线段树裸题。
data generate by cyaron.

参考代码：

// Solution from luoguP3373
#include <bits/stdc++.h>
#define int long long
#define ll long long
using namespace std;
int n, m, a[1000005], mod;

struct node {
    ll sum, l, r, mu, add;
} t[1000005];
ll read() {
    ll x = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9')
        ch = getchar();
    while (ch >= '0' && ch <= '9')
        x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
    return x;
}
void build(ll p, ll l, ll r) {
    t[p].l = l, t[p].r = r;
    t[p].mu = 1;
    if (l == r) {
        t[p].sum = a[l] % mod;
        return;
    }
    ll mid = (l + r) >> 1;
    build(p * 2, l, mid);
    build(p * 2 + 1, mid + 1, r);
    t[p].sum = (t[p * 2].sum + t[p * 2 + 1].sum) % mod;
}
void spread(ll p) {
    t[p * 2].sum = (ll)(t[p].mu * t[p * 2].sum +
                                            ((t[p * 2].r - t[p * 2].l + 1) * t[p].add) % mod) %
                                 mod;
    t[p * 2 + 1].sum =
            (ll)(t[p].mu * t[p * 2 + 1].sum +
                     (t[p].add * (t[p * 2 + 1].r - t[p * 2 + 1].l + 1)) % mod) %
            mod; // add已经乘过mu啦

    t[p * 2].mu = (ll)(t[p * 2].mu * t[p].mu) % mod;
    t[p * 2 + 1].mu = (ll)(t[p * 2 + 1].mu * t[p].mu) % mod;

    t[p * 2].add = (ll)(t[p * 2].add * t[p].mu + t[p].add) % mod;
    t[p * 2 + 1].add = (ll)(t[p * 2 + 1].add * t[p].mu + t[p].add) % mod;

    t[p].mu = 1, t[p].add = 0;
}
void add(ll p, ll l, ll r, ll k) {
    if (t[p].l >= l && t[p].r <= r) {
        t[p].add = (t[p].add + k) % mod;
        t[p].sum =
                (ll)(t[p].sum + k * (t[p].r - t[p].l + 1)) % mod; // 只要加上增加的就好
        return;
    }
    spread(p);
    t[p].sum = (t[p * 2].sum + t[p * 2 + 1].sum) % mod;
    ll mid = (t[p].l + t[p].r) >> 1;
    if (l <= mid)
        add(p * 2, l, r, k);
    if (mid < r)
        add(p * 2 + 1, l, r, k);
    t[p].sum = (t[p * 2].sum + t[p * 2 + 1].sum) % mod;
}
void mu(ll p, ll l, ll r, ll k) {
    if (t[p].l >= l && t[p].r <= r) {
        t[p].add =
                (t[p].add * k) %
                mod; // 比较重要的一步,add要在这里乘上k,因为后面可能要加其他的数而那些数其实是不用乘k的
        t[p].mu = (t[p].mu * k) % mod;
        t[p].sum = (t[p].sum * k) % mod;
        return;
    }
    spread(p);
    t[p].sum = t[p * 2].sum + t[p * 2 + 1].sum;
    ll mid = (t[p].l + t[p].r) >> 1;
    if (l <= mid)
        mu(p * 2, l, r, k);
    if (mid < r)
        mu(p * 2 + 1, l, r, k);
    t[p].sum = (t[p * 2].sum + t[p * 2 + 1].sum) % mod;
}
ll ask(ll p, ll l, ll r) {
    if (t[p].l >= l && t[p].r <= r) {
        return t[p].sum;
    }
    spread(p);
    ll val = 0;
    ll mid = (t[p].l + t[p].r) >> 1;
    if (l <= mid)
        val = (val + ask(p * 2, l, r)) % mod;
    if (mid < r)
        val = (val + ask(p * 2 + 1, l, r)) % mod;
    return val;
}
signed main() {
    // freopen(".in", "r", stdin);
    // freopen(".out", "w", stdout);
    cin >> n >> m; mod = 998244353;
    for (int i = 1; i <= n; i++) {
        a[i] = read();
    }
    build(1, 1, n);
    for (int i = 1; i <= m; i++) {
        int ty = read();
        if (ty == 1) {
            ll cn = read(), cm = read(), cw = read();
            mu(1, cn, cm, cw);
        } else if (ty == 2) {
            ll cn = read(), cm = read(), cw = read();
            add(1, cn, cm, cw);
        } else {
            ll cn = read(), cm = read();
            cout << ask(1, cn, cm) << endl;
        }
    }
}

generater:

from cyaron import *

_n = ati([8, 8, 8, 8, 8, 8, 1e3, 1e3, 1e3, 1e3, 1e3, 1e3, 1e3, 1e3, 1e5, 1e5, 1e5, 1e5, 1e5, 1e5])
_m = ati([10, 10, 10, 10, 10, 10, 1e4, 1e4, 1e4, 1e4, 1e4, 1e4, 1e4, 1e4, 1e5, 1e5, 1e5, 1e5, 1e5, 1e5])

print(len(_n), len(_m))

for i in range(0, 20):
    test_data = IO(file_prefix = "seq", data_id = i + 1)

    print(i)
    n = _n[i]
    m = _m[i]
    a = []
    q = []
    for j in range(0, n):
        a.append(randint(1, 1e9))
    for j in range(0, m):
        nowques = []
        nowques.append(randint(1, 3))
        left = randint(1, n - 1)
        right = randint(left, n)
        nowques.append(left)
        nowques.append(right)
        if nowques[0] == 1 or nowques[0] == 2:
            k = randint(1, 1e9)
            nowques.append(k)
        q.append(nowques)

    # output
    test_data.input_writeln(n, m)
    for j in range(0, n):
        test_data.input_write(a[j], "")
    test_data.input_writeln("")
    for j in range(0, m):
        test_data.input_writeln(q[j])

    test_data.output_gen("D:\\Robot\\OI\\Code\\Exercise\\23118\\P1006\\P1006.exe")