题意

链接

Sol

势能分析。

除法是不能打标记的，所以只能暴力递归。这里我们加一个剪枝：如果区间内最大最小值的改变量都相同的话，就变成区间减。

这样复杂度是\((n + mlogn) logV\)的。

简单的证明一下：如果没有加的话，每个节点会被除至多log次, 总会除4nlogn次,每次区间加会恢复log个点的势能函数，这样总递归次数就是\(nlog^2n\)。

下传标记的时候别忘了把min和max都更新一下

#include<bits/stdc++.h> 
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define LL long long 
#define ull unsigned long long 
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int MAXN = 1e6 + 10, INF = 1e9 + 1;
const double eps = 1e-9, pi = acos(-1);
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, Q, a[MAXN];
LL mx[MAXN], mn[MAXN], add[MAXN], sum[MAXN], ll[MAXN], rr[MAXN];
#define ls k << 1
#define rs k << 1 | 1
void update(int k) {
    mx[k] = max(mx[ls], mx[rs]);
    mn[k] = min(mn[ls], mn[rs]);
    sum[k] = sum[ls] + sum[rs];
}
void ps(int k, int v) {
    sum[k] += (rr[k] - ll[k] + 1) * v;
    mn[k] += v; mx[k] += v;
    add[k] += v;
}
void pushdown(int k) {
    if(!add[k]) return ;
    ps(ls, add[k]); ps(rs, add[k]);
    add[k] = 0;
}
void Build(int k, int l, int r) {
    ll[k] = l; rr[k] = r;
    if(l == r) {sum[k] = mx[k] = mn[k] = a[l]; return ;}
    int mid = l + r >> 1;
    Build(ls, l, mid); Build(rs, mid + 1, r);
    update(k);
}
void Add(int k, int l, int r, int ql, int qr, LL v) {
    if(ql <= l && r <= qr) {ps(k, v); return ;}
    int mid = l + r >> 1;
    pushdown(k);
    if(ql <= mid) Add(ls, l, mid, ql, qr, v);
    if(qr  > mid) Add(rs, mid + 1, r, ql, qr, v);
    update(k);
}
LL get(LL x, int d) {
    return (x >= 0 ? x / d : (x - d + 1) / d);
}
void Div(int k, int l, int r, int ql, int qr, LL v) {
    if(ql <= l && r <= qr && (mx[k] - get(mx[k], v) == mn[k] - get(mn[k], v))) {
        ps(k, get(mx[k], v) - mx[k]);
        return ;
    }
    pushdown(k);
    int mid = l + r >> 1;
    if(ql <= mid) Div(ls, l, mid, ql, qr, v);
    if(qr  > mid) Div(rs, mid + 1, r, ql, qr, v);
    update(k);
}
LL Min(int k, int l, int r, int ql, LL qr) {
    if(ql <= l && r <= qr) return mn[k];
    int mid = l + r >> 1;
    pushdown(k);
    if(ql > mid) return Min(rs, mid + 1, r, ql, qr);
    else if(qr <= mid) return Min(ls, l, mid, ql, qr);
    else return min(Min(ls, l, mid, ql, qr), Min(rs, mid + 1, r, ql, qr));
}
LL Sum(int k, int l, int r, int ql, int qr) {
    if(ql <= l && r <= qr) return sum[k];
    int mid = l + r >> 1;
    pushdown(k);
    if(ql > mid) return Sum(rs, mid + 1, r, ql, qr);
    else if(qr <= mid) return Sum(ls, l, mid, ql, qr);
    else return Sum(ls, l, mid, ql, qr) + Sum(rs, mid + 1, r, ql, qr);
}
signed main() {
    N = read(); Q = read();
    for(int i = 1; i <= N; i++) a[i] = read();
    Build(1, 1, N);
    while(Q--) {
        int opt = read(), l = read() + 1, r = read() + 1;
        if(opt == 1) {
            int c = read();
            Add(1, 1, N, l, r, c);
        } else if(opt == 2) {
            int d = read();
            Div(1, 1, N, l, r, d);
        } else if(opt == 3) {
            cout << Min(1, 1, N, l, r) << '\n';
        } else {
            cout << Sum(1, 1, N, l, r) << '\n';
        }
    }
    return 0;
}