帮我计算max[(x1+x2+x3)/(c1**2+c2**2+c3**2)]的复杂度
时间: 2023-10-02 13:05:47 浏览: 54
假设x1、x2、x3、c1、c2、c3均为n位数。
首先,计算c1**2、c2**2、c3**2的复杂度为3n(平方运算需要进行n次乘法)。
然后,计算x1 x2 x3的复杂度为3n(乘法运算需要进行n次乘法)。
最后,计算(x1 x2 x3)/(c1**2 c2**2 c3**2)的复杂度为1次除法和3次乘法,即4n。
因此,max[(x1 x2 x3)/(c1**2 c2**2 c3**2)]的总复杂度为3n + 3n + 4n = 10n。
因此,该计算的复杂度为O(n)。
相关问题
c++ 三维 2线段最小距离 源代码
以下是C++的三维两线段最小距离的源代码:
```c++
#include <bits/stdc++.h>
using namespace std;
double DotProduct(double x1, double y1, double z1, double x2, double y2, double z2) {
return x1 * x2 + y1 * y2 + z1 * z2;
}
double CrossProduct(double x1, double y1, double z1, double x2, double y2, double z2, double& x, double& y, double& z) {
x = y1 * z2 - y2 * z1;
y = z1 * x2 - z2 * x1;
z = x1 * y2 - x2 * y1;
return sqrt(x * x + y * y + z * z);
}
double Distance(double x1, double y1, double z1, double x2, double y2, double z2) {
return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2) + (z1 - z2) * (z1 - z2));
}
double SegmentAndSegment(double x1, double y1, double z1, double x2, double y2, double z2, double x3, double y3, double z3, double x4, double y4, double z4) {
double x, y, z;
double a1 = x2 - x1;
double b1 = y2 - y1;
double c1 = z2 - z1;
double a2 = x4 - x3;
double b2 = y4 - y3;
double c2 = z4 - z3;
double a3 = x1 - x3;
double b3 = y1 - y3;
double c3 = z1 - z3;
double d1 = DotProduct(a1, b1, c1, a2, b2, c2);
double d2 = DotProduct(a1, b1, c1, a1, b1, c1);
double d3 = DotProduct(a2, b2, c2, a2, b2, c2);
double d4 = DotProduct(a1, b1, c1, a3, b3, c3);
double d5 = DotProduct(a2, b2, c2, a3, b3, c3);
double d = d2 * d3 - d1 * d1;
double t1, t2;
if (d == 0) {
t1 = 0;
t2 = (d1 > d3 ? d1 / d2 : d3 / d2);
} else {
t1 = (d1 * d5 - d2 * d4) / d;
t2 = (d1 * t1 + d5) / d3;
}
t1 = max(0.0, min(t1, 1.0));
t2 = max(0.0, min(t2, 1.0));
double u1 = x1 + t1 * a1;
double v1 = y1 + t1 * b1;
double w1 = z1 + t1 * c1;
double u2 = x3 + t2 * a2;
double v2 = y3 + t2 * b2;
double w2 = z3 + t2 * c2;
return Distance(u1, v1, w1, u2, v2, w2);
}
int main() {
double x1, y1, z1, x2, y2, z2, x3, y3, z3, x4, y4, z4;
cin >> x1 >> y1 >> z1 >> x2 >> y2 >> z2 >> x3 >> y3 >> z3 >> x4 >> y4 >> z4;
cout << SegmentAndSegment(x1, y1, z1, x2, y2, z2, x3, y3, z3, x4, y4, z4) << endl;
return 0;
}
```
其中,`DotProduct`函数计算向量的点积,`CrossProduct`函数计算向量的叉积,`Distance`函数计算两点之间的距离,`SegmentAndSegment`函数计算两条线段之间的最小距离。
基于灰狼优化算法的2维边坡最危险滑面搜索代码
好的,针对您的问题,我可以提供一份基于灰狼优化算法的2维边坡最危险滑面搜索的代码。
不过在此之前,我需要先说明一下,灰狼优化算法(Grey Wolf Optimizer, GWO)是一种基于群体智能的优化算法,灵感来源于灰狼的捕猎行为。该算法在解决连续优化问题方面表现出色,并且具有全局搜索能力和高精度搜索能力,因此被广泛应用于各种实际问题中。
而2维边坡最危险滑面搜索则是指在已知边坡形状和土体力学参数的情况下,寻找该边坡上能够产生最大稳定力矩的滑动面。
下面是基于灰狼优化算法的2维边坡最危险滑面搜索代码,仅供参考:
```python
import numpy as np
# 定义待优化的函数
def obj_func(x, y):
# 计算滑动面的稳定力矩
# 这里的计算方法可以根据实际问题进行修改
return -1*(x**2 + y**2)
# 灰狼优化算法的实现
def GWO(obj_func, lb, ub, dim, SearchAgents_no, Max_iter):
Alpha_pos = np.zeros(dim)
Alpha_score = float("inf")
Beta_pos = np.zeros(dim)
Beta_score = float("inf")
Delta_pos = np.zeros(dim)
Delta_score = float("inf")
Positions = np.zeros((SearchAgents_no, dim))
for i in range(dim):
Positions[:, i] = np.random.uniform(0, 1, SearchAgents_no) * (ub[i] - lb[i]) + lb[i]
Convergence_curve = np.zeros(Max_iter)
for l in range(0, Max_iter):
for i in range(0, SearchAgents_no):
# 更新Alpha,Beta和Delta狼的位置和得分
if (obj_func(Positions[i, 0], Positions[i, 1]) < Alpha_score):
Alpha_score = obj_func(Positions[i, 0], Positions[i, 1])
Alpha_pos = Positions[i, :]
if (obj_func(Positions[i, 0], Positions[i, 1]) > Alpha_score and obj_func(Positions[i, 0], Positions[i, 1]) < Beta_score):
Beta_score = obj_func(Positions[i, 0], Positions[i, 1])
Beta_pos = Positions[i, :]
if (obj_func(Positions[i, 0], Positions[i, 1]) > Alpha_score and obj_func(Positions[i, 0], Positions[i, 1]) > Beta_score and obj_func(Positions[i, 0], Positions[i, 1]) < Delta_score):
Delta_score = obj_func(Positions[i, 0], Positions[i, 1])
Delta_pos = Positions[i, :]
a = 2 - 2 * l / (Max_iter - 1)
# 更新所有狼的位置
for i in range(0, SearchAgents_no):
for j in range(0, dim):
r1 = np.random.rand()
r2 = np.random.rand()
A1 = 2 * a * r1 - a
C1 = 2 * r2
D_alpha = abs(C1 * Alpha_pos[j] - Positions[i, j])
X1 = Alpha_pos[j] - A1 * D_alpha
r1 = np.random.rand()
r2 = np.random.rand()
A2 = 2 * a * r1 - a
C2 = 2 * r2
D_beta = abs(C2 * Beta_pos[j] - Positions[i, j])
X2 = Beta_pos[j] - A2 * D_beta
r1 = np.random.rand()
r2 = np.random.rand()
A3 = 2 * a * r1 - a
C3 = 2 * r2
D_delta = abs(C3 * Delta_pos[j] - Positions[i, j])
X3 = Delta_pos[j] - A3 * D_delta
Positions[i, j] = (X1 + X2 + X3) / 3
Convergence_curve[l] = Alpha_score
return Alpha_pos
# 设置边界和参数
lb = [-5, -5]
ub = [5, 5]
dim = 2
SearchAgents_no = 30
Max_iter = 100
# 调用灰狼优化算法
result = GWO(obj_func, lb, ub, dim, SearchAgents_no, Max_iter)
# 输出结果
print("最危险滑面的坐标为:", result)
print("最大稳定力矩为:", -1*obj_func(result[0], result[1]))
```
需要注意的是,这份代码仅为基本实现,具体的滑动面稳定力矩计算方法和边坡形状、土体力学参数的输入方式等都需要根据具体问题进行修改。