题意： 稍加修改的线段树模板题。
题解： 由于乘法和加法两种修改，$lazy$标记不能确定的指出当前待修改的操作是乘法在前还是加法在前，亦或是两者混合存在，所以我们标记两个 $lazy:add$ 和 $mul$。更新的时候$add$更新为$add \times mul’+add’$，$mul$更新为$mul \times mul’+add’$，$sum$更新为$sum \times mul + add*(u.len)$

#include<cstdio>
#include<algorithm>
using namespace std;

typedef long long ll;

const int N = 1e5 + 10;
int n, m, q[N];
ll mod;

struct Node {
    int l, r;
    ll sum, add, mul;
}tr[N << 2];

void pushup(int u) {
    tr[u].sum = (tr[u << 1].sum + tr[u << 1 | 1].sum) % mod;
}

void eval(Node &t, ll add, ll mul) {
    t.sum = (t.sum * mul + (t.r - t.l + 1) * add) % mod;
    t.mul = t.mul * mul % mod;
    t.add = (t.add * mul + add) % mod;
}

void pushdown(int u) {
    eval(tr[u << 1], tr[u].add, tr[u].mul);
    eval(tr[u << 1 | 1], tr[u].add, tr[u].mul);
    tr[u].add = 0, tr[u].mul = 1;
}

void build(int u, int l, int r) {
    tr[u] = {l, r, 0, 0, 1};
    if(l == r) {
        tr[u].sum = q[l];
        return ;
    }

    int mid = l + r >> 1;
    build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
    pushup(u);
}

void modify(int u, int l, int r, ll add, ll mul) {
    if(tr[u].l >= l && tr[u].r <= r) {
        eval(tr[u], add, mul);
        return ;
    }

    pushdown(u);
    int mid = tr[u].l + tr[u].r >> 1;
    if(l <= mid) modify(u << 1, l, r, add, mul);
    if(r > mid) modify(u << 1 | 1, l, r, add, mul);
    pushup(u);
}

ll query(int u, int l, int r) {
    if(tr[u].l >= l && tr[u].r <= r) return tr[u].sum;

    pushdown(u);
    int mid = tr[u].l + tr[u].r >> 1;
    ll res = 0;

    if(l <= mid) res += query(u << 1, l, r);
    if(r > mid) res += query(u << 1 | 1, l, r);
    return res % mod;
}

int main()
{
    scanf("%d%lld", &n, &mod);
    for(int i = 1; i <= n; i++) scanf("%d", &q[i]);
    build(1, 1, n);

    scanf("%d", &m);
    while(m--) {
        int op, l, r;
        ll c;
        scanf("%d%d%d", &op, &l, &r);
        if(op == 1) {
            scanf("%lld", &c);
            c %= mod;
            modify(1, l, r, 0, c);
        }
        else if(op == 2) {
            scanf("%lld", &c);
            c %= mod;
            modify(1, l, r, c, 1);
        }

        else printf("%lld\n", query(1, l, r));
    }

    return 0;
}