题目传送门

死因：不认真读题

首先，我们可以想到暴力枚举每段长度为\(k\)的区间，对于这一段区间，求将它变为同一高度的最小操作次数。显然，当我们取区间的中位数时，这段区间变为同一高度的次数最小。

操作次数为：
对于大于中位数的数，求和，减去中位数乘它们的个数
对于小于等于中位数的数，中位数乘它们的个数减去它们的和
最后两项求和即可

所以我们需要支持的操作有：

1. 求这段区间的中位数
2. 求小于中位数的数的个数
3. 求小于中位数的数的和

可以使用平衡树，在这里我是用的是权值线段树，比较方便。

最后要注意的是，高度可以为0

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#define LL long long
#define ls p << 1
#define rs p << 1 | 1
#define mid ((l + r) >> 1)
using namespace std;
LL read() {LL k = 0, f = 1; char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9')k = k * 10 + c - 48, c = getchar();return k * f;
}
LL a[100010], n, k, maxn, sum[100010];
struct zzz {LL cnt[1000010 << 2]; LL sum[1000010 << 2];void up(LL p) {cnt[p] = cnt[ls] + cnt[rs];sum[p] = sum[ls] + sum[rs];}void insert(LL x, LL p = 1, LL l = 0, LL r = maxn) {if(l == r) {++cnt[p]; sum[p] += l;//cout << cnt[p] << ' ' << l << endl;return ;}if(x <= mid) insert(x, ls, l, mid);else insert(x, rs, mid+1, r);up(p);}void earse(LL x, LL p = 1, LL l = 0, LL r = maxn) {if(l == r) {--cnt[p]; sum[p] -= l; return ;}if(x <= mid) earse(x, ls, l, mid);else earse(x, rs, mid+1, r);up(p);}LL k_th(LL x, LL p = 1, LL l = 0, LL r = maxn) {if(l == r) return l;if(x <= cnt[ls]) return k_th(x, ls, l, mid);else return k_th(x - cnt[ls], rs, mid+1, r);}LL summ(LL nr, LL p = 1, LL l = 0, LL r = maxn, LL nl = 0) {LL ans = 0;if(nl <= l && nr >= r) return sum[p];if(nl <= mid) ans += summ(nr, ls, l, mid, nl);if(nr > mid) ans += summ(nr, rs, mid+1, r, nl);return ans;}LL numm(LL nr, LL p = 1, LL l = 0, LL r = maxn, LL nl = 0) {LL ans = 0;if(nl <= l && nr >= r) return cnt[p];if(nl <= mid) ans += numm(nr, ls, l, mid, nl);if(nr > mid) ans += numm(nr, rs, mid+1, r, nl);return ans;}
}tree;
int main() {n = read(), k = read();LL anss = -1, pos, num;for(LL i = 1; i <= n; ++i) a[i] = read(), maxn = max(a[i], maxn), sum[i] = a[i] + sum[i-1];for(LL i = 1; i <= n; ++i) {tree.insert(a[i]);if(i > k) tree.earse(a[i-k]);if(i >= k) {LL x = tree.k_th((k+1) >> 1);LL y = tree.summ(x), summ = sum[i] - sum[i-k], numl = tree.numm(x);//cout << y << ' ' << numl << ' ' << summ << endl;if(summ-y-x*(k-numl)+x*numl-y < anss || anss == -1) {anss = summ-y-x*(k-numl)+x*numl-y, pos = i; num = x;}}}printf("%lld\n", anss);for(LL i = 1; i <= n; ++i) {if(i >= pos-k+1 && i <= pos) {printf("%lld\n", num);}else printf("%lld\n", a[i]);}return 0;
}