【UOJ #246】【UER #7】套路

http://uoj.ac/contest/35/problem/246
神奇！我这辈子是想不出这样的算法了。
对区间长度分类讨论：题解很好的~
我已经弱到爆了，看完题解后还想了一晚上。
题解中“利用(r_y)进行计算更新答案”的具体方法是记录以当前点为右端点，任意两个数的差值的最小值大于等于j的区间的左端点，记为(pos_j)。
就这个问题我想了一晚上啊TWT，我不滚粗谁滚粗QAQ

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 200003;
int in() {
	int k = 0; char c = getchar();
	for (; c < '0' || c > '9'; c = getchar());
	for (; c >= '0' && c <= '9'; c = getchar())
		k = k * 10 + c - 48;
	return k;
}

int S, n, m, k, a[N], ans = 0, f[N], up, last[N], pos[N];

void solve_1() {
	for (int i = 1; i < n; ++i) {
		f[i] = abs(a[i + 1] - a[i]);
		if (k <= 2) ans = max(ans, f[i]);
	}
	
	for (int p = 3; p <= S; ++p)
		for (int i = 1; i + p - 1 <= n; ++i) {
			f[i] = min(abs(a[i + p - 1] - a[i]), min(f[i], f[i + 1]));
			if (k <= p) ans = max(ans, f[i] * (p - 1));
		}
}

void solve_2() {
	int lo, bi;
	for (int i = 1; i <= n; ++i) {
		pos[0] = max(pos[0], last[a[i]]);
		for (int j = 1; j <= up; ++j) {
			pos[j] = max(pos[j], pos[j - 1]);
			lo = a[i] - j;
			bi = a[i] + j;
			if (lo >= 1) pos[j] = max(pos[j], last[lo]);
			if (bi <= m) pos[j] = max(pos[j], last[bi]);
			if (i - pos[j - 1] >= k)
				ans = max(ans, j * (i - (pos[j - 1] + 1)));
		}
		last[a[i]] = i;
	}
}

int main() {
	n = in(); m = in(); k = in();
	for (int i = 1; i <= n; ++i)
		a[i] = in();
	S = ceil(sqrt(n));
	
	solve_1();
	
	up = m / S;
	solve_2();
	
	printf("%d
", ans);
	return 0;
}