C语言利用指针做二元一次方程

可以利用指针来解二元一次方程组，具体步骤如下：

1. 定义二元一次方程组的系数矩阵和常数矩阵：

float coef[2][2] = {{a1, b1}, {a2, b2}};
float cons[2] = {c1, c2};

2. 定义指向这两个矩阵的指针：

float (*pcoef)[2] = coef;
float *pcons = cons;

3. 利用高斯-约旦消元法解出方程组的未知数：

float x, y;
float k = *(pcoef[1] + 0) / *(pcoef[0] + 0);
*(pcoef[1] + 1) -= k * *(pcoef[0] + 1);
*(pcons + 1) -= k * *(pcons + 0);
y = *(pcons + 1) / *(pcoef[1] + 1);
x = (*(pcons + 0) - *(pcoef[0] + 1) * y) / *(pcoef[0] + 0);

在上述代码中，利用指针访问系数矩阵和常数矩阵的元素，并利用高斯-约旦消元法解出未知数x和y。

完整代码如下：

#include <stdio.h>

int main() {
    float a1, b1, c1, a2, b2, c2;
    printf("Enter the coefficients of the first equation:\n");
    scanf("%f%f%f", &a1, &b1, &c1);
    printf("Enter the coefficients of the second equation:\n");
    scanf("%f%f%f", &a2, &b2, &c2);

    float coef[2][2] = {{a1, b1}, {a2, b2}};
    float cons[2] = {c1, c2};
    float (*pcoef)[2] = coef;
    float *pcons = cons;

    float x, y;
    float k = *(pcoef[1] + 0) / *(pcoef[0] + 0);
    *(pcoef[1] + 1) -= k * *(pcoef[0] + 1);
    *(pcons + 1) -= k * *(pcons + 0);
    y = *(pcons + 1) / *(pcoef[1] + 1);
    x = (*(pcons + 0) - *(pcoef[0] + 1) * y) / *(pcoef[0] + 0);

    printf("x: %f, y: %f\n", x, y);
    return 0;
}