输入方程组，用三角分解法解决c语言

以下是使用三角分解法解决线性方程组的C语言代码：

#include <stdio.h>

void crout(double **a, int n, double *b) {
    int i, j, k;
    double sum;
    double *c, *d;

    c = (double *)malloc(n * sizeof(double));
    d = (double *)malloc(n * sizeof(double));

    for (i = 0; i < n; i++) {
        d[i] = a[i][i];
        for (j = 0; j < i; j++) {
            sum = 0.0;
            for (k = 0; k < j; k++) {
                sum += a[i][k] * c[k] * d[j];
            }
            a[i][j] = (a[i][j] - sum) / d[j];
        }
        sum = 0.0;
        for (k = 0; k < i; k++) {
            sum += a[i][k] * c[k];
        }
        c[i] = b[i] - sum;
    }

    for (i = 0; i < n; i++) {
        for (j = 0; j < i; j++) {
            sum = 0.0;
            for (k = 0; k < j; k++) {
                sum += a[j][k] * a[i][k] * d[k];
            }
            a[j][i] = (a[j][i] - sum) / d[j];
        }
    }

    for (i = 0; i < n; i++) {
        sum = 0.0;
        for (k = 0; k < i; k++) {
            sum += a[i][k] * d[k] * c[k];
        }
        c[i] = (c[i] - sum) / d[i];
    }

    for (i = n - 1; i >= 0; i--) {
        sum = 0.0;
        for (k = i + 1; k < n; k++) {
            sum += a[i][k] * b[k];
        }
        b[i] = (c[i] - sum) / a[i][i];
    }

    free(c);
    free(d);
}

int main() {
    int n, i, j;
    double **a, *b;

    printf("Enter the number of equations: ");
    scanf("%d", &n);

    a = (double **)malloc(n * sizeof(double *));
    for (i = 0; i < n; i++) {
        a[i] = (double *)malloc(n * sizeof(double));
    }

    b = (double *)malloc(n * sizeof(double));

    printf("Enter the coefficients of the equations:\n");
    for (i = 0; i < n; i++) {
        for (j = 0; j < n; j++) {
            scanf("%lf", &a[i][j]);
        }
    }

    printf("Enter the constants of the equations:\n");
    for (i = 0; i < n; i++) {
        scanf("%lf", &b[i]);
    }

    crout(a, n, b);

    printf("\nThe solution is:\n");
    for (i = 0; i < n; i++) {
        printf("x[%d] = %lf\n", i, b[i]);
    }

    for (i = 0; i < n; i++) {
        free(a[i]);
    }
    free(a);
    free(b);

    return 0;
}

该代码使用了Crout方法进行三角分解，并用前、后代替法求解方程组的解。在运行代码之前，需要输入方程组的系数和常数。