线段树: 维护序列

原题

本题每个节点存在两个懒标记add和mul用来更新区间节点的答案.

我们把区间的值sum看成sum = a * mul + add

对于modify操作

如果我们需要对一段区间的所有数乘上d, 相当于:

sum = a * mul * d + add * d, 即, 我们只用修改对应mul和add的值:

mul = mul * d, add = add * d.

如果我们需要对一段区间的所有数加上d,相当于:

sum = a * mul + add + d, 即, 我们只需要修改add值:

add = add + d.

#include <bits/stdc++.h>
using namespace std;
#define IO ios::sync_with_stdio(false);cin.tie(0); cout.tie(0);
inline int lowbit(int x) { return x & (-x); }
#define ll long long
#define ull unsigned long long
#define pb push_back
#define PII pair<int, int>
#define VIT vector<int>
#define x first
#define y second
#define inf 0x3f3f3f3f
const int N = 1e5 + 10; 
int n, p;
int w[N];
struct Node {
    int l, r;
    ll sum, add, mul;
}tr[4 * N];


void pushup(int u) {
    tr[u].sum = (tr[u << 1].sum + tr[u << 1 | 1].sum) % p;
}
    
void build(int u, int l, int r) {
    if (l == r) tr[u] = {l, r, w[r], 0, 1};
    else {
        tr[u] = {l, r, 0, 0, 1};
        int mid = l + r >> 1;
        build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
        pushup(u);
    }
}

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

void pushdown(int u) {
    eval(tr[u << 1], tr[u].add, tr[u].mul);      // 父节点更新左儿子
    eval(tr[u << 1 | 1], tr[u].add, tr[u].mul);  // 父节点更新右儿子
    tr[u].add = 0, tr[u].mul = 1;
}

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

int 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;
    int sum = 0;
    if (l <= mid) sum = query(u << 1, l, r);
    if (r > mid) sum = (sum + query(u << 1 | 1, l, r)) % p;
    return sum;
}

int main() {
    //freopen("in.txt", "r", stdin);
    IO;
    cin >> n >> p;
    for (int i = 1; i <= n; ++i) cin >> w[i];
    build(1, 1, n);
    int m;
    cin >> m;
    while (m--) {
        int t, l, r, d;
        cin >> t >> l >> r;
        if (t == 1) {
            cin >> d;
            modify(1, l, r, 0, d);
        } else if (t == 2) {
            cin >> d;
            modify(1, l, r, d, 1);
        } else cout << query(1, l, r) << '\n';
        
    }
    return 0;
}