Ultimate Weirdness of an Array

寫不出來， 日常好菜啊。。

考慮枚舉GCD， 算出一共有多少個對 f(l, r) <= GCD， 我們用fuc[ i ] 表示的是在 l = i 這個位置開始, 最小的合法的 R，?

可以發現這個函數隨 i 單調不下降， 枚舉GCD 的時候， 找到GCD 的倍數的位置， 用線段樹更新最大值。

#include<bits/stdc++.h> #define LL long long #define LD long double #define ull unsigned long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ(x) ((int)x.size()) #define ALL(x) (x).begin(), (x).end() #define fio ios::sync_with_stdio(false); cin.tie(0);using namespace std;const int N = 2e5 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 998244353; const double eps = 1e-8; const double PI = acos(-1);template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;} template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;} template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;} template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}int n, a[N]; int maxPos[N][2]; int minPos[N][2]; int cnt[N]; LL H[N]; int mx[2], mn[2], tot;#define lson l, mid, rt << 1 #define rson mid + 1, r, rt << 1 | 1 struct segmentTree {LL sum[N << 2]; int lazy[N << 2], mn[N << 2], mx[N << 2];inline void pull(int rt) {sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];mx[rt] = max(mx[rt << 1], mx[rt << 1 | 1]);mn[rt] = min(mn[rt << 1], mn[rt << 1 | 1]);}inline void push(int rt, int l, int r) {if(~lazy[rt]) {int mid = l + r >> 1;sum[rt << 1] = 1LL * (mid - l + 1) * lazy[rt];sum[rt << 1 | 1] = 1LL * (r - mid) * lazy[rt];lazy[rt << 1] = lazy[rt << 1 | 1] = lazy[rt];mx[rt << 1] = mn[rt << 1] = lazy[rt];mx[rt << 1 | 1] = mn[rt << 1 | 1] = lazy[rt];lazy[rt] = -1;}}void build(int l, int r, int rt) {lazy[rt] = -1;if(l == r) {sum[rt] = l;mx[rt] = l;mn[rt] = l;return;}int mid = l + r >> 1;build(lson); build(rson);pull(rt);}void update(int L, int R, int val, int l, int r, int rt) {if(R < l || r < L || R < L) return;if(mn[rt] >= val) return;if(L <= l && r <= R && mx[rt] <= val) {sum[rt] = 1LL * (r - l + 1) * val;mx[rt] = val;mn[rt] = val;lazy[rt] = val;return;}if(l == r) return;push(rt, l, r);int mid = l + r >> 1;update(L, R, val, lson);update(L, R, val, rson);pull(rt);} } Tree;int main() {memset(maxPos, 0xc0, sizeof(maxPos));memset(minPos, 0x3f, sizeof(minPos));scanf("%d", &n);for(int i = 1; i <= n; i++) {scanf("%d", &a[i]);cnt[a[i]]++;if(i >= maxPos[a[i]][0]) {maxPos[a[i]][1] = maxPos[a[i]][0];maxPos[a[i]][0] = i;} else if(i > maxPos[a[i]][1]) maxPos[a[i]][1] = i;if(i <= minPos[a[i]][0]) {minPos[a[i]][1] = minPos[a[i]][0];minPos[a[i]][0] = i;} else if(i < minPos[a[i]][1]) minPos[a[i]][1] = i;}Tree.build(1, n, 1);for(int v = 200000; v >= 0; v--) {H[v] = 1LL * n * n - Tree.sum[1] + n;if(!v) break;mx[0] = mx[1] = -inf - 1;mn[0] = mn[1] = inf;tot = 0;for(int w = v; w <= 200000; w += v) {if(maxPos[w][0] == -inf - 1) continue;if(maxPos[w][0] >= mx[0]) mx[1] = mx[0], mx[0] = maxPos[w][0];else if(maxPos[w][0] > mx[1]) mx[1] = maxPos[w][0];if(maxPos[w][1] >= mx[0]) mx[1] = mx[0], mx[0] = maxPos[w][1];else if(maxPos[w][1] > mx[1]) mx[1] = maxPos[w][1];if(minPos[w][0] <= mn[0]) mn[1] = mn[0], mn[0] = minPos[w][0];else if(minPos[w][0] < mn[1]) mn[1] = minPos[w][0];if(minPos[w][1] <= mn[0]) mn[1] = mn[0], mn[0] = minPos[w][1];else if(minPos[w][1] < mn[1]) mn[1] = minPos[w][1];tot += cnt[w];}if(tot == 2) {int p1 = mn[0], p2 = mn[1];Tree.update(1, p1, p1, 1, n, 1);Tree.update(p1 + 1, p2, p2, 1, n, 1);Tree.update(p2 + 1, n, n + 1, 1, n, 1);} else if(tot == 3) {int p1 = mn[0], p2 = mn[1], p3 = mx[0];Tree.update(1, p1, p2, 1, n, 1);Tree.update(p1 + 1, p2, p3, 1, n, 1);Tree.update(p2 + 1, n, n + 1, 1, n, 1);} else if(tot > 3){int p1 = mn[0], p2 = mn[1], p3 = mx[1], p4 = mx[0];Tree.update(p2 + 1, n, n + 1, 1, n, 1);Tree.update(p1 + 1, p2, p4, 1, n, 1);Tree.update(1, p1, p3, 1, n, 1);}}LL ans = 0;for(int i = 1; i <= 200000; i++)ans += (H[i] - H[i - 1]) * i;printf("%lld\n", ans);return 0; }/* */

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轉載于:https://www.cnblogs.com/CJLHY/p/10978952.html

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