue;
    } //労働不可の日はスキップ
    tmp=0; //tmp日目
    cou=0;
    //s[i]=1; //労働不可と仮定

    for(j=0; j < n; j++)
      pt[j]=1;
    for(j=0; j < n; j++){
      if(s[j] == 1){
        //    printf("--continue-- i=%d\n\n", i);
        continue;
      }
      tmp=j;
      pt[j]=0;
      cou++;
      break;
    }

    while(tmp < n){
      tmp+=c + 1;
      while(tmp < n && s[tmp] == 1)
        tmp++;
      if(tmp < n){
        cou++;
        pt[tmp]=0;
      }
      if(cou >= k) break; //成立可能
    }
/*
    printf("i=%d\n", i);
    for(j=0; j < n; ++j)
      printf("%d", s[j]);
    printf("  <=s   j=%d  cou=%d\n", n, cou);
    for(j=0; j < n; ++j)
      printf("%d", p[j]);
    printf("  <=p   j=%d  tmp=%d\n", n, tmp);
    for(j=0; j < n; ++j)
      printf("%d", pt[j]);
    printf("  <=pt  j=%d   \n", n);
*/
    for(j=i; j <= tmp; ++j)
      p[j]=p[j] | pt[j];
    for(j=tmp+1; j < n; ++j)
      p[j]=1;

  //  for(j=0; j < n; ++j)
  //    printf("%d", p[j]);
  //  printf("  <=p\n\n");
  //  printf("  tmp:%d\n\n",tmp);
    break;
  }


  ////後ろから予定を詰めていく////

  memcpy(p2, s, sizeof(s));
  memcpy(pt, p2, sizeof(p2));

  for(i=0; i < n; i++){
    if(p2[n - i - 1] == 1){
      //  printf("XXcontinueXX i=%d\n", i);
      continue;
    } //労働不可の日はスキップ
    tmp=n - 1; //tmp日目
    cou=0;
    //s[i]=1; //労働不可と仮定

    for(j=0; j < n; j++)
      pt[n - j - 1]=1;
    for(j=0; j < n; j++){
      if(s[n - j - 1] == 1){
        //    printf("--continue-- i=%d\n\n", i);
        continue;
      }
      tmp=n - j - 1;
      pt[n - j - 1]=0;
      cou++;
      break;
    }

    while(tmp >= 0){
      tmp-=c + 1;
      while(tmp >= 0 && s[tmp] == 1)
        tmp--;
      if(tmp >= 0){
        cou++;
        pt[tmp]=0;
      }
      if(cou >= k) break; //労働不可で成立可能なら労働確定日でない
    }
/*
    printf("i=%d\n", i);
    for(j=0; j < n; ++j)
      printf("%d", s[j]);
    printf("  <=s   j=%d  cou=%d\n", n, cou);
    for(j=0; j < n; ++j)
      printf("%d", p2[j]);
    printf("  <=p2  j=%d  tmp=%d\n", n, tmp);
    for(j=0; j < n; ++j)
      printf("%d", pt[j]);
    printf("  <=pt  j=%d   t2=%d\n", n, t2);
*/
    for(j=i; j <= n - tmp; ++j)
      p2[j]=p2[j] | pt[j];
    for(j=n-tmp+1; j < n; ++j)
      p2[j]=1;

//   for(j=0; j < n; ++j)
//      printf("%d", p2[j]);
//    printf("  <=p2\n\n");
    break;
  }

  for(i=0; i < n; i++)
    if((p[i] | p2[i]) == 0) printf("%d\n", i+1);

}