http://codeforces.com/contest/738/problem/D

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 bconsecutive 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.

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.

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.

5 1 2 100100

24 2

13 3 2 31000000010001

27 11

题意：海战棋游戏，长度为n的01串，1代表炸过且没有船的位置，0代表没有炸过的位置。有a个船，长度都是b，求打到一艘船至少还需要多少炸弹，并输出炸的位置。

分析：每连续的b个0就要炸一次，不然不知道有没有是不是刚好一艘船在这b个位置上面。贪心可知炸这b个的最后一个最划算。因为只要炸到一艘即可，所以答案减去a-1，即有a-1艘可以不管它。

代码：

#include<cstdio>
#define N 200005
int a,b,n,k,d,ans,p[N];
char s[N];
int main(){scanf("%d%d%d%d%s",&n,&a,&b,&k,s);for(int i=0;s[i];i++){if(s[i]=='0')d++;if(s[i]=='1')d=0;if(d==b){d=0;ans++;p[ans]=i+1;}}ans-=a-1;printf("%d\n",ans);for(int i=1;i<=ans;i++)printf("%d ",p[i]);return 0;
}