方程组：x1a1-x2a2=0;x2a3-x3a4=0;x3a5-x4a6=0;x4a7-x1a8=0；约束方程x1+x2+x3+x4=4。其中a1到a8分别是：152.967061,177.396563,147.882339,174.300663,89.771655,119.448549,98.088508,114.954773。要求x1+x2+x3+x4的值相加为4，且解出来的x1、x2、x3、x4必须大于0。请用c++代码编写程序解出x1,x2,x3,x4的值。

以下是使用C++解方程组的代码：

#include <iostream>
#include <cmath>

using namespace std;

double a[8] = {152.967061, 177.396563, 147.882339, 174.300663, 89.771655, 119.448549, 98.088508, 114.954773};

double f(double x1, double x2, double x3, double x4) {
    return pow(x1 * a[0] / x2 / a[1], 0.5) * pow(x2 * a[2] / x3 / a[3], 0.5) * pow(x3 * a[4] / x4 / a[5], 0.5) * pow(x4 * a[6] / x1 / a[7], 0.5);
}

double g(double x1, double x2, double x3, double x4) {
    return x1 + x2 + x3 + x4 - 4;
}

int main() {
    double x1 = 1, x2 = 1, x3 = 1, x4 = 1;
    double eps = 1e-8;
    double step = 0.001;
    while (abs(g(x1, x2, x3, x4)) > eps || x1 <= 0 || x2 <= 0 || x3 <= 0 || x4 <= 0) {
        double fx1 = f(x1 + step, x2, x3, x4);
        double fx2 = f(x1, x2 + step, x3, x4);
        double fx3 = f(x1, x2, x3 + step, x4);
        double fx4 = f(x1, x2, x3, x4 + step);
        double gx1 = g(x1 + step, x2, x3, x4);
        double gx2 = g(x1, x2 + step, x3, x4);
        double gx3 = g(x1, x2, x3 + step, x4);
        double gx4 = g(x1, x2, x3, x4 + step);
        double dx1 = step / fx1 * f(x1, x2, x3, x4) * gx1;
        double dx2 = step / fx2 * f(x1, x2, x3, x4) * gx2;
        double dx3 = step / fx3 * f(x1, x2, x3, x4) * gx3;
        double dx4 = step / fx4 * f(x1, x2, x3, x4) * gx4;
        x1 -= dx1;
        x2 -= dx2;
        x3 -= dx3;
        x4 -= dx4;
    }
    cout << "x1 = " << x1 << endl;
    cout << "x2 = " << x2 << endl;
    cout << "x3 = " << x3 << endl;
    cout << "x4 = " << x4 << endl;
    return 0;
}

在该程序中，我们使用了牛顿迭代法求解方程组。我们定义了一个目标函数f(x1, x2, x3, x4)，它是方程组中每个方程的左右两边相乘起来的结果。我们还定义了一个约束条件g(x1, x2, x3, x4)，它是四个变量之和减4的结果，表示四个变量之和必须为4。我们使用牛顿迭代法不断更新x1、x2、x3、x4的值，直到约束条件g(x1, x2, x3, x4)满足要求并且x1、x2、x3、x4大于0为止。最终的解就是x1、x2、x3、x4的值。