user: 这段汇编代码用C语言表示是什么样的:

```asm
<floydAPSP>:
  endbr64
  push   %rbp
  mov    %rsi,%r8
  mov    %rsp,%rbp
  push   %r15
  push   %r14
  push   %r13
  mov    %edi,%r13d
  push   %r12
  push   %rbx
  sub    $0x38,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  movslq %edi,%rax
  mov    %rsp,%rcx
  lea    0x0(,%rax,4),%r10
  imul   %rax,%rax
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rsi
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rsi
  and    $0xfffffffffffffff0,%rdx
  sub    %rsi,%rcx
  cmp    %rcx,%rsp
  je     L140
L77:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L77
L140:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L34
L192:
  mov    %rax,%rdx
  mov    %rsp,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rbx
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L215
L92:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L92
L215:
  mov    %rdx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  jne    L116
L180:
  mov    %r10,%r14
  mov    %rsp,%r12
  xor    %esi,%esi
  xor    %edi,%edi
  shr    $0x2,%r14
  jmp    L48
  nopl   0x0(%rax,%rax,1)
L64:
  add    $0x1,%edi
  cmp    %edi,%r13d
  jle    L55
  mov    $0x1,%esi
  xor    %ecx,%ecx
L67:
  movslq %edi,%rdx
  mov    %rdx,%rax
  imul   %r10,%rdx
  imul   %r14,%rax
  add    %rcx,%rax
  lea    (%r8,%rcx,4),%rcx
  mov    (%rcx,%rdx,1),%edx
  movl   $0x0,(%r12,%rax,4)
  mov    %edx,(%rbx,%rax,4)
L48:
  cmp    %r13d,%esi
  jge    L64
  movslq %esi,%rcx
  add    $0x1,%esi
  jmp    L67
L34:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L192
L55:
  test   %r13d,%r13d
  jle    L129
  mov    %r13d,%eax
  mov    %r14,%rdx
  mov    %r14,-0x48(%rbp)
  xor    %r8d,%r8d
  lea    0x0(,%rax,4),%rdi
  neg    %rdx
  neg    %rax
  movq   $0x0,-0x50(%rbp)
  mov    %rdi,-0x60(%rbp)
  lea    0x0(,%rdx,4),%rdi
  lea    0x0(,%rax,4),%r15
  mov    %rdi,%r14
  nopl   (%rax)
L51:
  movslq %r8d,%r11
  mov    -0x50(%rbp),%rdi
  mov    -0x60(%rbp),%rsi
  add    $0x1,%r8d
  mov    %r11d,-0x54(%rbp)
  xor    %r9d,%r9d
  add    %rbx,%rdi
  nopl   0x0(%rax)
L138:
  movslq %r9d,%rcx
  imul   -0x48(%rbp),%rcx
  lea    (%r15,%rsi,1),%rax
  add    %r11,%rcx
  nop
L18:
  mov    (%rdi,%rax,1),%edx
  add    (%rbx,%rcx,4),%edx
  cmp    %edx,(%rbx,%rax,1)
  jle    L62
  mov    %edx,(%rbx,%rax,1)
  mov    %r8d,(%r12,%rax,1)
L62:
  add    $0x4,%rax
  cmp    %rax,%rsi
  jne    L18
  lea    0x1(%r9),%eax
  add    %r10,%rsi
  add    %r14,%rdi
  cmp    %eax,%r13d
  je     L27
  mov    %eax,%r9d
  jmp    L138
  nop
L27:
  add    %r10,-0x50(%rbp)
  cmp    %r9d,-0x54(%rbp)
  jne    L51
  mov    -0x48(%rbp),%r14
L129:
  xor    %r9d,%r9d
  mov    $0x1,%r15d
  mov    %r12,%rax
  xor    %edx,%edx
  mov    %r9d,%r12d
  mov    %r15d,%esi
  mov    %rax,%r9
  mov    %rbx,%r15
  mov    %edx,%ebx
  nopw   0x0(%rax,%rax,1)
L16:
  cmp    %r13d,%ebx
  jl     L131
L217:
  mov    %esi,%r12d
  cmp    %esi,%r13d
  jle    L13
  mov    $0x1,%ebx
  xor    %eax,%eax
  add    $0x1,%esi
L4:
  cmp    %eax,%r12d
  je     L16
  movslq %r12d,%rdx
  lea    D5(%rip),%rdi
  mov    %r9,-0x50(%rbp)
  imul   %r14,%rdx
  mov    %esi,-0x48(%rbp)
  add    %rdx,%rax
  mov    %ebx,%edx
  mov    (%r9,%rax,4),%r8d
  mov    (%r15,%rax,4),%ecx
  call   <printf@plt>
  mov    -0x50(%rbp),%r9
  mov    -0x48(%rbp),%esi
  cmp    %r13d,%ebx
  jge    L217
L131:
  movslq %ebx,%rax
  add    $0x1,%ebx
  jmp    L4
L13:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L177
  lea    -0x28(%rbp),%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L116:
  orq    $0x0,-0x8(%rsp,%rax,1)
  jmp    L180
L177:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_label: D5
datatype: string

tool: D5: Shortest path between %d and %d is %d via %d


assistant: ```c
int floydAPSP(int n, int G[n][n]) {
  int D[n][n], P[n][n];
  for (int i = 0, j = 0; j < n || (j = 0, ++i) < n; j++) {
    D[i][j] = G[i][j];
    P[i][j] = 0;
  }
  for (int k = 0; k < n; k++) {
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        int d = D[i][k] + D[k][j];
        if (D[i][j] > d) {
          D[i][j] = d;
          P[i][j] = k + 1;
        }
      }
    }
  }
  for (int i = 0, j = 0; j < n || (j = 0, ++i) < n; j++) {
    i ^ j &&printf("Shortest path between %d and %d is %d via %d\n", i + 1, j + 1, D[i][j], P[i][j]);
  }
}
```