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：

#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。