codeforces 660C C. Hard Process(二分)

题目链接：C. Hard Processtime limit per test
1 secondmemory limit per test
256 megabytesinput
standard inputoutput
standard outputYou 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.Examplesinput

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

题意：

最多把k个0变成1,变完后连续的最长的全是1的串的长度是多少,并且输出最后得串;

思路：

用一个数组记录当前位一共有多少个0,暴力枚举最长串的最后一位,二分查找最长串的第一个1的位置;更新结果并记录好最长串的开始和结束位置,最后再输出就好啦;

AC代码：

/*
2014300227    660C - 5    GNU C++11    Accepted    93 ms    4388 KB
*/
#include <bits/stdc++.h>
using namespace std;
const int N=3e5+4;
typedef long long ll;
const double PI=acos(-1.0);
int n,a
,k,b
;
int check(int x,int y)
{
if(b[y]-b[x-1]<=k)return 1;
return 0;
}
int bis(int num)
{
int l=1,r=num,mid;
while(l<=r)
{
mid=(l+r)>>1;
if(check(mid,num))r=mid-1;
else l=mid+1;
}
return r+1;
}
int main()
{
scanf("%d%d",&n,&k);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
if(!a[i])b[i]=b[i-1]+1;
else b[i]=b[i-1];
}
int ans=0,l=0,r=0;
for(int i=1;i<=n;i++)
{
int fx=bis(i);
if(i-fx+1>ans)
{
ans=i-fx+1;
l=fx;
r=i;
}
}
printf("%d\n",ans);
for(int i=1;i<=n;i++)
{
if(i>=l&&i<=r)
{
printf("1 ");
}
else printf("%d ",a[i]);
}
return 0;
}