rt，样例输出

6
17

都快改的跟题解一样了。。实在找不出来了 谢谢各位大佬（）

#include <memory.h>

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <iostream>
#include <queue>
#include <stack>
#include <string>
#include <vector>
using namespace std;
const int N = 1e5 + 10;

long long tr[4 * N], lazy_add[4 * N], lazy_mul[4 * N], a[N], n, m, p;

void push_up(int id) { tr[id] = (tr[id * 2] + tr[id * 2 + 1]) % p; }

void build(int id, int l, int r) {
    lazy_add[id] = 0, lazy_mul[id] = 1;
    if (l == r) {
        tr[id] = a[l] % p;
        return;
    }
    int mid = (l + r) >> 1;
    build(id * 2 + 1, l, mid);
    build(id * 2 + 1, mid + 1, r);
    push_up(id);
}

void push_down(int id, int l, int r) {
    int mid = (l + r) >> 1;
    tr[id * 2] = (tr[id * 2] * lazy_mul[id] + lazy_add[id] * (mid - l + 1)) % p;
    tr[id * 2 + 1] =
        (tr[id * 2 + 1] * lazy_mul[id] + lazy_add[id] * (r - mid)) % p;

    lazy_mul[id * 2] *= lazy_mul[id];
    lazy_mul[id * 2 + 1] *= lazy_mul[id];
    lazy_mul[id * 2] %= p, lazy_mul[id * 2 + 1] %= p;

    lazy_add[id * 2] = lazy_add[id * 2] * lazy_mul[id] + lazy_add[id];
    lazy_add[id * 2 + 1] = lazy_add[id * 2 + 1] * lazy_mul[id] + lazy_add[id];
    lazy_add[id * 2] %= p, lazy_add[id * 2 + 1] %= p;

    lazy_mul[id] = 1, lazy_add[id] = 0;
}

void add(int id, int l, int r, int x, int y, long long v) {
    if (x <= l && r <= y) {
        lazy_add[id] += v, lazy_add[id] %= p;
        tr[id] += (r - l + 1) * v, tr[id] %= p;
        return;
    }

    push_down(id, l, r);
    int mid = (l + r) >> 1;

    if (x <= mid) {
        add(id * 2, l, mid, x, y, v);
    }

    if (y > mid) {
        add(id * 2 + 1, mid + 1, r, x, y, v);
    }

    push_up(id);
}

void mul(int id, int l, int r, int x, int y, long long v) {
    if (x <= l && r <= y) {
        lazy_mul[id] *= v, lazy_mul[id] %= p;
        lazy_add[id] *= v, lazy_add[id] %= p;
        tr[id] *= v, tr[id] %= p;
        return;
    }

    push_down(id, l, r);
    int mid = (l + r) >> 1;

    if (x <= mid) {
        mul(id * 2, l, mid, x, y, v);
    }

    if (y > mid) {
        mul(id * 2 + 1, mid + 1, r, x, y, v);
    }

    push_up(id);
}

long long query(int id, int l, int r, int x, int y) {
    if (x <= l && r <= y) {
        return tr[id];
    }

    push_down(id, l, r);
    int mid = (l + r) >> 1;
    long long ans = 0;

    if (x <= mid) {
        ans += query(id * 2, l, mid, x, y) % p;
    }

    if (y > mid) {
        ans += query(id * 2 + 1, mid + 1, r, x, y) % p;
    }

    return ans % p;
}

int main() {
    cin >> n >> m >> p;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        a[i] %= p;
    }
    build(1, 1, n);
    while (m--) {
        int op, x, y, k;
        cin >> op >> x >> y;
        if (op == 1) {
            cin >> k;
            mul(1, 1, n, x, y, k);
        } else if (op == 2) {
            cin >> k;
            add(1, 1, n, x, y, k);
        } else {
            cout << query(1, 1, n, x, y) << endl;
        }
    }
    return 0;
}