已知y1 = z1 * (y1 + y2 + y3); y2 = z2 * (y1 + y2 + y3); y3 = z3 * (y1 + y2 + y3), 对于上面3条公式,已知z1, z2, z3,如何联合起来求y1, y2, y3 ? 请用C++代码实现求解
时间: 2024-01-18 16:04:35 浏览: 31
可以使用迭代的方式求解,具体实现如下:
```c++
#include <iostream>
using namespace std;
void solve(double z1, double z2, double z3, double& y1, double& y2, double& y3) {
double eps = 1e-6;
int max_iter = 1000;
double y1_new, y2_new, y3_new;
y1_new = y2_new = y3_new = 1.0;
for (int i = 0; i < max_iter; i++) {
double y1_old = y1_new, y2_old = y2_new, y3_old = y3_new;
y1_new = z1 * (y1_old + y2_old + y3_old);
y2_new = z2 * (y1_old + y2_old + y3_old);
y3_new = z3 * (y1_old + y2_old + y3_old);
if (abs(y1_new - y1_old) < eps && abs(y2_new - y2_old) < eps && abs(y3_new - y3_old) < eps)
break;
}
y1 = y1_new;
y2 = y2_new;
y3 = y3_new;
}
int main() {
double z1 = 0.2, z2 = 0.3, z3 = 0.5;
double y1, y2, y3;
solve(z1, z2, z3, y1, y2, y3);
cout << "y1 = " << y1 << endl;
cout << "y2 = " << y2 << endl;
cout << "y3 = " << y3 << endl;
return 0;
}
```
其中,输入的三个系数 `z1`、`z2`、`z3` 可以通过用户输入或其他方式获取。代码中的 `solve` 函数使用迭代的方式求解线性方程组。迭代停止的条件为三个未知数的相邻两次迭代值的差的绝对值小于给定的精度 `eps`。最大迭代次数 `max_iter` 可以根据实际情况进行调整。