Ultimate Weirdness of an Array

写不出来， 日常好菜啊。。

考虑枚举GCD， 算出一共有多少个对 f(l, r) <= GCD， 我们用fuc[ i ] 表示的是在 l = i 这个位置开始, 最小的合法的 R， 

可以发现这个函数随 i 单调不下降， 枚举GCD 的时候， 找到GCD 的倍数的位置， 用线段树更新最大值。

#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;
}

/*
*/