需要求解y1,y2,y3. 已知方程: z1=y1 / ( y1 + y2 + y3 ); z2=y2 / ( y1 + y2 + y3 ) ;z3=y3 / ( y1 + y2 + y3 ); 假设 z1、z2、z3是已知值,y1,y2,y3,z1,z2,z3均是大于零的正值。且z1 + z2 + z3 =1。请列举C++实现代码
时间: 2024-03-02 15:51:39 浏览: 23
以下是一个简单的 C++ 实现代码,用于求解 y1、y2、y3:
```cpp
#include <iostream>
using namespace std;
int main() {
double z1, z2, z3, y1, y2, y3;
cout << "请输入 z1、z2、z3 的值:" << endl;
cin >> z1 >> z2 >> z3;
double sum = z1 + z2 + z3;
if (sum == 0) {
cout << "z1 + z2 + z3 的值不能为 0!" << endl;
return 0;
}
y1 = z1 / sum * (y1 + y2 + y3);
y2 = z2 / sum * (y1 + y2 + y3);
y3 = z3 / sum * (y1 + y2 + y3);
cout << "y1 = " << y1 << endl;
cout << "y2 = " << y2 << endl;
cout << "y3 = " << y3 << endl;
return 0;
}
```
这个程序首先读入 z1、z2、z3 的值,然后计算它们的和。如果和为 0,则输出错误信息;否则,根据公式计算 y1、y2、y3 的值,并输出结果。
相关问题
z1=y1/(y1+y2+y3) z2=y2/(y1+y2+y3) z3=y3/(y1+y2+y3) 已知z1和z2和z3, 且上述方程已知数和未知数均是大于0的正值,求y1,y2,y3。 请问,怎么在C++代码用牛顿迭代法实现求解?
可以使用牛顿迭代法进行求解,步骤如下:
1. 定义目标函数f(y1, y2, y3) = [z1 - y1/(y1+y2+y3)]^2 + [z2 - y2/(y1+y2+y3)]^2 + [z3 - y3/(y1+y2+y3)]^2,目标是使得f(y1, y2, y3)最小。
2. 求解f(y1, y2, y3)的梯度向量g(y1, y2, y3),以及Hessian矩阵H(y1, y2, y3)。
3. 初始化y1, y2, y3为任意正数,然后进行迭代。在每次迭代中,计算g(y1, y2, y3)和H(y1, y2, y3),然后计算方程H(y1, y2, y3) * delta = -g(y1, y2, y3)的解delta,更新y1, y2, y3为y1+delta[0], y2+delta[1], y3+delta[2]。
4. 重复第3步,直到f(y1, y2, y3)收敛到足够小的值。
以下是使用牛顿迭代法求解的示例代码:
```c++
#include <iostream>
#include <cmath>
using namespace std;
double f(double y1, double y2, double y3, double z1, double z2, double z3) {
double denom = y1 + y2 + y3;
double f1 = z1 - y1 / denom;
double f2 = z2 - y2 / denom;
double f3 = z3 - y3 / denom;
return f1 * f1 + f2 * f2 + f3 * f3;
}
void g(double y1, double y2, double y3, double z1, double z2, double z3, double& g1, double& g2, double& g3) {
double denom = y1 + y2 + y3;
g1 = -2 * (z1 - y1 / denom) / denom + 2 * y1 / (denom * denom);
g2 = -2 * (z2 - y2 / denom) / denom + 2 * y2 / (denom * denom);
g3 = -2 * (z3 - y3 / denom) / denom + 2 * y3 / (denom * denom);
}
void H(double y1, double y2, double y3, double z1, double z2, double z3, double& h11, double& h12, double& h13,
double& h21, double& h22, double& h23, double& h31, double& h32, double& h33) {
double denom = y1 + y2 + y3;
double denom2 = denom * denom;
h11 = 2 * (2 * y1 * y1 - 2 * z1 * y1 + y2 * y2 + y3 * y3) / (denom2 * denom) - 2 * y1 * y1 / (denom2 * denom2);
h12 = 2 * y1 * y2 / (denom2 * denom) - 2 * y1 * y2 / (denom2 * denom2);
h13 = 2 * y1 * y3 / (denom2 * denom) - 2 * y1 * y3 / (denom2 * denom2);
h21 = h12;
h22 = 2 * (y1 * y1 + 2 * y2 * y2 - 2 * z2 * y2 + y3 * y3) / (denom2 * denom) - 2 * y2 * y2 / (denom2 * denom2);
h23 = 2 * y2 * y3 / (denom2 * denom) - 2 * y2 * y3 / (denom2 * denom2);
h31 = h13;
h32 = h23;
h33 = 2 * (y1 * y1 + y2 * y2 + 2 * y3 * y3 - 2 * z3 * y3) / (denom2 * denom) - 2 * y3 * y3 / (denom2 * denom2);
}
int main() {
double z1 = 0.3;
double z2 = 0.4;
double z3 = 0.3;
double y1 = 1.0;
double y2 = 1.0;
double y3 = 1.0;
double eps = 1e-6;
int max_iter = 1000;
int iter = 0;
double g1, g2, g3, h11, h12, h13, h21, h22, h23, h31, h32, h33;
double delta1, delta2, delta3;
do {
g(y1, y2, y3, z1, z2, z3, g1, g2, g3);
H(y1, y2, y3, z1, z2, z3, h11, h12, h13, h21, h22, h23, h31, h32, h33);
double det = h11 * h22 * h33 + h12 * h23 * h31 + h21 * h32 * h13
- h13 * h22 * h31 - h12 * h21 * h33 - h11 * h23 * h32;
delta1 = (-h23 * h32 * g1 + h22 * h33 * g2 - h22 * h33 * g3) / det;
delta2 = (h13 * h32 * g1 - h12 * h33 * g2 + h12 * h33 * g3) / det;
delta3 = (-h13 * h22 * g1 + h12 * h23 * g2 - h12 * h23 * g3) / det;
y1 += delta1;
y2 += delta2;
y3 += delta3;
iter++;
} while (iter < max_iter && sqrt(g1 * g1 + g2 * g2 + g3 * g3) > eps);
std::cout << "y1 = " << y1 << std::endl;
std::cout << "y2 = " << y2 << std::endl;
std::cout << "y3 = " << y3 << std::endl;
return 0;
}
```
在每次迭代中,使用g(y1, y2, y3, z1, z2, z3, g1, g2, g3)计算梯度向量g(y1, y2, y3),使用H(y1, y2, y3, z1, z2, z3, h11, h12, h13, h21, h22, h23, h31, h32, h33)计算Hessian矩阵H(y1, y2, y3)。然后使用牛顿迭代公式delta = -H^{-1}g更新y1, y2, y3的值。重复这个过程,直到收敛到足够小的值。
假设已知方程y1 = z1 * (y1 + y2 + y3); y2 = z2 * (y1 + y2 + y3); y3 = z3 * (y1 + y2 + y3),但是z1,z2,z3仅保留小数点后3位,如何反推出y1,y2,y3的最优解?请用C++代码实现
以下是使用迭代法求解方程组的C++代码实现:
```cpp
#include <iostream>
#include <cmath>
using namespace std;
const double eps = 1e-6; // 收敛阈值
double z1 = 0.123;
double z2 = 0.234;
double z3 = 0.345;
double y1 = 1.0; // 初始化y1,y2,y3的值
double y2 = 2.0;
double y3 = 3.0;
double calc_z1(double y1, double y2, double y3)
{
return round(z1 * (y1 + y2 + y3) * 1000) / 1000; // 保留小数点后3位
}
double calc_z2(double y1, double y2, double y3)
{
return round(z2 * (y1 + y2 + y3) * 1000) / 1000; // 保留小数点后3位
}
double calc_z3(double y1, double y2, double y3)
{
return round(z3 * (y1 + y2 + y3) * 1000) / 1000; // 保留小数点后3位
}
int main()
{
double y1_new, y2_new, y3_new;
int iter = 0;
do {
y1_new = calc_z1(y1, y2, y3) / (1 - z1);
y2_new = calc_z2(y1, y2, y3) / (1 - z2);
y3_new = calc_z3(y1, y2, y3) / (1 - z3);
double err = max(abs(y1_new - y1), max(abs(y2_new - y2), abs(y3_new - y3)));
y1 = y1_new;
y2 = y2_new;
y3 = y3_new;
iter++;
} while (err > eps && iter < 1000);
cout << "y1 = " << y1 << endl;
cout << "y2 = " << y2 << endl;
cout << "y3 = " << y3 << endl;
return 0;
}
```
注意,这里使用了`round`函数来保留小数点后3位,同时迭代次数不宜过多,避免死循环。