【线段树维护复杂状态】Ryuji doesn't want to study

https://nanti.jisuanke.com/t/31460

tree[rt].ans = tree[rt << 1].ans + tree[rt << 1 | 1].ans + tree[rt << 1].sum * tree[rt << 1 | 1].len;

注意询问时的合并操作

代码:

#include <iostream>
#include <cstdio>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const int maxn = (int)1e5 + 5;
typedef long long ll;
using namespace std;
int n, q;
ll arr[maxn];
struct Tree {
	ll sum, ans, len;
} tree[maxn << 2];
void pushup(int rt) {
	tree[rt].sum = tree[rt << 1].sum + tree[rt << 1 | 1].sum;
	tree[rt].ans = tree[rt << 1].ans + tree[rt << 1 | 1].ans + tree[rt << 1].sum * tree[rt << 1 | 1].len;
}
void build(int l, int r, int rt) {
	tree[rt].len = r - l + 1;
	if (l == r) {
		tree[rt].sum = tree[rt].ans = arr[l];
		return;
	}
	int m = (l + r) >> 1;
	build(lson);
	build(rson);
	pushup(rt);
}
void update(int p, ll val, int l, int r, int rt) {
	if (l == r) {
		tree[rt].ans = tree[rt].sum = val;
		return;
	}
	int m = (l + r) >> 1;
	if (p <= m) {
		update(p, val, lson);
	}
	else {
		update(p, val, rson);
	}
	pushup(rt);
}
ll _query(int L, int R, int l, int r, int rt) {
	if (L <= l && R >= r) {
		return tree[rt].sum;
	}
	int m = (l + r) >> 1;
	ll res = 0;
	if (L <= m) {
		res += _query(L, R, lson);
	}
	if (R > m) {
		res += _query(L, R, rson);
	}
	return res;
}
ll query(int L, int R, int l, int r, int rt) {
	if (L <= l && R >= r) {
		return tree[rt].ans;
	}
	int m = (l + r) >> 1;
	if (L <= m && R > m) {
		return query(L, m, lson) + query(m + 1, R, rson) + _query(L, m, lson) * (R - m);
	}
	else if (L <= m) {
		return query(L, R, lson);
	}
	else {
		return query(L, R, rson);
	}
}
int main() {
	scanf("%d%d", &n, &q);
	for (int i = 1; i <= n; i++) {
		scanf("%lld", &arr[i]);
	}
	build(1, n, 1);
	while (q--) {
		int op, a, b;
		scanf("%d%d%d", &op, &a, &b);
		if (op == 1) {
			printf("%lld\n", query(a, b, 1, n, 1));
		}
		else {
			update(a, b, 1, n, 1);
		}
	}
}
posted @ 2018-09-20 22:32  Stolf  阅读(173)  评论(0编辑  收藏  举报