BZOJ1798 [Ahoi2009]Seq 维护序列seq (线段树维护区间加法减法)

题目链接

思路：
$\ \ \ \ $线段树维护区间加法乘法，可以维护这样的形式：$a \cdot x+b$，假如当前位置的和表示形式是$a \cdot sum + b$，当一个区间乘以$d$的时候，形式变成了$d \cdot(a \cdot sum + b) = (a \cdot d) \cdot sum + (d \cdot b)$, 同样加法的话变成$a \cdot sum + b + d = a \cdot sum + (d + b)$，懒惰标记延迟下传$a$和$b$就行了。

#include<bits/stdc++.h>
typedef long long ll;
const int maxn = 3e5 + 10;
using namespace std;

ll sum[maxn * 4], add[maxn * 4], mul[maxn * 4];
ll n, m, mod;

void build(int o, int l, int r) {
    add[o] = 0; mul[o] = 1;
    if(l == r) { cin >> sum[o]; sum[o] %= mod; return ; }
    int mid = (l + r) >> 1;
    build(o << 1, l, mid);
    build(o << 1 | 1, mid + 1, r);
    sum[o] = (sum[o << 1] + sum[o << 1 | 1]) % mod;
}

void push_down(int o, int l, int r) {
    int o1 = o << 1, o2 = o1 | 1;
    if(mul[o] != 1) {
       mul[o1] = mul[o1] * mul[o] % mod;
       mul[o2] = mul[o2] * mul[o] % mod;
       add[o1] = (add[o1] * mul[o]) % mod;
       add[o2] = (add[o2] * mul[o]) % mod;
       mul[o] = 1; 
    }
    if(add[o]) {
        add[o1] = (add[o1] + add[o]) % mod;
        add[o2] = (add[o2] + add[o]) % mod;
        add[o] = 0;
    }
}

void push_up(int o, int l, int r) {
    int mid = (l + r) >> 1;
    int o1 = o << 1, o2 = o1 | 1;
    sum[o] = (mul[o1] * sum[o1] + add[o1] * (mid - l + 1)) % mod;
    sum[o] = (sum[o] + mul[o2] * sum[o2] + add[o2] * (r - mid)) % mod;
}

void update(int o, int l, int r, int ql, int qr, int x, int flag) {
    if(l > qr || r < ql) return ;
    if(l >= ql && r <= qr) {
        if(flag == 1) {
            mul[o] = mul[o] * x % mod;
            add[o] = add[o] * x % mod;
        } else {
            add[o] = (add[o] + x) % mod;
        }
        return ;
    }
    int mid = (l + r) >> 1;
    push_down(o, l, r);
    update(o << 1, l, mid, ql, qr, x, flag);
    update(o << 1 | 1, mid + 1, r, ql, qr, x, flag);
    push_up(o, l, r);
}

ll query(int o, int l, int r, int ql, int qr) {
    if(l > qr || r < ql) return 0;
    if(l >= ql && r <= qr) return (sum[o] * mul[o] + add[o] * (r - l + 1)) % mod;
    int mid = (l + r) >> 1;
    push_down(o, l, r);
    ll p1 = query(o << 1, l, mid, ql, qr);
    ll p2 = query(o << 1 | 1, mid + 1, r, ql, qr);
    push_up(o, l, r);
    return (p1 + p2) % mod; 
}

int main() {
    ios::sync_with_stdio(0);
    cin >> n >> mod;
    build(1, 0, n - 1);
    cin >> m;
    while(m--) {
        int op, l, r, x;
        cin >> op >> l >> r;
        l--; r--;
        if(op == 1) {
            cin >> x;
            update(1, 0, n - 1, l, r, x, 1); 
        } else if(op == 2) {
            cin >> x;
            update(1, 0, n - 1, l, r, x, 2);
        } else {
            ll ans = query(1, 0, n - 1, l, r);
            cout << ans << endl;
        }
    }
    return 0;
}