题干：
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 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
题目大意：
给定一个数组 a，含有 n 个元素。数组 a 中的每个元素要么是 0，要么是 1 。输入一个n，一个k，然后第二行是数组。
让我们假定，数组 a 中，仅由数字 1 组成的连续元素所构成的子分段，其最长的长度为 f(a)。您可以将不超过 k 个 0 更改为 1，使得 f(a) 最大化。
解题报告：
这题做法多样，可以nlogn二分长度然后遍历。还有一种方法是o(n)尺取，但是感觉难写一点？
AC代码：
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#define ll long long
using namespace std;
const int MAX = 3e5 + 5;
int n,k;
int a[MAX],sum[MAX];
bool ok(int len) {
for(int i = 1; i+len-1<=n; i++) {
if(sum[i+len-1] - sum[i-1] >= len-k) return 1;
}
return 0;
}
int main()
{
cin>>n>>k;
for(int i = 1; i<=n; i++) scanf("%d",a+i),sum[i] = sum[i-1] + a[i];
int l = 0,r = n;
int mid = (l+r)/2;
while(l<r) {
mid = (l+r+1)/2;
if(ok(mid)) l=mid;
else r=mid-1;
}
printf("%d\n",l);
for(int i = 1; i+l-1<=n; i++) {
if(sum[i+l-1] - sum[i-1] >= l-k) {
for(int j = 1; j<=n; j++) {
if(j < i || j > i+l-1) printf("%d",a[j]);
else printf("1");
if(j != n) putchar(' ');
}
return 0 ;
}
}
return 0 ;
} 总结：
注意这题是二分求最大值，所以需要mid=(l+r+1)/2调节一下。最后还是l是我们需要的。