网站是否备案怎么查询企业网搭建及应用

 

 

思路：

首先他是实数

1. 线段树结构：

   • 每个节点存储三个关键值：区间和sum、区间平方和sum2、延迟标记lazy。
   • sum用于计算区间平均数，sum2和sum一起用于计算区间方差。

2. 延迟标记优化：

   • 当需要更新一个区间时，不立即更新所有子节点，而是将更新操作存储在延迟标记中。
   • 当需要访问某个节点时，先将延迟标记下传到子节点，确保数据的正确性。

3. 方差计算：

   • 方差公式：
   • 利用线段树维护的sum和sum2，可以高效计算这两个部分。

#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<iostream>
#include<bits/stdc++.h>
using namespace std;
int n, m;
struct {double sum;double lazy;double sum2;
}  a[400005];
double b[100005];
void Chu(int x, int y, int i) {if (x == y) {a[i].sum = b[x];a[i].sum2 = pow(b[x], 2);a[i].lazy = 0;return;}int mid = (x + y) / 2;Chu(x, mid, i * 2);Chu(mid + 1, y, i * 2 + 1);a[i].sum2 = a[i * 2].sum2 + a[i * 2 + 1].sum2;a[i].sum = a[i * 2].sum + a[i * 2 + 1].sum;a[i].lazy = 0;
}
void Xiu(int x, int y, int i, int l, int r,  double k) {if (l <= x && y <= r) {a[i].sum2 += 2 * a[i].sum * k + (y - x + 1) * pow(k, 2);a[i].sum += k * (y - x + 1);a[i].lazy += k;return;}int mid = (x + y) / 2;if (a[i].lazy != 0) {a[i * 2].sum2 += 2 * a[i * 2].sum * a[i].lazy + (mid - x + 1) * pow(a[i].lazy, 2);a[i * 2].sum += a[i].lazy * (mid - x + 1);a[i * 2].lazy += a[i].lazy;a[i * 2 + 1].sum2 += 2 * a[i * 2 + 1].sum * a[i].lazy + (y - mid) * pow(a[i].lazy, 2);a[i * 2 + 1].sum += a[i].lazy * (y - mid);a[i * 2 + 1].lazy += a[i].lazy;a[i].lazy = 0;}if (mid >= l) {Xiu(x, mid, i * 2, l, r, k);}if (mid + 1 <= r) {Xiu(mid + 1, y, i * 2 + 1, l, r, k);}a[i].sum2 = a[i * 2].sum2 + a[i * 2 + 1].sum2;a[i].sum = a[i * 2].sum + a[i * 2 + 1].sum;
}
double Chasum(int x, int y, int i, int l, int r) {if (l <= x && y <= r) {return a[i].sum;}int mid = (x + y) / 2;if (a[i].lazy != 0) {a[i * 2].sum2 += 2 * a[i * 2].sum * a[i].lazy + (mid - x + 1) * pow(a[i].lazy, 2);a[i * 2].sum += a[i].lazy * (mid - x + 1);a[i * 2].lazy += a[i].lazy;a[i * 2+1].sum2 += 2 * a[i * 2+1].sum * a[i].lazy + (y-mid) * pow(a[i].lazy, 2);a[i * 2 + 1].sum += a[i].lazy * (y - mid);a[i * 2 + 1].lazy += a[i].lazy;a[i].lazy = 0;}double sum = 0;if (mid >= l) {sum += Chasum(x, mid, i * 2, l, r);}if (mid + 1 <= r) {sum += Chasum(mid + 1, y, i * 2 + 1, l, r);}return sum;
}
double Chasum2(int x, int y, int i, int l, int r) {if (l <= x && y <= r) {return a[i].sum2;}int mid = (x + y) / 2;if (a[i].lazy != 0) {a[i * 2].sum2 += 2 * a[i * 2].sum * a[i].lazy + (mid - x + 1) * pow(a[i].lazy, 2);a[i * 2].sum += a[i].lazy * (mid - x + 1);a[i * 2].lazy += a[i].lazy;a[i * 2 + 1].sum2 += 2 * a[i * 2 + 1].sum * a[i].lazy + (y - mid) * pow(a[i].lazy, 2);a[i * 2 + 1].sum += a[i].lazy * (y - mid);a[i * 2 + 1].lazy += a[i].lazy;a[i].lazy = 0;}double sum = 0;if (mid >= l) {sum += Chasum2(x, mid, i * 2, l, r);}if (mid + 1 <= r) {sum += Chasum2(mid + 1, y, i * 2 + 1, l, r);}return sum;
}
int main() {int l, r, h;double k;cin >> n >> m;for (int i = 1; i <= n; i++) {cin >> b[i];}Chu(1, n, 1);for (int i = 1; i <= m; i++) {cin >> h;if (h == 1) {cin >> l >> r >> k;Xiu(1, n, 1, l, r, k);}else if(h==2){cin >> l >> r;cout <<fixed<<setprecision(4)<< Chasum(1, n, 1, l, r)/(double)(r-l+1) << endl;}else {cin >> l >> r;double average = Chasum(1, n, 1, l, r) / (double)(r - l + 1);double sum = Chasum2(1, n, 1, l, r) / (double)(r - l + 1) - pow(average, 2);cout << fixed << setprecision(4) << sum << endl;}}return 0;
}