user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x28(%rbp)
  lea    D39(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x34(%rbp),%rax
  mov    %rax,%rsi
  lea    D24(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    -0x34(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x20(%rbp)
  movl   $0x0,-0x30(%rbp)
  jmp    L138
L96:
  mov    -0x34(%rbp),%eax
  add    $0x1,%eax
  cltq
  shl    $0x2,%eax
  mov    %eax,%edx
  mov    -0x30(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rcx
  mov    -0x20(%rbp),%rax
  lea    (%rcx,%rax,1),%rbx
  mov    %edx,%edi
  call   <malloc@plt>
  mov    %rax,(%rbx)
  addl   $0x1,-0x30(%rbp)
L138:
  mov    -0x34(%rbp),%eax
  cmp    %eax,-0x30(%rbp)
  jl     L96
  movl   $0x0,-0x30(%rbp)
  jmp    L113
L111:
  mov    -0x30(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%esi
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x2c(%rbp)
  jmp    L44
L105:
  mov    -0x30(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x2c(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D30(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  addl   $0x1,-0x2c(%rbp)
L44:
  mov    -0x34(%rbp),%eax
  cmp    %eax,-0x2c(%rbp)
  jle    L105
  addl   $0x1,-0x30(%rbp)
L113:
  mov    -0x34(%rbp),%eax
  cmp    %eax,-0x30(%rbp)
  jl     L111
  movl   $0x0,-0x30(%rbp)
  jmp    L23
L131:
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x24(%rbp)
  movl   $0x0,-0x2c(%rbp)
  jmp    L39
L141:
  mov    -0x30(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x2c(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  movd   %eax,%xmm0
  call   <fabs@plt>
  movss  -0x24(%rbp),%xmm1
  addss  %xmm1,%xmm0
  movss  %xmm0,-0x24(%rbp)
  addl   $0x1,-0x2c(%rbp)
L39:
  mov    -0x34(%rbp),%eax
  cmp    %eax,-0x2c(%rbp)
  jl     L141
  mov    -0x30(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x30(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  movd   %eax,%xmm0
  call   <fabs@plt>
  movaps %xmm0,%xmm1
  movss  -0x24(%rbp),%xmm0
  subss  %xmm1,%xmm0
  movss  %xmm0,-0x24(%rbp)
  mov    -0x30(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x30(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  movd   %eax,%xmm0
  call   <fabs@plt>
  movss  -0x24(%rbp),%xmm1
  comiss %xmm0,%xmm1
  jb     L142
  movl   $0x1,-0x28(%rbp)
  jmp    L34
L142:
  addl   $0x1,-0x30(%rbp)
L23:
  mov    -0x34(%rbp),%eax
  cmp    %eax,-0x30(%rbp)
  jl     L131
L34:
  cmpl   $0x0,-0x28(%rbp)
  je     L1
  lea    D45(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L12
L1:
  mov    -0x34(%rbp),%edx
  mov    -0x20(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <solve@plt>
L12:
  mov    $0x0,%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L129
  call   <__stack_chk_fail@plt>
L129:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: data_tag: D39
datatype: string
name: disassemble_data
arguments: data_tag: D24
datatype: string
name: disassemble_data
arguments: data_tag: D12
datatype: string
name: disassemble_data
arguments: data_tag: D30
datatype: string
name: disassemble_data
arguments: data_tag: D45
datatype: string

tool: D39: Enter no. of variables to be solved: 

tool: D24: %d

tool: D12: Enter the co-efficient of variables and constant term of eqn. %d one by one: 


tool: D30: %f

tool: D45: The method will not converge

assistant: ```c
int main(void) {
  float **arr;
  int i, j, n, flag = 0;
  float S;
  printf("Enter no. of variables to be solved: ");
  scanf("%d", &n);
  arr = (float **)malloc(n * sizeof(float *));
  for (i = 0; i < n; i++)
    arr[i] = malloc((n + 1) * sizeof(float));
  for (i = 0; i < n; i++) {
    printf("Enter the co-efficient of variables and constant term of eqn. %d one by one: \n", i + 1);
    for (j = 0; j < (n + 1); j++)
      scanf("%f", &arr[i][j]);
  }
  for (i = 0; i < n; i++) {
    S = 0;
    for (j = 0; j < n; j++)
      S += fabs(arr[i][j]);
    S -= fabs(arr[i][i]);
    if (fabs(arr[i][i]) <= S) {
      flag = 1;
      break;
    }
  }
  if (flag)
    printf("The method will not converge\n");
  else
    solve(arr, n);
  return 0;
}
```