上帝造题的七分钟 2 / 花神游历各国

lazy 标记是否还要开方

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int MAXN = 100000;
int n, m, a[MAXN + 5];
int t[MAXN * 4 + 5];
bool lazy[MAXN * 4 + 5]; // 是否还需要开方
// 当前节点为 now，对应的线段为 [l,r]
void build(int now, int l, int r)
{
	lazy[now] = true;
	if (l == r)
	{
		t[now] = a[l];
		return;
	}
	int mid = (l + r) / 2;
	build(now * 2, l, mid);
	build(now * 2 + 1, mid + 1, r);
	t[now] = t[now * 2] + t[now * 2 + 1];
}
// 做到了 now 节点，对应区间为 [l,r]
// 单点修改：a[x]~a[y] 开方
void update(int now, int l, int r, int x, int y)
{
	if (lazy[now] == false)
		return;
	if (l == r)
	{
		t[now] = sqrt(t[now]);
		if (t[now] <= 1)
			lazy[now] = false;
	}
	else
	{
		int mid = (l + r) / 2;
		if (x <= mid)
			update(now * 2, l, mid, x, y);
		if (y >= mid + 1)
			update(now * 2 + 1, mid + 1, r, x, y);
		t[now] = t[now * 2] + t[now * 2 + 1];
		lazy[now] = lazy[now * 2] || lazy[now * 2 + 1];
	}
}
int query(int now, int l, int r, int x, int y)
{
	if (x <= l && r <= y)
		return t[now];
	int mid = (l + r) / 2;
	int res = 0;
	if (x <= mid)
		res += query(now * 2, l, mid, x, y);
	if (y >= mid + 1)
		res += query(now * 2 + 1, mid + 1, r, x, y);
	return res;
}
signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin >> n;
	for (int i = 1; i <= n; i++)
		cin >> a[i];
	build(1, 1, n);
	cin >> m;
	while (m--)
	{
		int op, x, y, z;
		cin >> op;
		if (op == 0)
		{
			cin >> x >> y;
			if (x > y)
				swap(x, y);
			update(1, 1, n, x, y);
		}
		else
		{
			cin >> x >> y;
			if (x > y)
				swap(x, y);
			cout << query(1, 1, n, x, y) << "\n";
		}
	}
	return 0;
}

用区间和判断

这题的数不会变为 0，所以不需要开方的条件就是区间和等于区间长度（全都是 1）

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int MAXN = 100000;
int n, m, a[MAXN + 5];
int t[MAXN * 4 + 5];
int lazy[MAXN * 4 + 5];
void build(int now, int l, int r)
{
	if (l == r)
	{
		t[now] = a[l];
		return;
	}
	int mid = (l + r) / 2;
	build(now * 2, l, mid);
	build(now * 2 + 1, mid + 1, r);
	t[now] = t[now * 2] + t[now * 2 + 1];
}
void update(int now, int l, int r, int x, int y, int z)
{
	if (t[now] == r - l + 1)
		return;
	if (l == r)
	{
		t[now] = sqrt(t[now]);
		return;
	}
	int mid = (l + r) / 2;
	if (x <= mid)
		update(now * 2, l, mid, x, y, z);
	if (y >= mid + 1)
		update(now * 2 + 1, mid + 1, r, x, y, z);
	t[now] = t[now * 2] + t[now * 2 + 1];
}
int query(int now, int l, int r, int x, int y)
{
	if (x <= l && r <= y)
		return t[now];
	int mid = (l + r) / 2;
	int res = 0;
	if (x <= mid)
		res += query(now * 2, l, mid, x, y);
	if (y >= mid + 1)
		res += query(now * 2 + 1, mid + 1, r, x, y);
	return res;
}
signed main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin >> n;
	for (int i = 1; i <= n; i++)
		cin >> a[i];
	build(1, 1, n);
	cin >> m;
	while (m--)
	{
		int op, x, y, z;
		cin >> op;
		if (op == 0)
		{
			cin >> x >> y;
			if (x > y)
				swap(x, y);
			update(1, 1, n, x, y, z);
		}
		else
		{
			cin >> x >> y;
			if (x > y)
				swap(x, y);
			cout << query(1, 1, n, x, y) << "\n";
		}
	}
	return 0;
}