【洛谷 P3792】由乃与大母神原型和偶像崇拜:线段树 + 思维 +哈希-优快云博客

这篇博客介绍了如何利用平方和哈希以及线段树数据结构来高效地处理区间平方和查询。文章首先阐述了问题背景，然后详细解释了线段树的构建与更新过程，接着展示了查询最大值和最小值的实现。通过实例演示了如何在O(logn)的时间复杂度内完成修改和查询操作。最后，通过样例展示了如何判断区间内的平方和是否为完全平方数。

传送门

分析

用平方和哈希，线段树维护即可

代码

#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define debug(x) cout<<#x<<":"<<x<<endl;
#define dl(x) printf("%lld\n",x);
#define di(x) printf("%d\n",x);
#define _CRT_SECURE_NO_WARNINGS
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef vector<int> VI;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const int N = 5e5 + 10;
const ll mod = 1000000007;
const double eps = 1e-9;
const double PI = acos(-1);
template<typename T>inline void read(T &a) {
	char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}
	while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}
int gcd(int a, int b) {return (b > 0) ? gcd(b, a % b) : a;}
int n;
ll a[N];
struct Node {
	int l, r;
	ll max, min;
	ll sum;
} tr[N * 4];

void push(int u) {
	tr[u].sum = tr[u << 1].sum + tr[u << 1 | 1].sum;
	tr[u].max = max(tr[u << 1].max , tr[u << 1 | 1].max);
	tr[u].min = min(tr[u << 1].min , tr[u << 1 | 1].min);
}

void build(int u, int l, int r) {
	tr[u] = {l, r};
	if (l == r) {
		tr[u].sum = a[l] * a[l];
		tr[u].max = tr[u].min = a[l];
		return;
	}
	int mid = l + r >> 1;
	build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
	push(u);
}

void modify(int u, int p, ll x) {
	if (tr[u].l == tr[u].r) {
		tr[u].sum = x * x;
		tr[u].max = tr[u].min = x;
		tr[u].min = tr[u].min = x;
		return;
	}
	int mid = tr[u].l + tr[u].r >> 1;
	if (p <= mid) modify(u << 1, p, x);
	else modify(u << 1 | 1, p, x);
	push(u);
}

ll query_max(int u, int l, int r) {
	if (tr[u].l >= l && tr[u].r <= r) {
		return tr[u].max;
	}
	int mid = tr[u].l + tr[u].r >> 1;
	ll res = -INF;
	if (l <= mid) res = query_max(u << 1, l, r);
	if (r > mid) res = max(res, query_max(u << 1 | 1, l, r));
	return res;
}


ll query_min(int u, int l, int r) {
	if (tr[u].l >= l && tr[u].r <= r) {
		return tr[u].min;
	}
	int mid = tr[u].l + tr[u].r >> 1;
	ll res = INF;
	if (l <= mid) res = query_min(u << 1, l, r);
	if (r > mid) res = min(query_min(u << 1 | 1, l, r), res);
	return res;
}


ll getsum(ll l, ll r) {
	return r * (r + 1) * (2 * r + 1) - l * (l + 1) * (2 * l + 1) ;
}



ll query(int u, int l, int r) {
	if (tr[u].l >= l && tr[u].r <= r) {
		return tr[u].sum;
	}
	int mid = tr[u].l + tr[u].r >> 1;
	ll res = 0;
	if (l <= mid) res = query(u << 1, l, r);
	if (r > mid) res += query(u << 1 | 1, l, r);
	return res;
}

int main() {
	int n, m;
	read(n), read(m);
	for (int i = 1; i <= n; i++) read(a[i]);
		build(1, 1, n);
	while (m--) {
		int op, x, y;
		read(op), read(x), read(y);
		if (op == 1) modify(1, x, y);
		else {
			ll ppp = query(1, x, y);
			ll MIN = query_min(1, x, y), MAX = query_max(1, x, y);
			ll pppp = getsum(MIN - 1, MAX);
// 			cout << MIN << ' ' << MAX << ' ' << pppp << endl;
			if (ppp * 6 == pppp) puts("damushen");
			else puts("yuanxing");
		}
	}
	return 0;
}

/**
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*　┏┛┻━━━┛┻┓ + +
*　┃　　　　　　　┃
*　┃　　　━　　　┃ ++ + + +
*  ████━████+
*  ◥██◤　◥██◤ +
*　┃　　　┻　　　┃
*　┃　　　　　　　┃ + +
*　┗━┓　　　┏━┛
*　　　┃　　　┃ + + + +Code is far away from 　
*　　　┃　　　┃ + bug with the animal protecting
*　　　┃　 　 ┗━━━┓ 神兽保佑,代码无bug　
*　　　┃ 　　　　　　 ┣┓
*　　  ┃ 　　　　　 　┏┛
*　    ┗┓┓┏━┳┓┏┛ + + + +
*　　　　┃┫┫　┃┫┫
*　　　　┗┻┛　┗┻┛+ + + +
*/