ACM-ICPC 2018 徐州赛区网络预赛 H. Ryuji doesn't want to study

最新推荐文章于 2019-04-22 17:36:05 发布

CaprYang

最新推荐文章于 2019-04-22 17:36:05 发布

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本文链接：https://blog.youkuaiyun.com/CaprYang/article/details/82592224

本文介绍了一种利用线段树解决特定区间查询问题的方法。针对原数组的n个元素进行q次操作，包括查询指定区间内加权和及更新指定位置的值。通过维护两种特殊和值，实现高效查询。

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题目连接
#题意
题目大意原数组a有n个数字q次操作
1查询[l, r]的和s 要求s = ∑a[i] * (r - i + 1)
2修改b位置的值为c

使用线段树维护两个值s和w s = ∑a[i] w = ∑a[i] * (n - i + 1)
查询时 w - s * (n - r)即为答案注意爆int
#AC代码

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

const int INF = 0x3f3f3f3f;
const int MAXN = 1e5 + 10;
int a[MAXN], n, q;

struct node
{
	int l, r;
	ll s, w;
}tree[MAXN << 2];

inline void PushUp(int x)
{
	tree[x].s = tree[x << 1].s + tree[x << 1 | 1].s;
	tree[x].w = tree[x << 1].w + tree[x << 1 | 1].w;
}
void Build(int x, int l, int r)
{
	tree[x].l = l, tree[x].r = r, tree[x].s = tree[x].w = 0;
	if (l == r)
		tree[x].s = a[l], tree[x].w = 1LL * a[l] * (n - l + 1); //爆int
	else
	{
		int m = l + r >> 1;
		Build(x << 1, l, m);
		Build(x << 1 | 1, m + 1, r);
		PushUp(x);
	}
}
void Update(int x, int l, int r, ll v)
{
	int xl = tree[x].l, xr = tree[x].r;
	if (xl == xr)
		tree[x].s = v, tree[x].w = v * (n - xl + 1);
	else
	{
		int xm = xl + xr >> 1;
		if (xm >= l)
			Update(x << 1, l, r, v);
		else
			Update(x << 1 | 1, l, r, v);
		PushUp(x);
	}
}
void Query(int x, int l, int r, ll &s, ll &w)
{
	int xl = tree[x].l, xr = tree[x].r;
	if (l <= xl && r >= xr)
		s += tree[x].s, w += tree[x].w;
	else
	{
		int xm = xl + xr >> 1;
		if (xm >= l)
			Query(x << 1, l, r, s, w);
		if (xm < r)
			Query(x << 1 | 1, l, r, s, w);
	}
}
int main()
{
#ifdef LOCAL
	freopen("C:/input.txt", "r", stdin);
#endif
	cin >> n >> q;
	for (int i = 1; i <= n; i++)
		scanf("%d", &a[i]);
	Build(1, 1, n);
	for (int i = 0; i < q; i++)
	{
		ll s = 0, w = 0;
		int op, l, r;
		scanf("%d%d%d", &op, &l, &r);
		if (op == 1)
		{
			Query(1, l, r, s, w);
			printf("%lld\n", w - s * (n - r)); //公式
		}
		else
			Update(1, l, l, r);
	}

	return 0;
}