Hard Process(二分)

Hard Process

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 660C

Description

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 a_i (0 ≤ a_i ≤ 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 a_j — the elements of the array a after the changes.

If there are multiple answers, you can print any one of them.

Sample Input

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
题解：让改变k个数，使序列连续1的个数最大，我们可以求0的前缀和，然后通过二分来找位置；我竟然用暴力超时了好长时间。。。
二分：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int INF = 0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
const int MAXN = ;
int num[MAXN];
int a[MAXN];
int L, n, k;
 
int js(int x){
    for(int i = ; i + x <= n; i++){
        if(num[i + x] - num[i] <= k){
            L = i + ;
            return true;
        }
    }
    return false;
}
int erfen(int l, int r){
    int mid, ans;
    while(l <= r){
        mid = (l + r) >> ;
        if(js(mid)){
            ans = mid;
            l = mid + ;
        }
        else
            r = mid - ;
    }
    return ans;
}
int main(){
    while(~scanf("%d%d",&n, &k)){
        int temp;
        memset(num, , sizeof(num));
        for(int i = ; i <= n; i++){
            scanf("%d", a + i);
            num[i] = num[i - ] + (a[i] == );
        }
        int ans = erfen(,n);
        for(int i = L; i < L + ans; i++){
            a[i] = ;
        }
        printf("%d\n", ans);
        for(int i = ; i <= n; i++){
            if(i != )printf(" ");
            printf("%d",a[i]);
        }puts("");
    }
    return ;
}

暴力超时：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int INF = 0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
const int MAXN = ;
int num[MAXN];
int pos[MAXN];
int p[MAXN];
int main(){
    int n, k;
    while(~scanf("%d%d",&n, &k)){
        int gg = ;
        for(int i = ; i < n; i++){
            scanf("%d", num + i);
            if(num[i] == )gg = ;
        }
        if(!gg){
            printf("%d\n",k);
            for(int i = ; i < k; i++){
                if(i)printf(" ");
                printf("");
            }
            for(int i = k; i < n; i++){
                if(i)printf(" ");
                printf("");
            }puts("");
            continue;
        }
        int kg = , ans = , temp = , cnt = , tp = ;
        for(int i = ; i < n; i++){
            if(kg ==  && num[i] == ){
                    temp = ;
                    kg = ;
                    cnt = ;
                    for(int j = i; j < n; j++){
 
                        if(num[j] == ){
                            temp++;
                        }
                        else{
                            if(cnt +  > k)break;
                            pos[cnt++] = j;
                            temp++;
                        }
                    }
                        int j = i;
                        while(cnt < k && j > ){
                            pos[cnt++] = --j;
                            temp++;
                        }
 
                    if(ans < temp){
                        ans = temp;
                        tp = cnt;
                        for(int j = ; j < tp; j++){
                            p[j] = pos[j];
                        }
                    }
                }
            if(num[i] == )kg = ;
        }
        for(int i = ; i < tp; i++){
        //    printf("%d ",p[i]);
            num[p[i]] = ;
        }//puts("");
        printf("%d\n", ans);
        for(int i = ; i < n; i++){
            if(i)printf(" ");
            printf("%d",num[i]);
        }
        puts("");
    }
    return ;
}