user: 请从给定的汇编指令推导出C源代码:

```asm
<solveWithInverseMethod>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    -0x75000(%rsp),%r11
L15:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L15
  sub    $0x340,%rsp
  mov    %rdi,-0x75328(%rbp)
  mov    %rsi,-0x75330(%rbp)
  mov    %rdx,-0x75338(%rbp)
  mov    %ecx,-0x7533c(%rbp)
  mov    %r8d,-0x75340(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x75314(%rbp)
  jmp    L5
L19:
  mov    -0x75314(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x75330(%rbp),%rax
  add    %rdx,%rax
  movsd  (%rax),%xmm0
  mov    -0x75314(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  add    %rdx,%rax
  shl    $0x5,%rax
  add    %rbp,%rax
  sub    $0x61a88,%rax
  movsd  %xmm0,(%rax)
  addl   $0x1,-0x75314(%rbp)
L5:
  mov    -0x75314(%rbp),%eax
  cmp    -0x7533c(%rbp),%eax
  jle    L19
  mov    -0x75340(%rbp),%ecx
  mov    -0x7533c(%rbp),%edx
  lea    -0x4e210(%rbp),%rsi
  mov    -0x75328(%rbp),%rax
  mov    %rax,%rdi
  call   <transpose@plt>
  mov    -0x75340(%rbp),%r8d
  mov    -0x7533c(%rbp),%edi
  mov    -0x75340(%rbp),%ecx
  lea    -0x27110(%rbp),%rdx
  mov    -0x75328(%rbp),%rsi
  lea    -0x4e210(%rbp),%rax
  mov    %r8d,%r9d
  mov    %edi,%r8d
  mov    %rax,%rdi
  call   <multiplyMatrix@plt>
  mov    -0x75340(%rbp),%edx
  lea    -0x3a990(%rbp),%rcx
  lea    -0x27110(%rbp),%rax
  mov    %rcx,%rsi
  mov    %rax,%rdi
  call   <inverseOfMatrix@plt>
  mov    -0x7533c(%rbp),%edi
  mov    -0x75340(%rbp),%ecx
  lea    -0x13890(%rbp),%rdx
  lea    -0x61a90(%rbp),%rsi
  lea    -0x4e210(%rbp),%rax
  mov    $0x1,%r9d
  mov    %edi,%r8d
  mov    %rax,%rdi
  call   <multiplyMatrix@plt>
  mov    -0x75340(%rbp),%edi
  mov    -0x75340(%rbp),%ecx
  lea    -0x75310(%rbp),%rdx
  lea    -0x13890(%rbp),%rsi
  lea    -0x3a990(%rbp),%rax
  mov    $0x1,%r9d
  mov    %edi,%r8d
  mov    %rax,%rdi
  call   <multiplyMatrix@plt>
  movl   $0x1,-0x75314(%rbp)
  jmp    L43
L49:
  mov    -0x75314(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x75338(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x75314(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  add    %rdx,%rax
  shl    $0x5,%rax
  add    %rbp,%rax
  sub    $0x75308,%rax
  movsd  (%rax),%xmm0
  movsd  %xmm0,(%rcx)
  addl   $0x1,-0x75314(%rbp)
L43:
  mov    -0x75314(%rbp),%eax
  cmp    -0x7533c(%rbp),%eax
  jle    L49
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L41
  call   <__stack_chk_fail@plt>
L41:
  leave
  ret
```

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];
}
```