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   %r15
  push   %r14
  lea    -0x3c(%rbp),%rsi
  push   %r13
  lea    D15(%rip),%r13
  push   %r12
  mov    %r13,%rdi
  push   %rbx
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <scanf@plt>
  movslq -0x3c(%rbp),%rax
  mov    %rsp,%rsi
  mov    %rax,%rcx
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdi
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rdi
  and    $0xfffffffffffffff0,%rdx
  sub    %rdi,%rsi
  cmp    %rsi,%rsp
  je     L159
L33:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rsi,%rsp
  jne    L33
L159:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L90
L62:
  mov    %rax,%rdx
  mov    %rsp,%rsi
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%r14
  and    $0xfffffffffffffff0,%rdx
  sub    %rax,%rsi
L144:
  cmp    %rsi,%rsp
  je     L85
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L144
L90:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L62
L85:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L115
L1:
  mov    %rsp,%r12
  test   %ecx,%ecx
  jle    L93
  mov    %r14,%r15
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L130:
  mov    %r15,%rsi
  mov    %r13,%rdi
  add    $0x1,%ebx
  add    $0x4,%r15
  call   <scanf@plt>
  mov    -0x3c(%rbp),%eax
  cmp    %ebx,%eax
  jg     L130
  test   %eax,%eax
  jle    L93
  mov    %r12,%r15
  xor    %ebx,%ebx
  cs nopw 0x0(%rax,%rax,1)
L88:
  mov    %r15,%rsi
  mov    %r13,%rdi
  add    $0x1,%ebx
  add    $0x4,%r15
  call   <scanf@plt>
  mov    -0x3c(%rbp),%eax
  cmp    %ebx,%eax
  jg     L88
  test   %eax,%eax
  jle    L93
  pxor   %xmm3,%xmm3
  mov    $0x1,%r13d
  movss  %xmm3,-0x44(%rbp)
L163:
  cmp    %r13d,%eax
  jle    L48
  mov    %r13,%rbx
  nopl   0x0(%rax,%rax,1)
L22:
  mov    -0x4(%r14,%r13,4),%edi
  sub    (%r14,%rbx,4),%edi
  call   <abs@plt>
  mov    -0x4(%r12,%r13,4),%edi
  sub    (%r12,%rbx,4),%edi
  add    $0x1,%rbx
  movss  %xmm0,-0x48(%rbp)
  call   <abs@plt>
  movss  -0x48(%rbp),%xmm1
  mulss  %xmm0,%xmm0
  mulss  %xmm1,%xmm1
  addss  %xmm1,%xmm0
  call   <sqrt@plt>
  addss  -0x44(%rbp),%xmm0
  mov    -0x3c(%rbp),%eax
  movss  %xmm0,-0x44(%rbp)
  cmp    %ebx,%eax
  jg     L22
  add    $0x1,%r13
  lea    -0x1(%r13),%edx
  cmp    %edx,%eax
  jg     L163
  nopl   0x0(%rax,%rax,1)
L48:
  movss  -0x44(%rbp),%xmm0
  lea    D9(%rip),%rdi
  call   <printf@plt>
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  lea    -0x28(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L93:
  pxor   %xmm4,%xmm4
  movss  %xmm4,-0x44(%rbp)
  jmp    L48
L115:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L1
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D15
type_infered: string
name: extract_symbolic_data
arguments: label: D9
type_infered: string

tool: D15: %d

tool: D9: %.4f

assistant: ```c
int main() {
  int n;
  scanf("%d", &n);
  int x[n], y[n];
  for (int i = 0; i < n; i++) {
    scanf("%d", &x[i]);
  }
  for (int i = 0; i < n; i++) {
    scanf("%d", &y[i]);
  }
  float distance = 0;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      float sq_x = abs(x[i] - x[j]);
      float sq_y = abs(y[i] - y[j]);
      distance = distance + sqrt(sq_x * sq_x + sq_y * sq_y);
    }
  }
  printf("%.4f", distance);
}
```