线段树多种操作模板（区间加，区间乘，区间赋值）

做这种题目呢，我们只需要分清楚每种操作的优先级就行了。在这里，我们规定优先级：区间赋值>区间乘>区间加。

也没啥好说而呀，我要咋办= =。
记得 $query$query 和 $update$update 时要 标记下饭(放就行了。
放代码吧。

代码如下

#include <bits/stdc++.h>
#define LL long long
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
#define N 100005
using namespace std;
int sum[N * 4], add[N * 4], col[N * 4], mul[N * 4];
void pushup(int rt){
	sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}
void pushdown(int rt, int x){
	if(col[rt] != -1){
		add[rt<<1] = add[rt<<1|1] = 0;
		mul[rt<<1] = mul[rt<<1|1] = 1;
		sum[rt<<1] = col[rt] * (x - x / 2);
		sum[rt<<1|1] = col[rt] * (x / 2);
		col[rt<<1] = col[rt<<1|1] = col[rt];
		col[rt] = -1;
	}
	if(mul[rt] != 1){
		int c = mul[rt];
		mul[rt<<1] *= c;
		mul[rt<<1|1] *= c;
		sum[rt<<1] *= c;
		sum[rt<<1|1] *= c;
		add[rt<<1] *= c;
		add[rt<<1|1] *= c;
		mul[rt] = 1;
	}
	if(add[rt]){
		add[rt<<1] += add[rt];
		add[rt<<1|1] += add[rt];
		sum[rt<<1] += add[rt] * (x - x / 2);
		sum[rt<<1|1] += add[rt] * (x / 2);
		add[rt] = 0;
	}
}
void build(int l, int r, int rt){
	if(l == r){
		scanf("%d", &sum[rt]);
		return;
	}
	int m = l + r >> 1;
	build(lson);
	build(rson);
	pushup(rt);
}
void update(int l, int r, int rt, int a, int b, int c, int o){
	if(l >= a && r <= b){
		if(o == 1){//set
			add[rt] = 0;
			mul[rt] = 1;
			sum[rt] = c * (r - l + 1);
			col[rt] = c;
		}
		else if(o == 2){//cheng
			mul[rt] *= c;
			sum[rt] *= c;
			add[rt] *= c;
		}
		else{//jia
			add[rt] += c;
			sum[rt] += c * (r - l + 1);
		}
		return;
	}
	pushdown(rt, r-l+1);
	int m = l + r >> 1;
	if(a <= m) update(lson, a, b, c, o);
	if(b > m) update(rson, a, b, c, o);
	pushup(rt);
}
int query(int l, int r, int rt, int a, int b){
	if(l >= a && r <= b) return sum[rt];
	pushdown(rt, r-l+1);
	int m = l + r >> 1, ans = 0;
	if(a <= m) ans += query(lson, a, b);
	if(b > m) ans += query(rson, a, b);
	return ans;
}
int main(){
	int i, j, n, m, o, a, b, c;
	scanf("%d%d", &n, &m);
	for(i = 1; i <= n * 4; i++){
		mul[i] = 1;
		col[i] = -1;
	}
	build(1, n, 1);
	while(m--){
		scanf("%d%d%d", &o, &a, &b);
		if(o == 4) printf("%d\n", query(1, n, 1, a, b));
		else{
			scanf("%d", &c);
			update(1, n, 1, a, b, c, o);
		}
	}
	return 0;
}