Codeforces 660C Hard Process[2점]

제목 링크: 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
제목: 최대 k 개 0 을 수정하고 가장 긴 연속 1 열을 물어보세요.
2분이면 됩니다.
AC 코드:

#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
", 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("
");
    }
    return 0;
}