Educational Codeforces Round 11C. Hard Process two pointer

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

题目：

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.

Examples

input

7 11 0 0 1 1 0 1

output

41 0 0 1 1 1 1

input

10 21 0 0 1 0 1 0 1 0 1

output

51 0 0 1 1 1 1 1 0 1

思路：一开始我用的是n^2的算法，一直tle，后来才知道有种算法叫尺取法：就是动态维护一个长度为x的区间，并同时记录起始位置和终点位置。

　　对这题而言，就是维护含0数为k的0,1区间，记录长度最大值，和起始位置和终点位置;

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cstdio>
 4 #include <cmath>
 5 #include <cstring>
 6 #include <queue>
 7 #include <stack>
 8 #include <map>
 9 #include <vector>
10
11 #define PI acos((double)-1)
12 #define E exp(double(1))
13 using namespace std;
14 int a[1000000];
15 int main (void)
16 {
17     int n,k,s=0,e=0,sum=0,len=0;
18     cin>>n>>k;
19     for(int i = 1; i<=n; i++)
20         {
21             scanf("%d",&a[i]);
22             sum += (a[i] == 0);
23             while(sum > k)
24                 {
25                     sum -= (a[++s] == 0);
26                 }
27             if(len < i - s)
28                 {
29                     e = i;
30                     len = i - s;
31                 }
32         }
33     cout<<len<<endl;
34     for(int i = 1; i<=n; i++)
35         if(e>= i && i> e - len )
36             {
37                     printf("1 ");
38             }
39         else
40             {
41                     printf("%d ",a[i]);
42             }
43     return 0;
44 }

View Code

转载于:https://www.cnblogs.com/weeping/p/5371900.html

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