Transformation HDU - 4578

题目链接 $1$ 题目链接 $2$

题解

设一个区间的和、平方和、立方和分别是 $s u m_{0}, s u m_{1}, s u m_{2}$
对于 $a d d$ 操作，推推公式可知 ${\begin{cases} n e w s u m_{2} = s u m_{2} + v a l^{3} \times l e n + 3 \times v a l \times s u m_{1} + 3 \times v a l^{2} \times s u m_{0} \\ n e w s u m_{1} = s u m_{1} + v a l^{2} \times l e n + 2 \times v a l \times s u m_{0} \\ n e w s u m_{0} = s u m_{0} + v a l \times l e n \end{cases}$

对于 $t i m, t a g$ 的操作很好维护。

注意在修改时，修改区间要加懒标记，尤其是 $t i m$ 操作时要记得 $a d d \times = v a l$ 。

代码

#include <iostream>
using namespace std;
typedef long long LL;
const int N = 100010,MOD = 10007,INF = 2e9;
int n,m;
struct segment_tree_node {
	int l,r;
	LL sum[3];
	LL add,tim,tag;
}tr[4 * N];
void push_up (int u) {
	for (int i = 0;i <= 2;i++) tr[u].sum[i] = (tr[u << 1].sum[i] + tr[u << 1 | 1].sum[i]) % MOD;
}
LL power (LL x,int d) {
	LL ans = 1;
	while (d--) ans = ans * x % MOD;
	return ans;
}
void opt_tim (int u,LL val) {
	for (int i = 0;i <= 2;i++) tr[u].sum[i] = tr[u].sum[i] * power (val,i + 1) % MOD;
}
void opt_add (int u,LL val) {
	int len = tr[u].r - tr[u].l + 1;
	tr[u].sum[2] = (tr[u].sum[2] + power (val,3) * len % MOD + 3 * val % MOD * (tr[u].sum[1] + tr[u].sum[0] * val % MOD) % MOD) % MOD;
	tr[u].sum[1] = (tr[u].sum[1] + power (val,2) * len % MOD + 2 * val * tr[u].sum[0] % MOD) % MOD;
	tr[u].sum[0] = (tr[u].sum[0] + val * len % MOD) % MOD;
}
void opt_tag (int u,LL val) {
	int len = tr[u].r - tr[u].l + 1;
	for (int i = 0;i <= 2;i++) tr[u].sum[i] = power (val,i + 1) * len % MOD;
}
void push_down (int u) {
	if (tr[u].tag != INF) {
		tr[u << 1].tag = tr[u << 1 | 1].tag = tr[u].tag;
		tr[u << 1].add = tr[u << 1 | 1].add = 0;
		tr[u << 1].tim = tr[u << 1 | 1].tim = 1;
		opt_tag (u << 1,tr[u].tag),opt_tag (u << 1 | 1,tr[u].tag);
		tr[u].tag = INF;
	}
	if (tr[u].tim != 1) {
		tr[u << 1].add = tr[u << 1].add * tr[u].tim % MOD,tr[u << 1 | 1].add = tr[u << 1 | 1].add * tr[u].tim % MOD;
		tr[u << 1].tim = tr[u << 1].tim * tr[u].tim % MOD,tr[u << 1 | 1].tim = tr[u << 1 | 1].tim * tr[u].tim % MOD;
		opt_tim (u << 1,tr[u].tim),opt_tim (u << 1 | 1,tr[u].tim);
		tr[u].tim = 1;
	}
	if (tr[u].add) {
		tr[u << 1].add = (tr[u << 1].add + tr[u].add) % MOD,tr[u << 1 | 1].add = (tr[u << 1 | 1].add + tr[u].add) % MOD;
		opt_add (u << 1,tr[u].add),opt_add (u << 1 | 1,tr[u].add);
		tr[u].add = 0;
	}
}
void build_segment_tree (int u,int l,int r) {
	if (l == r) {
		tr[u] = {l,r,{0,0,0},0,1,INF};
		return ;
	}
	tr[u] = {l,r,{0,0,0},0,1,INF};
	int mid = l + r >> 1;
	build_segment_tree (u << 1,l,mid),build_segment_tree (u << 1 | 1,mid + 1,r);
}
void modify_add (int u,int l,int r,int d) {
	if (l <= tr[u].l && tr[u].r <= r) {
		opt_add (u,d);
		tr[u].add = (tr[u].add + d) % MOD;
		return ;
	}
	push_down (u);
	int mid = tr[u].l + tr[u].r >> 1;
	if (l <= mid) modify_add (u << 1,l,r,d);
	if (r >= mid + 1) modify_add (u << 1 | 1,l,r,d);
	push_up (u);
}
void modify_tim (int u,int l,int r,int d) {
	if (l <= tr[u].l && tr[u].r <= r) {
		opt_tim (u,d);
		tr[u].tim = tr[u].tim * d % MOD;
		tr[u].add = tr[u].add * d % MOD;
		return ;
	}
	push_down (u);
	int mid = tr[u].l + tr[u].r >> 1;
	if (l <= mid) modify_tim (u << 1,l,r,d);
	if (r >= mid + 1) modify_tim (u << 1 | 1,l,r,d);
	push_up (u);
}
void modify_tag (int u,int l,int r,int d) {
	if (l <= tr[u].l && tr[u].r <= r) {
		opt_tag (u,d);
		tr[u].tag = d,tr[u].add = 0,tr[u].tim = 1;
		return ;
	}
	push_down (u);
	int mid = tr[u].l + tr[u].r >> 1;
	if (l <= mid) modify_tag (u << 1,l,r,d);
	if (r >= mid + 1) modify_tag (u << 1 | 1,l,r,d);
	push_up (u);
}
LL query (int u,int l,int r,int p) {
	if (l <= tr[u].l && tr[u].r <= r) return tr[u].sum[p] % MOD;
	push_down (u);
	int mid = tr[u].l + tr[u].r >> 1;
	LL ans = 0;
	if (l <= mid) ans = (ans + query (u << 1,l,r,p)) % MOD;
	if (r >= mid + 1) ans = (ans + query (u << 1 | 1,l,r,p)) % MOD;
	return ans;
}
int main () {
	ios :: sync_with_stdio (false),cin.tie (0),cout.tie (0);
	while (cin >> n >> m,n || m) {
		build_segment_tree (1,1,n);
		while (m--) {
			int op,l,r,d; 
			cin >> op >> l >> r >> d;
			if (op == 1) modify_add (1,l,r,d);
			else if (op == 2) modify_tim (1,l,r,d);
			else if (op == 3) modify_tag (1,l,r,d);
			else cout << query (1,l,r,d - 1) << endl;
		}
	}
	return 0;
}