第一行包含三个整数 n,m,pn,m,pn,m,p,分别表示该数列数字的个数、操作的总个数和模数。
第二行包含 nnn 个用空格分隔的整数,其中第 iii 个数字表示数列第 iii 项的初始值。
接下来 mmm 行每行包含若干个整数,表示一个操作,具体如下:
操作 111: 格式:1 x y k 含义:将区间 [x,y][x,y][x,y] 内每个数乘上 kkk
操作 222: 格式:2 x y k 含义:将区间 [x,y][x,y][x,y] 内每个数加上 kkk
操作 333: 格式:3 x y 含义:输出区间 [x,y][x,y][x,y] 内每个数的和对 ppp 取模所得的结果

#include <bits/stdc++.h>
using namespace std;
using ll = long long;

const int N = 2e5 + 10;
int tot = 0;
int n, m;
ll p;
struct node
{
   
    int l, r;
    ll sum, add, mul;
    node() {
   l = r = sum = add = 0; mul = 1;};
}tree[N * 4];

inline int ls(int cur) 
{
   
    return cur << 1;
}

inline int rs (int cur)
{
   
    return (cur << 1) | 1;
}

inline void push_up(int cur)
{
   
    tree[cur].sum = (tree[ls(cur)].sum + tree[rs(cur)].sum) % p;
}

inline void push_down(int cur)
{
   
    if (tree[cur].mul != 1 or tree[cur].add != 0) {
   
        tree[ls(cur)].sum = (tree[ls(cur)].sum * tree[cur].mul + (tree[ls(cur)].r - tree[ls(cur)].l + 1) * tree[cur].add) % p;
        tree[rs(cur)].sum = (tree[rs(cur)].sum * tree[cur].mul + (tree[rs(cur)].r - tree[rs(cur)].l + 1) * tree[cur].add) % p;
        tree[ls(cur)].mul = tree[ls(cur)].mul * tree[cur].mul % p;
        tree[rs(cur)].mul = tree[rs(cur)].mul * tree[cur].mul % p;
        tree[ls(cur)].add = (tree[ls(cur)].add * tree[cur].mul + tree[cur].add) % p;
        tree[rs(cur)].add = (tree[rs(cur)].add * tree[cur].mul + tree[cur].add) % p;
        tree[cur].add = 0;
        tree[cur].mul = 1;
    }
}


void build(int cur, int l, int r) 
{
   
    if (l == r) {
   
        ll a;
        scanf("%lld", &a);
        tree[cur].sum = a;
        tree[cur].l = l;
        tree[cur].r = r;
        return;
    }
    int mid = (l + r) >> 1;
    tree[cur].l = l; tree[cur].r = r;
    build(ls(cur), l, mid);
    build(rs(cur), mid + 1, r);
    push_up(cur);
}

void add(int cur, int l, int r, int v)
{
   
    if (tree[cur].l == l and tree[cur].r == r){
   
        tree[cur].sum = (tree[cur].sum + (r - l + 1) * v) %p;
        tree[cur].add = (tree[cur].add + v) % p;
        return;
    }
    push_down(cur);
    //继续往下
    //更新接下来的区间,保证他们是最新的
    int mid = (tree[cur].l + tree[cur].r) >> 1;
    if (r <= mid) {
   
        add(ls(cur), l, r, v);
    }
    else if (l > mid) {
   
        add(rs(cur), l, r, v);
    }
    else {
   
        add(ls(cur), l, mid, v);
        add(rs(cur), mid + 1, r, v);
    }
    //所有区间都从自己的子区间获取值的更新
    push_up(cur);
}

void mul(int cur, int l, int r, int v)
{
   
    if (tree[cur].l == l and tree[cur].r == r){
   
        tree[cur].sum = (tree[cur].sum * v) % p;
        tree[cur].mul = (tree[cur].mul * v) % p;
        tree[cur].add = (tree[cur].add * v) % p;
        return;
    }
    push_down(cur);
    int mid = (tree[cur].l + tree[cur].r) >> 1;
    if (r <= mid) {
   
        mul(ls(cur), l, r, v);
    }
    else if (l > mid) {
   
        mul(rs(cur), l, r, v);
    }
    else {
   
        mul(ls(cur), l, mid, v);
        mul(rs(cur), mid + 1, r, v);
    }
    //所有区间都从自己的子区间获取值的更新
    push_up(cur);
}

ll query(int cur, int l, int r)
{
   
    if (tree[cur].l == l and tree[cur].r == r){
   
        return tree[cur].sum;
    }
    push_down(cur);
    int mid = (tree[cur].l + tree[cur].r) >> 1;
    if (r <= mid) {
   
        return query(ls(cur), l, r);
    }
    else if (l > mid) {
   
        return query(rs(cur), l, r);
    }
    else {
   
        return (query(ls(cur), l, mid) + query(rs(cur), mid + 1, r)) % p;
    }
    //没有更新操作,所以不需要push_up
}

int main() {
   
#ifndef ONLINE_JUDGE
    freopen("D:/VS CODE/C++/in.txt", "r", stdin);
    freopen("D:/VS CODE/C++/out.txt", "w", stdout);
#endif
    cin >> n >> m >> p;
    build(1, 1, n);
    while (m--) {
   
        int o, x, y, k;
        scanf("%d", &o);
        if (o == 1) {
   
            cin >> x >> y >> k;
            mul(1, x, y, k);
        }
        else if (o == 2) {
   
            cin >> x >> y >> k;
            add(1, x, y, k);
        }
        else {
   
            cin >> x >> y;
            cout << query(1, x, y) << endl;
        }
    }

    fclose(stdin);
    fclose(stdout);
    return 0;
}