user: 请从给定的汇编指令推导出C源代码:

```asm
<main1>:
  endbr64
  push   %r14
  lea    D41(%rip),%rdi
  push   %r13
  lea    D23(%rip),%r13
  push   %r12
  push   %rbp
  push   %rbx
  xor    %ebx,%ebx
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x4(%rsp),%rdx
  mov    %rsp,%rsi
  call   <scanf@plt>
  movslq (%rsp),%rdi
  mov    %rdi,%r14
  shl    $0x2,%rdi
  call   <malloc@plt>
  mov    %rax,%r12
  mov    %rax,%rbp
  test   %r14d,%r14d
  jle    L39
  nopl   0x0(%rax,%rax,1)
L2:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <scanf@plt>
  mov    (%rsp),%eax
  add    $0x4,%rbp
  cmp    %ebx,%eax
  jg     L2
  test   %eax,%eax
  jle    L39
  xor    %ebx,%ebx
  lea    D34(%rip),%rbp
  nopl   0x0(%rax,%rax,1)
L59:
  mov    (%r12,%rbx,4),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,(%rsp)
  jg     L59
L39:
  mov    %r12,%rdi
  call   <free@plt>
  mov    $0xa,%edi
  call   <putchar@plt>
  movslq (%rsp),%rbp
  lea    0x0(,%rbp,8),%rdi
  call   <malloc@plt>
  mov    0x4(%rsp),%ebx
  mov    %ebp,%edi
  mov    %rax,%r12
  imul   %ebx,%edi
  movslq %edi,%rdi
  shl    $0x2,%rdi
  call   <malloc@plt>
  mov    %rax,(%r12)
  cmp    $0x1,%ebp
  jle    L124
  lea    -0x1(%rbp),%esi
  lea    0x0(,%rbp,4),%rcx
  lea    0x8(%r12),%rdx
  lea    (%r12,%rsi,8),%rsi
  jmp    L142
  nopl   0x0(%rax)
L11:
  add    $0x8,%rdx
L142:
  add    %rcx,%rax
  mov    %rax,(%rdx)
  cmp    %rsi,%rdx
  jne    L11
L154:
  xor    %r14d,%r14d
  lea    D23(%rip),%r13
  nopl   0x0(%rax)
L8:
  test   %ebx,%ebx
  jle    L16
  mov    (%r12,%r14,8),%rbp
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L134:
  mov    %rbp,%rsi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%ebx
  call   <scanf@plt>
  add    $0x4,%rbp
  cmp    %ebx,0x4(%rsp)
  jg     L134
L16:
  add    $0x1,%r14
  cmp    %r14d,(%rsp)
  jle    L129
  mov    0x4(%rsp),%ebx
  jmp    L8
L124:
  je     L154
  nopl   0x0(%rax)
L129:
  lea    D25(%rip),%rdi
  xor    %r14d,%r14d
  lea    D34(%rip),%r13
  call   <puts@plt>
  mov    (%rsp),%edx
  test   %edx,%edx
  jle    L75
  nopl   (%rax)
L104:
  mov    0x4(%rsp),%eax
  test   %eax,%eax
  jle    L5
  mov    (%r12,%r14,8),%rbp
  xor    %ebx,%ebx
  cs nopw 0x0(%rax,%rax,1)
L112:
  mov    0x0(%rbp,%rbx,4),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,0x4(%rsp)
  jg     L112
L5:
  mov    $0xa,%edi
  add    $0x1,%r14
  call   <putchar@plt>
  cmp    %r14d,(%rsp)
  jg     L104
L75:
  mov    (%r12),%rdi
  call   <free@plt>
  mov    %r12,%rdi
  call   <free@plt>
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L148
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L148:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D41
datatype: string
name: unpack_data
arguments: data_label: D23
datatype: string
name: unpack_data
arguments: data_label: D34
datatype: string
name: unpack_data
arguments: data_label: D25
datatype: string

tool: D41: %d%d

tool: D23: %d

tool: D34: %d 

tool: D25: 


assistant: ```c
int main1() {
  int *a, i, j, n, **mat, m;
  scanf("%d%d", &n, &m);
  a = (int *)malloc(n * sizeof(int));
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  for (i = 0; i < n; i++) {
    printf("%d ", a[i]);
  }
  free(a);
  printf("\n");
  mat = (int **)malloc(n * sizeof(int *));
  mat[0] = (int *)malloc(n * m * sizeof(int));
  for (i = 1; i < n; i++) {
    mat[i] = mat[i - 1] + n;
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      scanf("%d", &mat[i][j]);
    }
  }
  printf("\n\n");
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      printf("%d ", mat[i][j]);
    }
    printf("\n");
  }
  free(mat[0]);
  free(mat);
  return 0;
}
```