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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)



import java.util.Scanner;
import java.util.Stack;

public class Main {

	public static void main(String[] args) {
		// TODO 自動生成されたメソッド・スタブ
		Scanner sc = new Scanner(System.in);

		int n = sc.nextInt();
		int k = sc.nextInt();
		int c = sc.nextInt();

		char[] str = sc.next().toCharArray();
		int dp[] = new int[n];

		for(int i = 0; i < n; i++){
			if(str[i] == 'o'){
				if(i <= c){
					dp[i] = 1;
				}
				else {
					dp[i] = dp[i - c - 1] + 1;
				}
			}
			else {
				if(i == 0){
					dp[i] = 0;
				}
				else {
					dp[i] = dp[i - 1];
				}
			}

//			System.out.print(dp[i]+" ");
		}
//		System.out.println();

		if(dp[n - 1] > k){
			return;
		}

		Stack<Integer> stack = new Stack<Integer>();

		int current;
		int count = 0;
		int latestIndex = -1;
		for(int i = n - 1; i >= 0; i--){
			current = dp[i];
//			System.out.println("i = "+i+ " number "+current);

			if(str[i] == 'o'){
				count++;
				latestIndex = i;
			}

			if(i == 0 || dp[i - 1] != current){
				if(count == 1){
					stack.add(latestIndex + 1);
					if(i != 0){
//						i -= c;

						i = Math.min(i, latestIndex - c);
					}
				}
				count = 0;
			}
		}

		while(!stack.isEmpty()){
			System.out.println(stack.pop());
		}
	}

}