两种修改用一个 update

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int MAXN = 100000;
int n, m, mod;
int a[MAXN + 5];
int t[MAXN * 4 + 5];
int lazy_add[MAXN * 4 + 5], lazy_mul[MAXN * 4 + 5];
void down(int now, int l, int r)
{
    int mid = (l + r) / 2;
    int add = lazy_add[now];
    int mul = lazy_mul[now];
    // 传 mul
    if (mul != 1)
    {
        t[now * 2] *= mul;
        lazy_mul[now * 2] *= mul;
        lazy_add[now * 2] *= mul;
        t[now * 2 + 1] *= mul;
        lazy_mul[now * 2 + 1] *= mul;
        lazy_add[now * 2 + 1] *= mul;
    }
    // 传 add
    if (add != 0)
    {
        t[now * 2] += (mid - l + 1) * add;
        lazy_add[now * 2] += add;
        t[now * 2 + 1] += (r - mid) * add;
        lazy_add[now * 2 + 1] += add;
    }
    // mod
    t[now * 2] %= mod;
    lazy_mul[now * 2] %= mod;
    lazy_add[now * 2] %= mod;
    t[now * 2 + 1] %= mod;
    lazy_mul[now * 2 + 1] %= mod;
    lazy_add[now * 2 + 1] %= mod;
    // clear
    lazy_add[now] = 0;
    lazy_mul[now] = 1;
}
void build(int now, int l, int r)
{
    lazy_mul[now] = 1;
    lazy_add[now] = 0;
    if (l == r)
    {
        t[now] = a[l];
        return;
    }
    int mid = (l + r) / 2;
    build(now * 2, l, mid);
    build(now * 2 + 1, mid + 1, r);
    t[now] = (t[now * 2] + t[now * 2 + 1]) % mod;
}
void update(int now, int l, int r, int x, int y, int mul, int add)
{
    if (x <= l && r <= y)
    {
        t[now] *= mul;
        lazy_mul[now] *= mul;
        lazy_add[now] *= mul;
        t[now] += (r - l + 1) * add;
        lazy_add[now] += add;
        t[now] %= mod;
        lazy_mul[now] %= mod;
        lazy_add[now] %= mod;
        return;
    }
    down(now, l, r);
    int mid = (l + r) / 2;
    if (x <= mid)
        update(now * 2, l, mid, x, y, mul, add);
    if (y > mid)
        update(now * 2 + 1, mid + 1, r, x, y, mul, add);
    t[now] = (t[now * 2] + t[now * 2 + 1]) % mod;
}
int query(int now, int l, int r, int x, int y)
{
    if (x <= l && r <= y)
        return t[now];
    down(now, l, r);
    int mid = (l + r) / 2;
    int res = 0;
    if (x <= mid)
        res = (res + query(now * 2, l, mid, x, y)) % mod;
    if (y >= mid + 1)
        res = (res + query(now * 2 + 1, mid + 1, r, x, y)) % mod;
    return res;
}
signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> m >> mod;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    build(1, 1, n);
    for (int i = 1; i <= m; i++)
    {
        int op, x, y, z;
        cin >> op;
        if (op == 1)
        {
            cin >> x >> y >> z;
            update(1, 1, n, x, y, z, 0);
        }
        if (op == 2)
        {
            cin >> x >> y >> z;
            update(1, 1, n, x, y, 1, z);
        }
        if (op == 3)
        {
            cin >> x >> y;
            cout << query(1, 1, n, x, y) << "\n";
        }
    }
    return 0;
}

两个修改用两个 update

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int MAXN = 100000;
int n, m, p;
int a[MAXN + 5];
// 线段树：单点修改、区间查询（区间和）
int t[4 * MAXN + 5];
int lazyAdd[4 * MAXN + 5], lazyMul[4 * MAXN + 5];
// 基于a[]建线段树
void build(int now, int l, int r)
{
    lazyAdd[now] = 0;
    lazyMul[now] = 1;
    if (l == r)
    {
        t[now] = a[l];
        return;
    }
    int mid = (l + r) / 2;
    build(now * 2, l, mid);
    build(now * 2 + 1, mid + 1, r);
    t[now] = t[now * 2] + t[now * 2 + 1];
    t[now] %= p;
}
void down(int now, int l, int r)
{
    int mid = (l + r) / 2;
    // 左子树 now*2,l,mid
    // 右子树 now*2+1,mid+1,r
    int add = lazyAdd[now];
    int mul = lazyMul[now];

    lazyMul[now * 2] *= mul;
    lazyMul[now * 2] %= p;
    lazyAdd[now * 2] *= mul;
    lazyAdd[now * 2] %= p;
    t[now * 2] *= mul;
    t[now * 2] %= p;
    lazyAdd[now * 2] += add;
    lazyAdd[now * 2] %= p;
    t[now * 2] += (mid - l + 1) * add;
    t[now * 2] %= p;

    lazyMul[now * 2 + 1] *= mul;
    lazyMul[now * 2 + 1] %= p;
    lazyAdd[now * 2 + 1] *= mul;
    lazyAdd[now * 2 + 1] %= p;
    t[now * 2 + 1] *= mul;
    t[now * 2 + 1] %= p;
    lazyAdd[now * 2 + 1] += add;
    lazyAdd[now * 2 + 1] %= p;
    t[now * 2 + 1] += (r - mid) * add;
    t[now * 2 + 1] %= p;

    lazyAdd[now] = 0;
    lazyMul[now] = 1;
}
// 当前做到了树上的now号节点，当前节点对应的区间为 [l,r]
// 需要给 [x,y] 这些数都加 z
void update_add(int now, int l, int r, int x, int y, int z)
{
    if (x <= l && r <= y)
    {
        t[now] += (r - l + 1) * z;
        t[now] %= p;
        lazyAdd[now] += z;
        lazyAdd[now] %= p;
        return;
    }
    int mid = (l + r) / 2;
    // 左子树 now*2,l,mid
    // 右子树 now*2+1,mid+1,r
    down(now, l, r); // 把之前欠的结清
    if (x <= mid)
        update_add(now * 2, l, mid, x, y, z);
    if (y > mid)
        update_add(now * 2 + 1, mid + 1, r, x, y, z);
    t[now] = t[now * 2] + t[now * 2 + 1];
    t[now] %= p;
}
void update_mul(int now, int l, int r, int x, int y, int z)
{
    if (x <= l && r <= y)
    {
        t[now] *= z;
        t[now] %= p;
        lazyMul[now] *= z;
        lazyMul[now] %= p;
        lazyAdd[now] *= z;
        lazyAdd[now] %= p;
        return;
    }
    int mid = (l + r) / 2;
    // 左子树 now*2,l,mid
    // 右子树 now*2+1,mid+1,r
    down(now, l, r); // 把之前欠的结清
    if (x <= mid)
        update_mul(now * 2, l, mid, x, y, z);
    if (y > mid)
        update_mul(now * 2 + 1, mid + 1, r, x, y, z);
    t[now] = t[now * 2] + t[now * 2 + 1];
    t[now] %= p;
}
int query(int now, int l, int r, int x, int y)
{
    //[l,r] 是 [x,y] 的子区间
    if (x <= l && r <= y)
        return t[now];
    int mid = (l + r) / 2;
    // 左子树 now*2,l,mid
    // 右子树 now*2+1,mid+1,r
    down(now, l, r); // 把之前欠的结清
    int res = 0;
    if (x <= mid)
        res += query(now * 2, l, mid, x, y);
    if (y > mid)
        res += query(now * 2 + 1, mid + 1, r, x, y);
    return res % p;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> m >> p;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
    }
    build(1, 1, n);
    while (m--)
    {
        int op, x, y, z;
        cin >> op;
        if (op == 1)
        {
            cin >> x >> y >> z;
            update_mul(1, 1, n, x, y, z);
        }
        else if (op == 2)
        {
            cin >> x >> y >> z;
            update_add(1, 1, n, x, y, z);
        }
        else if (op == 3)
        {
            cin >> x >> y; // cout sum(a[x]~a[y])
            cout << query(1, 1, n, x, y) << "\n";
        }
    }
    return 0;
}