【二分】Hard Process
題目連結 --> https://codeforces.com/contest/660/problem/C
題目描述
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
題目大意
- 給定一個 n 表示01陣列的長度
- 改變其中的不超過k個0
- 使得這個序列中連續的1的長度達到最長
- 輸出最長長度並輸出改變後的序列
思路及程式碼實現
使用二分的思想，列舉每個可能的左端點，二分右端點，逐個比較,直到找到最長的連續的1的序列，並不斷記錄更新最長連續序列的起點下標和終點下標C++程式碼實現
#include <iostream>
#include <algorithm>
using namespace std;
typedef pair<int, int> PII;
const int N = 3e5 + 10;
int n, k;
int s[N], a[N];
int main(){
cin >> n >> k;
for (int i = 1; i <= n; i++){
scanf("%d", &a[i]);
s[i] = s[i - 1] + (a[i] == 0);
}
PII ans = {0, 0};
int res = 0;
// int l, r, mid;
for (int i = 1; i <= n; i++){
int l = i, r = n;
while (l <= r){
int mid = (l + r) >> 1;
if (s[mid] - s[i - 1] <= k){
if (mid - i + 1 > res){ //注意二分的邊界問題
res = mid - i + 1;
ans.first = i, ans.second = mid;
}
l = mid + 1;
}
else r = mid - 1;
}
}
cout << res << endl;
for (int i = 1; i <= n; i++){
if (i >= ans.first && i <= ans.second)
cout << 1 << ' ';
else
cout << a[i] << ' ';
}
cout << endl;
return 0;
}
/*
7 1
1 0 0 1 1 0 1
4
1 0 0 1 1 1 1
*/