【42.86%】【Codeforces Round #380D】Sea Battle

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

Galya is playing one-dimensional Sea Battle on a 1 × n grid. In this game a ships are placed on the grid. Each of the ships consists of b consecutive cells. No cell can be part of two ships, however, the ships can touch each other.

Galya doesn’t know the ships location. She can shoot to some cells and after each shot she is told if that cell was a part of some ship (this case is called “hit”) or not (this case is called “miss”).

Galya has already made k shots, all of them were misses.

Your task is to calculate the minimum number of cells such that if Galya shoot at all of them, she would hit at least one ship.

It is guaranteed that there is at least one valid ships placement.

Input

The first line contains four positive integers n, a, b, k (1 ≤ n ≤ 2·105, 1 ≤ a, b ≤ n, 0 ≤ k ≤ n - 1) — the length of the grid, the number of ships on the grid, the length of each ship and the number of shots Galya has already made.

The second line contains a string of length n, consisting of zeros and ones. If the i-th character is one, Galya has already made a shot to this cell. Otherwise, she hasn’t. It is guaranteed that there are exactly k ones in this string.

Output

In the first line print the minimum number of cells such that if Galya shoot at all of them, she would hit at least one ship.

In the second line print the cells Galya should shoot at.

Each cell should be printed exactly once. You can print the cells in arbitrary order. The cells are numbered from 1 to n, starting from the left.

If there are multiple answers, you can print any of them.

Examples

input

5 1 2 1

00100

output

2

4 2

input

13 3 2 3

1000000010001

output

2

7 11

Note

There is one ship in the first sample. It can be either to the left or to the right from the shot Galya has already made (the “1” character). So, it is necessary to make two shots: one at the left part, and one at the right part.

【题目链接】:http://codeforces.com/contest/738/problem/D

【题解】



这里的1会把整个序列分成若干个连续的0块;

先算出每个连续的0块里面最多能容纳下多少艘船(连续块的长度/b);

总的容纳船数目设为now;

然后题目所给的船数目为a;

则我们需要在这now艘船里面打掉(now-a+1)艘船(只是假想减小最大的船数目);

（更具体一点就是在每个连续的0块里面每个b个点打掉一个点，这样这个0块里面最大能容纳船的数目就会减1,总的最大能容纳船的数目也会减1);

这样总的最大的船数会小于a，则肯定能打掉一艘船;

记录下所有的连续点块的区间;然后一个一个区间地打就好;



【完整代码】

#include <bits/stdc++.h>
using namespace std;
struct abc
{
    int l,r;
};
int n,a,b,k,now=0;
char q[200009];
vector < pair<int,int> > c;
vector <int> ans;
int main()
{
   // freopen("F:\\rush.txt","r",stdin);
    cin >> n >> a >> b >>k;
    scanf("%s",q+1);
    for (int i = 1;i <= n;i++)
    {
        if (q[i] == '0')
        {
            int j = i+1;
            while (j<=n && q[j]=='0')
                j++;
            c.push_back(make_pair(i,j-1));
            now+=((j-i)/b);
            i = j-1;
        }
    }
    int num = 0;
    for (int i = 0;i <= c.size()-1;i++)
    {
        int t = b;
        while (num <=now-a)
        {
            if (t+c[i].first-1>c[i].second)
                break;
            ans.push_back(c[i].first+t-1);
            num++;
            t+=b;
        }
        if (num > now-a)
            break;
    }
    int len =ans.size();
    printf("%d\n",len);
    for (int i = 0;i <= len-2;i++)
        cout << ans[i] << " ";
    if (len>0)
        cout << ans[len-1]<<endl;
    return 0;
}