请写一个c语言完整代码，要求输入矩阵，输出该矩阵的逆矩阵

我可以听懂您的问题，下面是C语言的完整代码，实现输入矩阵，输出该矩阵的逆矩阵。

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

/* 求矩阵的逆 */
int matrix_inverse(double *input, int n, double *output) {
    double *a, *b;
    int *is, *js, l, k, i, j, m;
    double d, p;

    b = (double*) malloc(n * n * sizeof(double));
    a = (double*) malloc(n * n * sizeof(double));
    is = (int*) malloc(n * sizeof(int));
    js = (int*) malloc(n * sizeof(int));

    for (i = 0; i < n; i++) {
        for (j = 0; j < n; j++) {
            *(a + i * n + j) = *(input + i * n + j);
        }
    }

    for (k = 0; k < n; k++) {
        d = 0.0;
        for (i = k; i < n; i++) {
            for (j = k; j < n; j++) {
                l = i * n + j;
                p = fabs(*(a + l));
                if (p > d) {
                    d = p;
                    is[k] = i;
                    js[k] = j;
                }
            }
        }

        if (d == 0.0) {
            free(a);
            free(b);
            free(is);
            free(js);
            return 0;
        }

        if (is[k] != k) {
            for (j = 0; j < n; j++) {
                l = k * n + j;
                m = is[k] * n + j;
                p = *(a + l);
                *(a + l) = *(a + m);
                *(a + m) = p;
            }
        }

        if (js[k] != k) {
            for (i = 0; i < n; i++) {
                l = i * n + k;
                m = i * n + js[k];
                p = *(a + l);
                *(a + l) = *(a + m);
                *(a + m) = p;
            }
        }

        l = k * n + k;
        *(b + l) = 1.0 / *(a + l);
        for (j = 0; j < n; j++) {
            if (j != k) {
                m = k * n + j;
                *(a + m) = *(a + m) * (*(b + l));
            }
        }

        for (i = 0; i < n; i++) {
            if (i != k) {
                for (j = 0; j < n; j++) {
                    if (j != k) {
                        l = i * n + j;
                        *(a + l) = *(a + l) - *(a + i * n + k) * (*(a + k * n + j));
                    }
                }
            }
        }

        for (i = 0; i < n; i++) {
            if (i != k) {
                l = i * n + k;
                *(a + l) = -*(a + l) * (*(b + k * n + k));
            }
        }
    }

    for (k = n - 1; k >= 0; k--) {
        if (js[k] != k) {
            for (j = 0; j < n; j++) {
                l = k * n + j;
                m = js[k] * n + j;
                p = *(b + l);
                *(b + l) = *(b + m);
                *(b + m) = p;
            }
        }

        if (is[k] != k) {
            for (i = 0; i < n; i++) {
                l = i * n + k;
                m = i * n + is[k];
                p = *(b + l);
                *(b + l) = *(b + m);
                *(b + m) = p;
            }
        }
    }

    for (i = 0; i < n; i++) {
        for (j = 0; j < n; j++) {
            *(output + i * n + j) = *(b + i * n + j);
        }
    }

    free(a);
    free(b);
    free(is);
    free(js);
    return 1;
}

/* 测试 */
int main() {
    int i, j, n;
    double *input, *output;

    printf("请输入矩阵的阶数：");
    scanf("%d", &n);

    input = (double*) malloc(n * n * sizeof(double));
    output = (double*) malloc(n * n * sizeof(double));

    printf("请输入矩阵的元素：\n");
    for (i = 0; i < n; i++) {
        for (j = 0; j < n; j++) {
            scanf("%lf", input + i * n + j);
        }
    }

    if (matrix_inverse(input, n, output)) {
        printf("逆矩阵为：\n");
        for (i = 0; i < n; i++) {
            for (j = 0; j < n; j++) {
                printf("%lf ", *(output + i * n + j));
            }
            printf("\n");
        }
    } else {
        printf("该矩阵没有逆矩阵！\n");
    }

    free(input);
    free(output);
    return 0;
}

当您输入矩阵及其阶数后，程序会将其存储，并调用matrix_inverse()函数计算逆矩阵，最后输出结果。注意，这段代码并没有处理除数为 0 的异常情况，所以请确保您输入的矩阵是可逆的。