user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r15
  lea    D77(%rip),%rdi
  push   %r14
  push   %r13
  push   %r12
  lea    D94(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x348,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x338(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    D79(%rip),%rdi
  call   <puts@plt>
  lea    D62(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D23(%rip),%rdi
  call   <puts@plt>
  lea    0xc(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  lea    D1(%rip),%rdi
  call   <puts@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L40
  lea    0x1a0(%rsp),%r14
  xor    %r13d,%r13d
  nopl   (%rax)
L19:
  mov    %r14,%rbp
  xor    %ebx,%ebx
  test   %eax,%eax
  jle    L35
  nopl   0x0(%rax)
L96:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  add    $0x1,%ebx
  add    $0x4,%rbp
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  cmp    %ebx,%eax
  jg     L96
L35:
  add    $0x1,%r13
  add    $0x28,%r14
  cmp    %r13d,%eax
  jg     L19
L40:
  lea    D2(%rip),%rdi
  call   <puts@plt>
  mov    0xc(%rsp),%edx
  test   %edx,%edx
  jle    L43
  lea    0x1a0(%rsp),%r12
  xor    %r13d,%r13d
  lea    D42(%rip),%rbp
  nopl   (%rax)
L76:
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L84:
  mov    (%r12,%rbx,4),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,0xc(%rsp)
  jg     L84
  mov    $0xa,%edi
  add    $0x1,%r13d
  add    $0x28,%r12
  call   <putchar@plt>
  cmp    %r13d,0xc(%rsp)
  jg     L76
L43:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L41
  lea    0x0(,%rax,4),%r13
  lea    -0x1(%rax),%r15d
  xor    %r12d,%r12d
  add    %r13,%rax
  lea    0x1a0(%rsp),%r14
  lea    0x10(%rsp),%rbp
  lea    0x0(,%rax,8),%rbx
  cs nopw 0x0(%rax,%rax,1)
L134:
  lea    0x0(%rbp,%r12,1),%rdi
  lea    (%r14,%r12,1),%rsi
  mov    %r13,%rdx
  add    $0x28,%r12
  call   <memcpy@plt>
  cmp    %rbx,%r12
  jne    L134
  mov    %rbx,%r11
  lea    0x14(%rsp,%r15,4),%r12
  xor    %r10d,%r10d
  xor    %r9d,%r9d
  neg    %r11
  nopl   0x0(%rax,%rax,1)
L44:
  mov    %rbp,%rdi
  mov    %r12,%r8
  xor    %esi,%esi
  sub    %r10,%rdi
  nopl   0x0(%rax,%rax,1)
L89:
  lea    (%rdi,%r10,1),%rdx
  nopl   0x0(%rax)
L142:
  lea    (%rdx,%rsi,1),%rcx
  mov    (%rdi),%eax
  add    (%rcx,%r9,1),%eax
  mov    (%rdx),%ecx
  cmp    %ecx,%eax
  cmovg  %ecx,%eax
  add    $0x4,%rdx
  mov    %eax,-0x4(%rdx)
  cmp    %r8,%rdx
  jne    L142
  sub    $0x28,%rsi
  add    $0x28,%rdi
  lea    0x28(%rdx),%r8
  cmp    %rsi,%r11
  jne    L89
  add    $0x28,%r9
  sub    $0x4,%r10
  cmp    %rbx,%r9
  jne    L44
L41:
  lea    D19(%rip),%rdi
  call   <puts@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L121
  lea    0x10(%rsp),%r12
  xor    %r13d,%r13d
  lea    D42(%rip),%rbp
  cs nopw 0x0(%rax,%rax,1)
L109:
  xor    %ebx,%ebx
  nopw   0x0(%rax,%rax,1)
L140:
  mov    (%r12,%rbx,4),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %ebx,0xc(%rsp)
  jg     L140
  mov    $0xa,%edi
  add    $0x1,%r13d
  add    $0x28,%r12
  call   <putchar@plt>
  cmp    %r13d,0xc(%rsp)
  jg     L109
L121:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x338(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L112
  add    $0x348,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L112:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D77
type_infered: string
name: retrieve_data
arguments: tag: D94
type_infered: string
name: retrieve_data
arguments: tag: D79
type_infered: string
name: retrieve_data
arguments: tag: D62
type_infered: string
name: retrieve_data
arguments: tag: D23
type_infered: string
name: retrieve_data
arguments: tag: D1
type_infered: string
name: retrieve_data
arguments: tag: D2
type_infered: string
name: retrieve_data
arguments: tag: D42
type_infered: string
name: retrieve_data
arguments: tag: D19
type_infered: string

tool: D77: 
*********************************************************

tool: D94: %d

tool: D79: 
*	PROGRAM TO IMPLEMENT FLOYD'S ALGORITHM	*

tool: D62: *********************************************************

tool: D23: 
Enter the iNber of vertices

tool: D1: 
Enter the Cost adjacency Matrix

tool: D2: 
Input Graph

tool: D42: %d	

tool: D19: 
All Pair Shortest Path Matrix

assistant: ```c
int main(void) {
  int iN, i, j, k;
  int iaFloyd[10][10], iaCost[10][10];
  printf("\n*********************************************************");
  printf("\n*\tPROGRAM TO IMPLEMENT FLOYD'S ALGORITHM\t*\n");
  printf("*********************************************************");
  printf("\nEnter the iNber of vertices\n");
  scanf("%d", &iN);
  printf("\nEnter the Cost adjacency Matrix\n");
  for (i = 0; i < iN; i++)
    for (j = 0; j < iN; j++)
      scanf("%d", &iaCost[i][j]);
  printf("\nInput Graph\n");
  for (i = 0; i < iN; i++) {
    for (j = 0; j < iN; j++) {
      printf("%d\t", iaCost[i][j]);
    }
    printf("\n");
  }
  printf("\n");
  for (i = 0; i < iN; i++) {
    for (j = 0; j < iN; j++) {
      iaFloyd[i][j] = iaCost[i][j];
    }
  }
  for (k = 0; k < iN; k++) {
    for (i = 0; i < iN; i++) {
      for (j = 0; j < iN; j++) {
        iaFloyd[i][j] = (iaFloyd[i][j] < (iaFloyd[i][k] + iaFloyd[k][j])) ? (iaFloyd[i][j]) : (iaFloyd[i][k] + iaFloyd[k][j]);
      }
    }
  }
  printf("\nAll Pair Shortest Path Matrix\n");
  for (i = 0; i < iN; i++) {
    for (j = 0; j < iN; j++) {
      printf("%d\t", iaFloyd[i][j]);
    }
    printf("\n");
  }
  printf("\n");
  return 0;
}
```