P3373 【模板】线段树 2（加，乘懒标记）

三种操作：

1.区间加

2.区间乘

3.求区间和

用两个懒标记记录加(add)和乘(mul)，注意mul初始为1。

当乘上一个数时，要将他的add标记也乘，这样下传标记时直接加上就可以了。

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N = 1e5 + 10;
int n, q, m, a[N];
struct T {
	int l, r, sum, mul, add;
	#define l(p) t[p].l
	#define r(p) t[p].r
	#define sum(p) t[p].sum
	#define mul(p) t[p].mul
	#define add(p) t[p].add
}t[N << 2];
void pushup(int p) {
	sum(p) = (sum(p << 1) + sum(p << 1 | 1)) % m;
}
void build(int p, int l, int r) {
	l(p) = l, r(p) = r, mul(p) = 1;//***注意mul标记初始为1 
	if (l == r) {
		sum(p) = a[l] % m;
		return ;
	}
	int mid = (l + r) >> 1;
	build(p << 1, l, mid);
	build(p << 1 | 1, mid + 1, r);
	pushup(p);
}
void pushdown(int p) {//下传标记 
	sum(p << 1) = (sum(p << 1) * mul(p) % m + ((r(p << 1) - l(p << 1) + 1) * add(p)) % m) % m;
	sum(p << 1 | 1) = (sum(p << 1 | 1) * mul(p) % m + ((r(p << 1 | 1) - l(p << 1 | 1) + 1) * add(p)) % m) % m;
	
	mul(p << 1) = (mul(p << 1) * mul(p)) % m;
	mul(p << 1 | 1) = (mul(p << 1 | 1) * mul(p)) % m;
	
	add(p << 1) = (add(p) + add(p << 1) * mul(p)) % m;
	add(p << 1 | 1) = (add(p) + add(p << 1 | 1) * mul(p)) % m;
	
	mul(p) = 1, add(p) = 0;//***清除标记 
}
void modify1(int p, int l, int r, int val) {//乘 
	if (l(p) >= l && r(p) <= r) {
		sum(p) = (sum(p) * val) % m;
		add(p) = (add(p) * val) % m;
		mul(p) = (mul(p) * val) % m;
		return ;
	}
	pushdown(p);
	int mid = (l(p) + r(p)) >> 1;
	if (l <= mid) modify1(p << 1, l, r, val);
	if (r > mid) modify1(p << 1 | 1, l, r, val);
	pushup(p);
}
void modify2(int p, int l, int r, int val) {
	if (l(p) >= l && r(p) <= r) {
		sum(p) = (sum(p) + (r(p) - l(p) + 1) * val) % m;
		add(p) = (add(p) + val) % m;
		return ;
	}
	pushdown(p);
	int mid = (l(p) + r(p)) >> 1;
	if (l <= mid) modify2(p << 1, l, r, val);
	if (r > mid) modify2(p << 1 | 1, l, r, val);
	pushup(p);
}
int query(int p, int l, int r) {
	if (l(p) >= l && r(p) <= r) return sum(p);
	pushdown(p);
	int res = 0;
	int mid = (l(p) + r(p)) >> 1;
	if (l <= mid) res = (res + query(p << 1, l, r)) % m;
	if (r > mid) res = (res + query(p << 1 | 1, l, r)) % m;
	return res; 
}
signed main() {
	cin >> n >> q >> m;
	for (int i = 1; i <= n; i ++) cin >> a[i];
	build(1, 1, n);
	while (q --) {
		int op, x, y, k;
		cin >> op >> x >> y;
		if (op == 1) {
			cin >> k;
			modify1(1, x, y, k);
		}
		else if (op == 2) {
			cin >> k;
			modify2(1, x, y, k);
		}
		else cout << query(1, x, y) << endl;
	}
	return 0;
}