c语言写“. 求两个多项式的和并输出；求两个多项式的差并输出；4. 求两个多项式的积并输出；思考： 如何求两个多项式的商?

在C语言中，处理多项式通常涉及到结构体、数组以及一些基本数学运算。以下是简单的示例，演示如何实现多项式的加法、减法和乘法：

1. **多项式加法**:

typedef struct {
    int coef[100]; // 多项式的系数数组
    int exp;       // 最高指数
} Polynomial;

void add_poly(Polynomial *poly1, Polynomial *poly2) {
    Polynomial sum;
    int i = min(poly1->exp, poly2->exp);

    for (i = 0; i <= poly1->exp && i <= poly2->exp; i++) {
        sum.coef[i] = poly1->coef[i] + poly2->coef[i];
    }
    if (poly1->exp > poly2->exp)
        sum.exp = poly1->exp;
    else if (poly2->exp > poly1->exp)
        sum.exp = poly2->exp;
    else
        sum.exp++;

    *poly1 = sum;
}

// 使用示例
Polynomial p1 = {...}, p2 = {...};
add_poly(&p1, &p2);

2. **多项式减法**:

void subtract_poly(Polynomial *poly1, Polynomial *poly2) {
    Polynomial diff;
    int i = min(poly1->exp, poly2->exp);

    for (i = 0; i <= poly1->exp && i <= poly2->exp; i++) {
        diff.coef[i] = poly1->coef[i] - poly2->coef[i];
    }
    if (poly1->exp > poly2->exp)
        diff.exp = poly1->exp;
    else if (poly2->exp > poly1->exp)
        diff.exp = poly2->exp;
    else
        diff.exp++;

    *poly1 = diff;
}

3. **多项式乘法**:

void multiply_poly(Polynomial *poly1, Polynomial *poly2, Polynomial *result) {
    result->exp = poly1->exp + poly2->exp;
    int len = 1 << (poly1->exp + poly2->exp); // 计算结果长度
    result->coef = malloc(len * sizeof(result->coef[0]));

    for (int i = 0; i < len; ++i) {
        result->coef[i] = 0;
        for (int j = 0; j <= poly1->exp; ++j) {
            for (int k = 0; k <= poly2->exp; ++k) {
                if ((i - j - k) >= 0)
                    result->coef[i] += poly1->coef[j] * poly2->coef[k] * pow(10, (i - j - k));
            }
        }
    }
}

// 示例
Polynomial p3;
multiply_poly(&p1, &p2, &p3);

**关于多项式除法**，由于多项式除法涉及到更复杂的算法，比如长除法或拉格朗日插值，直接在C语言基础结构上实现可能会变得复杂。在实际应用中，你可能需要利用库函数或者编写专门的算法来处理这个任务，例如在数值计算库如GMP或MPFR中就有相关的支持。