Codeforces 660C Hard Process 【二分】

题目链接：Codeforces 660C Hard Process

C. Hard Process

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

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

题意：让你修改最多k个0，问最长连续的1串。

二分即可。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
#include <map>
#define ll o<<1
#define rr o<<1|1
#define CLR(a, b) memset(a, (b), sizeof(a))
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int INF = 0x3f3f3f3f;
const int MAXN = 3*1e5 + 10;
int sum[MAXN], a[MAXN];
int L, n, k;
bool judge(int o) {
for(int i = 1; i <= n; i++) {
int R = i + o - 1;
if(R > n) break;
if(sum[R] - sum[i-1] <= k) {
L = i;
return true;
}
}
return false;
}
int main()
{
while(scanf("%d%d", &n, &k) != EOF) {
sum[0] = 0;
for(int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
sum[i] = sum[i-1] + (a[i] == 0);
}
int l = 1, r = n; int ans = 0;
while(r >= l) {
int mid = (l + r) >> 1;
if(judge(mid)) {
ans = mid;
l = mid + 1;
}
else {
r = mid - 1;
}
}
printf("%d\n", ans);
for(int i = L; i <= L + ans - 1; i++) {
a[i] = 1;
}
for(int i = 1; i <= n; i++) {
if(i > 1) printf(" ");
printf("%d", a[i]);
}
printf("\n");
}
return 0;
}