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拜求大佬，这代码里面是什么意思，这是一个求一元高次方程的代码，用的什么方法，每个函数的作用是什么，可否帮忙解答一下！感谢

#include

#include

#include

using namespace std;

double x,error=1e-8;

int sign(double number) {

if (abs(number)

if (number>0) return 1;

else return -1;

}

void QD(int n, double a1[], double b1[][120])

{

int i, t = 0;

for (i = n - 2; i>=0; i--)

{

if (i == n - 2)

{

for (t = 0; t <= n; t++)

b1[i][t] = a1[t];

}

else {

for (t = 0; t <=i + 2; t++)

b1[i][t] = b1[i + 1][t] * (i - t + 3);

}

}

}

void QE(double h, double j, double k,int n,double r[][120])

{

double x1, x2,t;

t = j*j - 4 * h*k;

if (t == 0)

{

x1 = -j / (2 * h);

r[0][0] = 1;

r[0][1] = x1;

}

else if (t > 0)

{

x1 = (-j - sqrt(t)) / (2 * h);

x2 = (-j + sqrt(t)) /( 2 * h);

r[0][0] = 2;

r[0][1] = x1;

r[0][2] = x2;

}

else if (t <0)

{

r[0][0] = 0;

}

}

double ABX(double a, double b, double c[], int n)

{

double y1, y2, y3;

int i;

for (i = 0, y1 = y2 = 0; i

{

y1 = y1 + c[i] * pow(a, i);

y2 = y2 + c[i] * pow(b, i);

}

switch (sign(y1)*sign(y2))

{

case 0: return (abs(y1) < error) ? y1 : y2;

case 1: return 0;

case -1:

{ while (abs(y1)>error) {

double x = (a + b) / 2;

for (i = 0, y3 = 0; i < n + 1; i++)

{

y3 = y3 + c[i] * pow(x, i);

}

if (sign(y3) == 0) return x;

if (sign(y3)*sign(y1) == -1) {

b = x;

y2 = y3;

}

else {

a= x;

y1 = y3;

}

}

}

}

}

double QJ1(double min,double c[],int n)

{

double y1, y2,y=0, t = 0;

int i,j=1;

for (i = 0, y1=0; i

{

y1 = y1 + c[i] * pow(min, i);

}

t = min - j;

for (i = 0, y2 = 0; i

{

y2 = y2 + c[i] * pow(t, i);

}

if (abs(y2) < error)

{

return t;

}

if (y1<0 && y1 < y2)

{

while(y2<0)

{

j = 2 * j;

t = min - j;

for (i = 0, y1 = y2 = 0; i

{

y2 = y2 + c[i] * pow(t, i);

}

}

ABX(t,min,c,n);

return ABX(t, min, c, n);

}

else if (y1>0 && y1 > y2)

{

while (y2>0)

{

j = 2 * j;

t = min - j;

for (i = 0, y1 = y2 = 0; i

{

y2 = y2 + c[i] * pow(t, i);

}

}

ABX(t, min, c, n);

return ABX(t, min, c, n);

}

else return 0;

}

double QJ2(double max, double c[], int n)

{

double y1, y2, y=0, t = 0;

int i, j = 1;

for (i = 0, y1 = 0; i

{

y1 = y1 + c[i] * pow(max, i);

}

t = max +j;

for (i = 0, y2 = 0; i

{

y2 = y2 + c[i] * pow(t, i);

}

if (abs(y2) < 0.00000001)

{

return t;

}

if (y1<0 && y1 < y2)

{

while (y2<0)

{

j = 2 * j;

t = max + j;

for (i = 0, y1 = y2 = 0; i

{

y2 = y2 + c[i] * pow(t, i);

}

}

ABX(t,max, c, n);

return ABX(t, max, c, n);

}

else if (y1>0 && y1 > y2)

{

while (y2>0)

{

j = 2 * j;

t = max + j;

for (i = 0, y1 = y2 = 0; i

{

y2 = y2 + c[i] * pow(t, i);

}

}

ABX(t, max, c, n);

return ABX(t, max, c, n);

}

else return 0;

}

void CJ(double r[][120], int n, double h, double j, double k, double a1[120], double b1[][120])

{

double c[120];

double q;

int i, t = 1, z, p=0,j1=0;

QE(h, j, k, n, r);

if (r[0][0] != 0)

{

for (i = 1; i <= n - 2; i++)

{

j1 = r[i - 1][0] - 1;

p = r[i - 1][0];

if (p != 0)

{

p = 0;

for (z = 0; z <= i + 2; z++)

{

c[z] = *(*(b1 + (i)) + i + 2 - z);

}

q = QJ1(r[i - 1][1], c, i + 2);

if (q != 0)

{

p++;

r[i][p] = q;

}

else { p = p + 0; }

for (t = 1; t <= j1; t++)

{

ABX(*(*(r + i - 1) + t), *(*(r + i - 1) + t + 1), c, i + 2);

q = ABX(*(*(r + i - 1) + t), *(*(r + i - 1) + t + 1), c, i + 2);

if (q != 0)

{

p++;

r[i][p] = q;

}

else { p = p + 0; }

}

QJ2(*(*(r + (i - 1)) + t), c, i + 2);

q = QJ2(*(*(r + i - 1) + t), c, i + 2);

if (q != 0)

{

p++;

r[i][p] = q;

}

else { p = p + 0; }

r[i][0] = p;

}

else { cout << "此方程无解" << endl; }

}

}

else {

r[n-2][0] = 0;

cout << "此方程无解" << endl;

}

}

int main()

{

int n, i;

double c[120], a1[120], b1[120][120],r[120][120];

double h=0,j=0,k=0;

cout << "请输入方程最高次数：";

cin >> n;

for (i = 0; i <= n; i++)

{

cout << "请输入方程系数：";

cin >>a1[i];

}

QD(n, a1, b1);

h = *(*(b1 + 0) + 0);

j = *(*(b1 + 0) + 1);

k = *(*(b1 + 0) + 2);

CJ(r, n,h,j,k,a1,b1);

cout << "此方程共有" << *(*(r + n - 2) + 0) << "个解" << endl;

for (i = 1; i <= *(*(r + n - 2) + 0); i++)

{

cout << "x" << i << "=" << setiosflags(ios::fixed) << setprecision(14) << *(*(r + n - 2) + i) << endl;

}

system("pause");

return 0;

}