洛谷 P3373 【模板】线段树 2

一、思路

• 区间乘 + 区间加
• 用一个数组lazyM保存区间乘， 一个数组lazyP保存区间加。 如何进行pushdown呢？
• 假设一个区间的总和值是sum。 ((sum * b) + c ) * d = sum *bd + cd。 所以lazyM数组里面保存bd, lazyP数组里面保存cd。
• 若当前的操作是+。 我们只需要让lazyP 数组加上该数， 若是 *，让lazyM数组 * 上该数，并上lazyP = lazyP * 该数。
• 子区间的lazyP = 子区间lazyP * 父亲lazyM + 父亲lazyP。
• 子区间的lazyM = 子区间lazyM * 父亲lazyM
• 具体的可以自己设置值来推一下就行了。每一步都进行一下取余操作。

二、代码

#include<cstdio>
#include<cstring>
#include<cmath>
#define LL long long
#define N 100005

int n, m, x, y, code;
LL p, k, sum[N << 2], lazyP[N << 2], lazyM[N << 2], a[N];

void build(int id, int l, int r) {
	lazyM[id] = 1;
	lazyP[id] = 0;
	if (l == r) {
		sum[id] = a[l];
		return;
	}
	int mid= (l + r) >> 1;
	build(id << 1, l, mid);
	build(id << 1 | 1, mid + 1, r);
	sum[id] = (sum[id << 1] + sum[id << 1 | 1]) % p;
}
 
void pushdown(int id, int l, int r) {
	if (lazyP[id] == 0 && lazyM[id] == 1) {
		return;
	}
	int ll = id << 1;
	int rr = id << 1 | 1; 
	int mid = (l + r) >> 1;
	sum[ll] = (sum[ll] * lazyM[id] + lazyP[id] * (mid - l + 1)) % p;
	sum[rr] = (sum[rr] * lazyM[id] + lazyP[id] * (r - mid)) % p;
	lazyP[ll] = (lazyP[ll] * lazyM[id] + lazyP[id]) % p;
	lazyP[rr] = (lazyP[rr] * lazyM[id] + lazyP[id]) % p;
	lazyM[ll] = (lazyM[ll] * lazyM[id]) % p;
	lazyM[rr] = (lazyM[rr] * lazyM[id]) % p;
	lazyM[id] = 1;
	lazyP[id] = 0;
}
void pushtag(int id, int code, LL k, int l, int r) {
	if (code == 1) {
		sum[id] = (sum[id] * k) % p;
		lazyP[id] = (lazyP[id] * k)% p;
		lazyM[id] = (lazyM[id] * k) % p;
	} else {
		sum[id] = (sum[id] + (r - l + 1) * k) % p;
		lazyP[id] = (lazyP[id] + k)% p;
	} 
}
// * k  +k
void update(int id, int l, int r, int x, int y, LL k, int code) {
	if (x <= l && r <= y) {
		pushtag(id, code, k, l ,r);
		return;
	}
	pushdown(id, l, r);
	int mid = (l + r) >> 1;
	if (x <= mid) {
		update(id << 1, l, mid, x, y, k, code);
	}
	if (y > mid) {
		update(id << 1 | 1, mid + 1, r, x, y, k, code); 
	}
	sum[id] = (sum[id << 1] + sum[id << 1 | 1]) % p;
}
LL query(int id, int l, int r, int x, int y) {
	if (x <= l && r <= y) {
		return sum[id];
	}
	pushdown(id, l , r);
	LL ans = 0;
	int mid = (l + r) >> 1;
	if (x <= mid) ans += query(id << 1, l, mid, x, y);
	if (y > mid) ans += query(id << 1 | 1, mid + 1, r, x, y); 
	return ans;
}
int main() {
	scanf("%d %d %lld", &n, &m, &p);
	for (int i = 1; i <= n; i++) {
		scanf("%lld", &a[i]);
	}
	build(1, 1, n);
	for (int i = 1; i <= m; i++) {
		scanf("%d", &code);
		if (code == 1 || code == 2) {
			scanf("%d %d %lld", &x, &y, &k);
			update(1, 1, n, x, y, k, code);
		} else {
			scanf("%d %d", &x, &y);
			printf("%lld\n", query(1, 1, n, x, y) % p);	
		}
	}
	return 0;
}