【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\n", ans);    return 0;}