user: 这段汇编代码用C语言表示是什么样的?

```asm
<solveWithInverseMethod>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x75000(%rsp),%r11
L43:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L43
  sub    $0x328,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x75318(%rsp)
  xor    %eax,%eax
  mov    %rdi,%r14
  mov    %ecx,%r13d
  mov    %r8d,%r12d
  test   %ecx,%ecx
  jle    L50
  lea    -0x1(%rcx),%ebp
  mov    %rdx,%rbx
  lea    0x8(%rsi),%rax
  lea    0x0(,%rbp,8),%r10
  lea    0x13bb8(%rsp),%rdx
  lea    0x10(%rsi,%r10,1),%rcx
  nopl   0x0(%rax)
L56:
  movsd  (%rax),%xmm0
  add    $0x8,%rax
  add    $0x320,%rdx
  movsd  %xmm0,-0x320(%rdx)
  cmp    %rcx,%rax
  jne    L56
  mov    %r14,%rdi
  mov    %r12d,%ecx
  mov    %r13d,%edx
  mov    %r10,0x8(%rsp)
  lea    0x27110(%rsp),%rbp
  lea    0x4e210(%rsp),%r15
  mov    %rbp,%rsi
  call   <transpose@plt>
  mov    %r12d,%r9d
  mov    %r13d,%r8d
  mov    %r12d,%ecx
  mov    %r15,%rdx
  mov    %r14,%rsi
  mov    %rbp,%rdi
  call   <multiplyMatrix@plt>
  lea    0x3a990(%rsp),%r14
  mov    %r15,%rdi
  mov    %r12d,%edx
  mov    %r14,%rsi
  lea    0x61a90(%rsp),%r15
  call   <inverseOfMatrix@plt>
  mov    %r13d,%r8d
  mov    %r12d,%ecx
  mov    %r15,%rdx
  lea    0x13890(%rsp),%rsi
  mov    $0x1,%r9d
  mov    %rbp,%rdi
  call   <multiplyMatrix@plt>
  lea    0x10(%rsp),%rdx
  mov    %r12d,%ecx
  mov    %r12d,%r8d
  mov    $0x1,%r9d
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   <multiplyMatrix@plt>
  mov    0x8(%rsp),%r10
  lea    0x8(%rbx),%rax
  lea    0x338(%rsp),%rdx
  lea    0x10(%rbx,%r10,1),%rcx
L33:
  movsd  (%rdx),%xmm0
  add    $0x8,%rax
  add    $0x320,%rdx
  movsd  %xmm0,-0x8(%rax)
  cmp    %rcx,%rax
  jne    L33
L35:
  mov    0x75318(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L36
  add    $0x75328,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L50:
  lea    0x27110(%rsp),%rbp
  lea    0x4e210(%rsp),%r15
  mov    %r8d,%ecx
  mov    %r13d,%edx
  mov    %rbp,%rsi
  call   <transpose@plt>
  mov    %r12d,%r9d
  mov    %r13d,%r8d
  mov    %r12d,%ecx
  mov    %r15,%rdx
  mov    %r14,%rsi
  mov    %rbp,%rdi
  call   <multiplyMatrix@plt>
  lea    0x3a990(%rsp),%r14
  mov    %r12d,%edx
  mov    %r15,%rdi
  mov    %r14,%rsi
  lea    0x61a90(%rsp),%r15
  call   <inverseOfMatrix@plt>
  mov    %r13d,%r8d
  mov    %r12d,%ecx
  mov    %r15,%rdx
  lea    0x13890(%rsp),%rsi
  mov    %rbp,%rdi
  mov    $0x1,%r9d
  call   <multiplyMatrix@plt>
  lea    0x10(%rsp),%rdx
  mov    %r12d,%r8d
  mov    %r12d,%ecx
  mov    $0x1,%r9d
  mov    %r15,%rsi
  mov    %r14,%rdi
  call   <multiplyMatrix@plt>
  jmp    L35
L36:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void solveWithInverseMethod(double A[][100], double b[], double x[], int m, int n) {
  double tempx[100][100];
  double tempb[100][100];
  double transp[100][100];
  double inverse[100][100];
  double result1[100][100];
  double result2[100][100];
  int i, j;
  for (i = 1; i <= m; i++)
    tempb[i][1] = b[i];
  transpose(A, transp, m, n);
  multiplyMatrix(transp, A, result1, n, m, n);
  inverseOfMatrix(result1, inverse, n);
  multiplyMatrix(transp, tempb, result2, n, m, 1);
  multiplyMatrix(inverse, result2, tempx, n, n, 1);
  for (i = 1; i <= m; i++)
    x[i] = tempx[i][1];
}
```