c++一元多项式计算器

以下是一个简单的C++一元多项式计算器的实现，可以实现多项式的加减乘和求根功能：

#include <iostream>
#include <cmath>
using namespace std;

const int MAXN = 1005;
const double EPS = 1e-8;

struct Poly {
    double coef[MAXN];
    int deg;
    Poly() {
        memset(coef, 0, sizeof(coef));
        deg = 0;
    }
};

Poly read_poly() {
    Poly p;
    cin >> p.deg;
    for (int i = p.deg; i >= 0; i--) {
        cin >> p.coef[i];
    }
    return p;
}

void print_poly(Poly p) {
    cout << p.deg;
    for (int i = p.deg; i >= 0; i--) {
        cout << " " << p.coef[i];
    }
    cout << endl;
}

Poly add_poly(Poly a, Poly b) {
    Poly c;
    c.deg = max(a.deg, b.deg);
    for (int i = c.deg; i >= 0; i--) {
        c.coef[i] = a.coef[i] + b.coef[i];
    }
    while (c.deg > 0 && fabs(c.coef[c.deg]) < EPS) {
        c.deg--;
    }
    return c;
}

Poly sub_poly(Poly a, Poly b) {
    Poly c;
    c.deg = max(a.deg, b.deg);
    for (int i = c.deg; i >= 0; i--) {
        c.coef[i] = a.coef[i] - b.coef[i];
    }
    while (c.deg > 0 && fabs(c.coef[c.deg]) < EPS) {
        c.deg--;
    }
    return c;
}

Poly mul_poly(Poly a, Poly b) {
    Poly c;
    c.deg = a.deg + b.deg;
    for (int i = 0; i <= a.deg; i++) {
        for (int j = 0; j <= b.deg; j++) {
            c.coef[i+j] += a.coef[i] * b.coef[j];
        }
    }
    while (c.deg > 0 && fabs(c.coef[c.deg]) < EPS) {
        c.deg--;
    }
    return c;
}

double calc_poly(Poly p, double x) {
    double ans = 0;
    for (int i = p.deg; i >= 0; i--) {
        ans = ans * x + p.coef[i];
    }
    return ans;
}

double solve_poly(Poly p, double l, double r) {
    double mid;
    while (r - l > EPS) {
        mid = (l + r) / 2;
        if (calc_poly(p, mid) > 0) {
            l = mid;
        } else {
            r = mid;
        }
    }
    return mid;
}

int main() {
    Poly a = read_poly();
    Poly b = read_poly();
    Poly c = add_poly(a, b);
    Poly d = sub_poly(a, b);
    Poly e = mul_poly(a, b);
    print_poly(c);
    print_poly(d);
    print_poly(e);
    double root = solve_poly(e, -100, 100);
    printf("%.2f\n", root);
    printf("%.2f\n", calc_poly(e, 1.23));
    printf("%.2f\n", calc_poly(e, -4.56));
    return 0;
}