CF1677E Tokitsukaze and Beautiful Subsegments

好题。

对于区间最大值，考虑单调栈维护，记一个数 aia_iai 左右边第一个比他大的数分别为 LiL_iLi 和 RiR_iRi。

则对于区间 [l,r][l,r][l,r] 满足 Li<l≤r<RiL_i<l\leq r<R_iLi<l≤r<Ri，有 max⁡k=lrak=ai\max_{k=l}^{r}a_k=a_imaxk=lrak=ai。

对于每个 aia_iai，我们可以尝试确定区间左端点范围 [L,R][L,R][L,R]，初始 L=Li+1L=L_i+1L=Li+1，R=LiR=L_iR=Li。

接着，枚举 aia_iai 的因数 x,yx,yx,y，判断是否 Li<posx,posy≤iL_i<pos_x,pos_y \leq iLi<posx,posy≤i，即在范围内，然后更新 R=min⁡(posx,posy)R=\min(pos_x,pos_y)R=min(posx,posy)，总时间复杂度 O(∑ai)\mathcal O(\sum\sqrt{a_i})O(∑ai)。

然后枚举区间右端点，算出与它乘积为 aia_iai 的数的位置，判断是否在范围内，更新 RRR，记录。

这样就记录下了 (L,R,r)(L,R,r)(L,R,r)，表示右端点为 rrr，左端点为 [L,R][L,R][L,R] 的区间是美丽的。

考虑扫描线，将询问离线，右端点升序，把右端点为 iii 的所有三元组加入线段树，即区间加 111。

时间复杂度 O(n2log⁡n)\mathcal O(n^2\log n)O(n2logn)。

考虑优化，每次枚举 [i,Ri][i,R_i][i,Ri] 和 [Li,i][L_i,i][Li,i] 中长度小的，记录三元组，这样时间复杂度均摊 O(nlog⁡2n)\mathcal O(n\log^2n)O(nlog2n)。

#include<bits/stdc++.h>using namespace std;#define int long long
typedef long long ll;#define ha putchar(' ')
#define he putchar('\n')inline int read() {int x = 0, f = 1;char c = getchar();while (c < '0' || c > '9') {if (c == '-')f = -1;c = getchar();}while (c >= '0' && c <= '9')x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();return x * f;
}inline void write(ll x) {if (x < 0) {putchar('-');x = -x;}if (x > 9)write(x / 10);putchar(x % 10 + 48);
}const int _ = 1e6 + 10;int n, q, s[_], t, a[_], pos[_], L[_], R[_], sz[_ << 2], tag[_ << 2], tr[_ << 2], as[_];struct abc {int l, r, id;
} qr[_];bool cmp(abc a, abc b) {return a.r < b.r;
}bool cmp2(abc a, abc b) {return a.l > b.l;
}vector<pair<int, int>> d[_], d2[_];void build(int o, int l, int r) {sz[o] = (r - l + 1), tr[o] = tag[o] = 0;if (l == r) return;int mid = (l + r) >> 1;build(o << 1, l, mid), build(o << 1 | 1, mid + 1, r);
}void pushdown(int o) {if (tag[o]) {tr[o << 1] += sz[o << 1] * tag[o];tr[o << 1 | 1] += sz[o << 1 | 1] * tag[o];tag[o << 1] += tag[o];tag[o << 1 | 1] += tag[o];tag[o] = 0;}
}void upd(int o, int l, int r, int L, int R, int k) {if (L <= l && r <= R) {tr[o] += sz[o] * k, tag[o] += k;return;}pushdown(o);int mid = (l + r) >> 1;if (L <= mid) upd(o << 1, l, mid, L, R, k);if (R > mid) upd(o << 1 | 1, mid + 1, r, L, R, k);tr[o] = tr[o << 1] + tr[o << 1 | 1];
}int qry(int o, int l, int r, int L, int R) {if (L <= l && r <= R) return tr[o];pushdown(o);int mid = (l + r) >> 1, res = 0;if (L <= mid) res = qry(o << 1, l, mid, L, R);if (R > mid) res += qry(o << 1 | 1, mid + 1, r, L, R);tr[o] = tr[o << 1] + tr[o << 1 | 1];return res;
}signed main() {n = read(), q = read();for (int i = 1; i <= n; ++i) a[i] = read(), pos[a[i]] = i;t = 0;for (int i = 1; i <= n; ++i) {while (t > 0 && a[s[t]] < a[i]) t--;L[i] = s[t];s[++t] = i;}t = 0, s[0] = n + 1;for (int i = n; i >= 1; --i) {while (t > 0 && a[s[t]] < a[i]) t--;R[i] = s[t];s[++t] = i;}for (int i = 1; i <= q; ++i)qr[i].l = read(), qr[i].r = read(), qr[i].id = i;for (int i = 1; i <= n; ++i) {int l, r;if (R[i] - i <= i - L[i]) {r = L[i];for (int j = 1; j * j <= a[i]; ++j) {if (a[i] % j != 0) continue;int fx = pos[j], fy = pos[a[i] / j];if (fx == fy) continue;if (fx > fy) swap(fx, fy);if (fx > L[i] && fy <= i) r = max(r, fx);}for (int j = i; j < R[i]; ++j) {if (a[i] % a[j] == 0) {int nw = a[i] / a[j];if (pos[nw] == j) continue;if (pos[nw] > L[i] && pos[nw] < j) {r = max(r, pos[nw]);r = min(r, i);}}if (L[i] < r) d[j].push_back({L[i] + 1, r});}} else {l = R[i];for (int j = 1; j * j <= a[i]; ++j) {if (a[i] % j != 0) continue;int fx = pos[j], fy = pos[a[i] / j];if (fx == fy) continue;if (fx > fy) swap(fx, fy);if (fy < R[i] && fx >= i) l = min(l, fy);}for (int j = i; j > L[i]; --j) {if (a[i] % a[j] == 0) {int nw = a[i] / a[j];if (pos[nw] == j) continue;if (pos[nw] > j && pos[nw] < R[i]) {l = min(l, pos[nw]);l = max(l, i);}}if (l < R[i]) d2[j].push_back({l, R[i] - 1});}}}sort(qr + 1, qr + q + 1, cmp);build(1, 1, n);int k = 1;for (int i = 1; i <= n; ++i) {for (auto v : d[i]) upd(1, 1, n, v.first, v.second, 1);for (; qr[k].r == i && k <= q; ++k)as[qr[k].id] += qry(1, 1, n, qr[k].l, qr[k].r);}sort(qr + 1, qr + q + 1, cmp2);build(1, 1, n);k = 1;for (int i = n; i >= 1; --i) {for (auto v : d2[i]) upd(1, 1, n, v.first, v.second, 1);for (; qr[k].l == i && k <= q; ++k)as[qr[k].id] += qry(1, 1, n, qr[k].l, qr[k].r);}for (int i = 1; i <= q; ++i) write(as[i]), he;return 0;
}

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CF1677E Tokitsukaze and Beautiful Subsegments

CF1677E Tokitsukaze and Beautiful Subsegments

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