cf-12B Correct Solution[水] 660C Hard Process[尺取法]

12B-Correct Solution

Description

One cold winter evening Alice and her older brother Bob was sitting at home near the fireplace and giving each other interesting problems to solve. When it was Alice’s turn, she told the number n to Bob and said:

—Shuffle the digits in this number in order to obtain the smallest possible number without leading zeroes.

—No problem! — said Bob and immediately gave her an answer.

Alice said a random number, so she doesn’t know whether Bob’s answer is correct. Help her to find this out, because impatient brother is waiting for the verdict.

Input

The first line contains one integer n (0 ≤ n ≤ 109) without leading zeroes. The second lines contains one integer m (0 ≤ m ≤ 109) — Bob’s answer, possibly with leading zeroes.

Output

Print OK if Bob’s answer is correct and WRONG_ANSWER otherwise.

Sample Input 
Input

3310 
1033

Output

OK

Input

4 
5

Output

WRONG_ANSWER 
sort一下之後直接找第一個0和第一個非0就OK了

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
char a[10010];
char b[10010], c[1010], d[1010];
int main() {
    while (scanf("%s%s", a, b) != EOF) {
            int len = strlen(a);
    sort(a, a + len);
    int flag = 0;
    for (int i = 0; i < len; i++) {
            if (a[i] != '0') {
                flag = i;
    char t = a[i];
    a[i] = a[0];
    a[0] = t;
    break;
    }
    }
    if (strcmp(a, b) == 0) {
            printf("OK\n");
    }
    else
        {
            printf("WRONG_ANSWER\n");
            }
}
return 0;
}

 660C Hard Process

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.

Example

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

#include<iostream>
#include<cstdio> 
#include<cstdlib>
#include<cstring>
#include<algorithm> 
using namespace std;
const int maxn=300005;
int n,k;
int a[maxn],sum[maxn]={0};
int main(){
	cin>>n>>k;
	for(int i=1;i<=n;i++){//10010  01223  5 1
						 // 01101  01223  mid=3
		cin>>a[i];
		
		if(a[i]==1){
			sum[i]=sum[i-1];
//			ans=max(ans,sum[i]);
		}
		else{
			sum[i]=sum[i-1]+1;
		}
		a[i]=1-a[i];
	}
//	for(int i=1;i<=n;i++){
//		sum[i]=sum[i-1]+a[i];
//	}
	int ans=0,res=0;
	for(int i=1;i<=n;i++){
		int L=i,R=n;
		int temp=0;
		while(L<=R){
			int mid=(L+R)/2;
			if(sum[mid]-sum[i-1]-1>k-1){
				R=mid-1;
			}
			else{
				L=mid+1;
				temp=mid-i+1;
			}
		} 
		if(temp>ans){
			ans=temp;
			res=i;
		//	cout<<"ans="<<ans<<"  "<<"res="<<res<<endl;
		}
	}
	cout<<ans<<endl;
	for(int i=res;i<=n;i++){
		if(!ans) break;
		if(a[i]==1) a[i]=0;
		if(!a[i]) ans--;
		//cout<<"ans--="<<ans<<endl;//10010  01101 00101
	}
	for(int i=1;i<=n;i++){
		cout<<1-a[i]<<" ";
	}
	cout<<endl;
	return 0;
}