[尺取法] CF660 C.Hard Process

C. Hard Process
题意：
给一个01串，最多可以把k个0变成1，求最长连续1的长度，并且要输出最后的串。
题解：
two pointers尺取，$O(N)$。

#include<bits/stdc++.h>
using namespace std;
const int N = 3e5+5;
int num[N];
int main(){
    int n, k;
    scanf("%d%d", &n, &k);
    for(int i = 1; i <= n; ++i) scanf("%d", num+i);
    int l = 1, r = 1, zr = 0, ans = 0, ansl = 1, ansr = 0;
    while(r <= n){
        while(r <= n && zr <= k){
            if(!num[r]){
                if(zr == k) break;
                else ++zr;
            }
            ++r;
        }
        if(r-l > ans) ans = r-l, ansl = l, ansr = r-1;
        while(l <= n && num[l]) ++l;
        --zr, ++l;
    }
    printf("%d\n", ans);
    for(int i = 1; i <= n; ++i){
        if(ansl <= i && i <= ansr) printf("1 ");
        else printf("%d ", num[i]);
    }
}