_bzoj1798 [Ahoi2009]Seq 维护序列seq【线段树 lazy tag】

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=1798

注意，应保证当前节点维护的值是正确的，lazy tag只是一个下传标记，在下传时应即时更新儿子的维护值，在修改时也应即时更新当前节点的维护值。

#include <cstdio>

const int maxn = 100005;

int n, mod, a[maxn], m, t1, t2, t3, opr;
struct Node {
	int ql, qr;
	long long sm, mul, add;
} tree[maxn << 2];

inline void pushup(int p) {
	tree[p].sm = (tree[p << 1].sm + tree[p << 1 | 1].sm) % mod;
}
inline void pushdown(int p) {
	tree[p << 1].mul = tree[p << 1].mul * tree[p].mul % mod;
	tree[p << 1].add = (tree[p << 1].add * tree[p].mul + tree[p].add) % mod;
	tree[p << 1].sm = (tree[p << 1].sm * tree[p].mul + tree[p].add * (tree[p << 1].qr - tree[p << 1].ql + 1)) % mod;

	tree[p << 1 | 1].mul = tree[p << 1 | 1].mul * tree[p].mul % mod;
	tree[p << 1 | 1].add = (tree[p << 1 | 1].add * tree[p].mul + tree[p].add) % mod;
	tree[p << 1 | 1].sm = (tree[p << 1 | 1].sm * tree[p].mul + tree[p].add * (tree[p << 1 | 1].qr - tree[p << 1 | 1].ql + 1)) % mod;

	tree[p].mul = 1;
	tree[p].add = 0;
}
void make_tree(int p, int left, int right) {
	tree[p].ql = left;
	tree[p].qr = right;
	tree[p].mul = 1;
	if (left == right) {
		tree[p].sm = (long long)(a[left] % mod);
		return;
	}
	int mid = (left + right) >> 1;
	make_tree(p << 1, left, mid);
	make_tree(p << 1 | 1, mid + 1, right);
	pushup(p);
}
void mull(int p, int left, int right, int c) {
	if (tree[p].ql == left && tree[p].qr == right) {
		tree[p].mul = tree[p].mul * (long long)c % mod;
		tree[p].add = tree[p].add * (long long)c % mod;
		tree[p].sm = tree[p].sm * (long long)c % mod;
		return;
	}
	pushdown(p);
	int mid = (tree[p].ql + tree[p].qr) >> 1;
	if (right <= mid) {
		mull(p << 1, left, right, c);
	}
	else if (left > mid) {
		mull(p << 1 | 1, left, right, c);
	}
	else {
		mull(p << 1, left, mid, c);
		mull(p << 1 | 1, mid + 1, right, c);
	}
	pushup(p);
}
void addd(int p, int left, int right, int c) {
	if (tree[p].ql == left && tree[p].qr == right) {
		tree[p].add = (tree[p].add + (long long)c) % mod;
		tree[p].sm = (tree[p].sm + (long long)c * (long long)(tree[p].qr - tree[p].ql + 1)) % mod;
		return;
	}
	pushdown(p);
	int mid = (tree[p].ql + tree[p].qr) >> 1;
	if (right <= mid) {
		addd(p << 1, left, right, c);
	}
	else if (left > mid) {
		addd(p << 1 | 1, left, right, c);
	}
	else {
		addd(p << 1, left, mid, c);
		addd(p << 1 | 1, mid + 1, right, c);
	}
	pushup(p);
}
int qry(int p, int left, int right) {
	if (tree[p].ql == left && tree[p].qr == right) {
		return (int)tree[p].sm;
	}
	pushdown(p);
	int mid = (tree[p].ql + tree[p].qr) >> 1, rt;
	if (right <= mid) {
		rt = qry(p << 1, left, right);
	}
	else if (left > mid) {
		rt = qry(p << 1 | 1, left, right);
	}
	else {
		rt = qry(p << 1, left, mid);
		rt = (rt + qry(p << 1 | 1, mid + 1, right)) % mod;
	}
	pushup(p);
	return rt;
}

int main(void) {
	//freopen("in.txt", "r", stdin);
	//freopen("out.txt", "w", stdout);
	scanf("%d%d", &n, &mod);
	for (int i = 1; i <= n; ++i) {
		scanf("%d", a + i);
	}
	make_tree(1, 1, n);
	scanf("%d", &m);
	while (m--) {
		scanf("%d%d%d", &opr, &t1, &t2);
		if (opr == 1) {
			scanf("%d", &t3);
			mull(1, t1, t2, t3);
		}
		else if (opr == 2) {
			scanf("%d", &t3);
			addd(1, t1, t2, t3);
		}
		else {
			printf("%d\n", qry(1, t1, t2));
		}
	}
	return 0;
}