Educational Codeforces Round 11 C. Hard Process 二分

C. Hard Process

题目连接：

http://www.codeforces.com/contest/660/problem/C

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

7 1
1 0 0 1 1 0 1

Sample Output

4
1 0 0 1 1 1 1

Hint

题意

你有n个非0就是1的数字，你可以修改最多k个，使得0变成1

然后问你修改之后，最长的连续1的串是多长？

题解：

维护一个前缀0的个数

然后对于每个位置，直接暴力二分就好了，二分这个位置最远能够延展到哪儿

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6;
int n,k;
int a[maxn],sum[maxn];
int main()
{
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        a[i]=1-a[i];
    }
    for(int i=1;i<=n;i++)
        sum[i]=sum[i-1]+a[i];
    int ans1=0,ans2=0;
    for(int i=1;i<=n;i++)
    {
        int l = i,r = n,ans=0;
        while(l<=r)
        {
            int mid=(l+r)/2;
            if(sum[mid]-sum[i-1]>k)r=mid-1;
            else l=mid+1,ans=mid-i+1;
        }
        if(ans>ans1)
        {
            ans1=ans;
            ans2=i;
        }
    }
    cout<<ans1<<endl;
    for(int i=ans2;i<=n;i++)
    {
        if(ans1==0)break;
        if(a[i]==1)a[i]=0;
        if(a[i]==0)ans1--;
    }
    for(int i=1;i<=n;i++)
        cout<<1-a[i]<<" ";
}