[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]);    }}