user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r14
  lea    D5(%rip),%r14
  push   %r13
  lea    D46(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    D15(%rip),%rbx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%r12
  mov    %rsp,%rbp
  jmp    L38
  nop
L165:
  mov    %r14,%rdi
  call   <puts@plt>
L38:
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  xor    %eax,%eax
  mov    %r12,%rdx
  mov    %rbp,%rsi
  mov    %rbx,%rdi
  call   <scanf@plt>
  cmpl   $0x1,(%rsp)
  jle    L165
  mov    0x4(%rsp),%edi
  test   %edi,%edi
  jle    L165
  shl    $0x3,%edi
  call   <malloc@plt>
  mov    0x4(%rsp),%edx
  test   %edx,%edx
  jle    L133
  mov    (%rsp),%edi
  mov    %rax,%r13
  mov    %rax,%rbp
  xor    %r14d,%r14d
  lea    D54(%rip),%r12
L98:
  shl    $0x2,%edi
  call   <malloc@plt>
  mov    (%rsp),%edi
  mov    %rax,0x0(%rbp)
  mov    %edi,%esi
  test   %edi,%edi
  jle    L111
  xor    %ebx,%ebx
  jmp    L94
  nopl   0x0(%rax)
L164:
  mov    (%rsp),%esi
  add    $0x1,%rbx
  mov    %esi,%edi
  cmp    %ebx,%esi
  jle    L111
L94:
  lea    (%rax,%rbx,4),%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0x0(%rbp),%rax
  pxor   %xmm0,%xmm0
  ucomiss (%rax,%rbx,4),%xmm0
  jp     L164
  jne    L164
  lea    D23(%rip),%rdi
  call   <puts@plt>
L133:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L101
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
  cs nopw 0x0(%rax,%rax,1)
L111:
  mov    0x4(%rsp),%eax
  add    $0x1,%r14d
  add    $0x8,%rbp
  cmp    %r14d,%eax
  jg     L98
  test   %eax,%eax
  jle    L133
  xor    %ebx,%ebx
  lea    D32(%rip),%rbp
  jmp    L156
L129:
  add    $0x8,%r13
  cmp    %ebx,0x4(%rsp)
  jle    L133
L26:
  mov    (%rsp),%esi
L156:
  mov    0x0(%r13),%rdi
  add    $0x1,%ebx
  call   <checkProgression@plt>
  test   %rax,%rax
  je     L129
  mov    %ebx,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x8,%r13
  call   <printf@plt>
  cmp    %ebx,0x4(%rsp)
  jg     L26
  jmp    L133
L101:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D5
data_type: string
name: disassemble_data
arguments: tag: D46
data_type: string
name: disassemble_data
arguments: tag: D15
data_type: string
name: disassemble_data
arguments: tag: D54
data_type: string
name: disassemble_data
arguments: tag: D23
data_type: string
name: disassemble_data
arguments: tag: D32
data_type: string

tool: D5: Error! Y must be positive and X must be > 1

tool: D46: Enter size (x y): 

tool: D15: %i%i

tool: D54: %f

tool: D23: Error. Member of geometry progression can't be null. Exiting...

tool: D32: Progression in %i line


assistant: ```c
int main(void) {
  int sizeX, sizeY;
  float **array;
  int i, j;
  int done = 0;
  while (!done) {
    printf("Enter size (x y): ");
    scanf("%i%i", &sizeX, &sizeY);
    (sizeX > 1 && sizeY > 0) ? done = 1 : printf("Error! Y must be positive and X must be > 1\n");
  }
  array = malloc(sizeY * sizeof(float *));
  for (i = 0; i < sizeY; i++) {
    array[i] = malloc(sizeX * sizeof(float));
    for (j = 0; j < sizeX; j++) {
      scanf("%f", &array[i][j]);
      if (array[i][j] == 0) {
        printf("Error. Member of geometry progression can't be null. Exiting...\n");
        return 0;
      }
    }
  }
  for (i = 0; i < sizeY; i++)
    if (checkProgression(array[i], sizeX))
      printf("Progression in %i line\n", i + 1);
  return 0;
}
```