UOJ46. 【清华集训2014】玄学

传送门

Sol

考虑对于操作时间建立线段树，二进制分组

那么现在主要的问题就是怎么合并信息

你发现一个性质，就是每个修改只会在整个区间内增加两个端点

那么我们二进制分组可以得到每个区间内最多只有区间长度级别段，每一段的修改都是一样的

那么可以直接一层层归并上来

最后询问就是二分每一个线段树的节点的询问段即可

修改复杂度 \(\Theta(n log n)\) 询问复杂度 \(\Theta(n log^2 n)\)

# include <bits/stdc++.h>
using namespace std;
typedef long long ll;

namespace IO {
	const int maxn((1 << 21) + 1);

	char ibuf[maxn], *iS, *iT, obuf[maxn], *oS = obuf, *oT = obuf + maxn - 1, c, st[55];
	int f, tp;

	char Getc() {
		return (iS == iT ? (iT = (iS = ibuf) + fread(ibuf, 1, maxn, stdin), (iS == iT ? EOF : *iS++)) : *iS++);
	}

	void Flush() {
		fwrite(obuf, 1, oS - obuf, stdout);
		oS = obuf;
	}

	void Putc(char x) {
		*oS++ = x;
		if (oS == oT) Flush();
	}

	template <class Int> void In(Int &x) {
		for (f = 1, c = Getc(); c < '0' || c > '9'; c = Getc()) f = c == '-' ? -1 : 1;
		for (x = 0; c <= '9' && c >= '0'; c = Getc()) x = (x << 3) + (x << 1) + (c ^ 48);
		x *= f;
	}

	template <class Int> void Out(Int x) {
		if (!x) Putc('0');
		if (x < 0) Putc('-'), x = -x;
		while (x) st[++tp] = x % 10 + '0', x /= 10;
		while (tp) Putc(st[tp--]);
	}
}

using IO :: In;
using IO :: Out;
using IO :: Putc;
using IO :: Flush;

const int maxn(1e5 + 5);

int n, q, m, a[maxn], type, cnt, nowl[maxn << 2], nowr[maxn << 2], tot, ans;

struct Mdy {
	int r, a, b;
} mdy[maxn * 30];

void Modify(int x, int l, int r, int ql, int qr, int a, int b) {
	if (l == r) {
		nowl[x] = tot + 1;
		if (ql > 1) mdy[++tot] = (Mdy){ql - 1, 1, 0};
		mdy[++tot] = (Mdy){qr, a, b};
		if (qr < n) mdy[++tot] = (Mdy){n, 1, 0};
		nowr[x] = tot;
		return;
	}
	register int mid = (l + r) >> 1, l1, r1, l2, r2;
	cnt <= mid ? Modify(x << 1, l, mid, ql, qr, a, b) : Modify(x << 1 | 1, mid + 1, r, ql, qr, a, b);
	if (cnt < r) return;
	nowl[x] = tot + 1, l1 = nowl[x << 1], r1 = nowr[x << 1], l2 = nowl[x << 1 | 1], r2 = nowr[x << 1 | 1];
	while (l1 <= r1 && l2 <= r2) {
		mdy[++tot] = (Mdy){min(mdy[l1].r, mdy[l2].r), (ll)mdy[l1].a * mdy[l2].a % m, ((ll)mdy[l1].b * mdy[l2].a + mdy[l2].b) % m};
		if (mdy[l1].r == mdy[l2].r) ++l1, ++l2;
		else mdy[l1].r < mdy[l2].r ? ++l1 : ++l2;
	}
	nowr[x] = tot;
}

inline void Calc(int x, int p) {
	register int l = nowl[x], r = nowr[x], cur = 0, mid;
	while (l <= r) mdy[mid = (l + r) >> 1].r >= p ? cur = mid, r = mid - 1 : l = mid + 1;
	ans = ((ll)ans * mdy[cur].a + mdy[cur].b) % m;
}

void Query(int x, int l, int r, int ql, int qr, int p) {
	if (ql <= l && qr >= r) {
		Calc(x, p);
		return;
	}
	register int mid = (l + r) >> 1;
	if (ql <= mid) Query(x << 1, l, mid, ql, qr, p);
	if (qr > mid) Query(x << 1 | 1, mid + 1, r, ql, qr, p);
}

int main() {
	register int i, j, op, p, l, r;
	In(type), In(n), In(m);
	for (i = 1; i <= n; ++i) In(a[i]);
	In(q);
	while (q--) {
		In(op);
		if (op == 1) {
			In(l), In(r), In(i), In(j);
			if (type & 1) l ^= ans, r ^= ans;
			++cnt, Modify(1, 1, 100000, l, r, i, j);
		}
		else {
			In(l), In(r), In(p);
			if (type & 1) l ^= ans, r ^= ans, p ^= ans;
			ans = a[p], Query(1, 1, 100000, l, r, p);
			Out(ans), Putc('\n');
		}
	}
	return Flush(), 0;
}