Hard Process 没找到原题。。。

M - Hard Process
Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u
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Status
Description
You are given an array a with n elements. Each element of a is either 0 or 1.

Let's denote the length of the longest subsegment of consecutive elements in a, consisting of only numbers one, as f(a). You can change no more than k zeroes to ones to maximize f(a).

Input
The first line contains two integers n and k (1 ≤ n ≤ 3·105, 0 ≤ k ≤ n) — the number of elements in a and the parameter k.

The second line contains n integers ai (0 ≤ ai ≤ 1) — the elements of a.

Output
On the first line print a non-negative integer z — the maximal value of f(a) after no more than k changes of zeroes to ones.

On the second line print n integers aj — the elements of the array a after the changes.

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

Sample Input
Input
7 1
1 0 0 1 1 0 1
Output
4
1 0 0 1 1 1 1
Input
10 2
1 0 0 1 0 1 0 1 0 1
Output
5

1 0 0 1 1 1 1 1 0 1

模拟，，，，，

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,k,ks,sum,ss,lp,ji;
int shu[400000];
int main()
{
	int i,jj;
	while (~scanf("%d%d",&n,&k))
	{
		memset(shu,0,sizeof(shu));
		for (i=1;i<=n;i++)
		scanf("%d",&shu[i]);
		if (k==0)
		{
			sum=0,ss=0;
			for (i=1;i<=n;i++)
			{
				if (shu[i])
				sum++;
				else
				sum=0;
				ss=max(ss,sum);	
			}
			printf("%d\n%d",ss,shu[1]);
			for (i=2;i<=n;i++)
			printf(" %d",shu[i]);
			printf("\n");
			continue;
		}
		ks=0;sum=0;ji=1;lp=0;ss=0;
		for (i=1;i<=n;i++)
		{
			if (shu[i]==0)
			{
				if (ks)
				{
					if (ks==k)
					{
						for (jj=ji;jj<=i;jj++)
						{
							sum--;
							if (shu[jj]==0)
							{
								ji=jj+1;
								ks--;
								break;
							}
						}
						sum++;
						ks++;
		//				printf("%d    %d    %d   %d   99\n",i,jj,ji,sum);
					}
					else
					{
					
						ks++;
						sum++;
					}
				}
				else
				{
					lp=i;
					ks=1;
					sum++;
				}	
			}
			else
			sum++;
			if (sum>ss)
			{
		//		printf("%d  %d   66\n",ji,i);
				lp=ji;
				ss=sum;
			}
		}
		if (lp)
		{   
		//    printf("guo   %d\n",lp);
			for (i=lp;i<=n;i++)
			{
				if (shu[i]==0)
			    {
				    shu[i]=1;
				    k--;
				}
				if (k==0)
				break;
			}
		}
		printf("%d\n%d",ss,shu[1]);
		for (i=2;i<=n;i++)
		printf(" %d",shu[i]);
		printf("\n");
	}
	return 0;
}