user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x13000(%rsp),%r11
L109:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L109
  sub    $0x8d8,%rsp
  lea    D21(%rip),%rdi
  lea    D60(%rip),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x138c8(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x34(%rsp),%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  lea    D6(%rip),%rdi
  call   <puts@plt>
  mov    0x34(%rsp),%esi
  test   %esi,%esi
  jle    L190
  lea    0x40(%rsp),%r15
  lea    0x9c80(%rsp),%rax
  movq   $0x0,0x20(%rsp)
  mov    %r15,0x28(%rsp)
  movl   $0xa,0x18(%rsp)
  movl   $0x0,0x10(%rsp)
  mov    %rax,0x8(%rsp)
L177:
  mov    0x18(%rsp),%ebx
  mov    0x20(%rsp),%rax
  xor    %r14d,%r14d
  lea    0x1(%rbx),%ebp
  mov    0x28(%rsp),%rbx
  mov    %rax,%r13
  sub    %rax,%rbx
  nopl   (%rax)
L81:
  lea    (%rbx,%r13,1),%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    (%r15,%r13,1),%ecx
  test   %ecx,%ecx
  jne    L112
  movl   $0x270f,(%r15,%r13,1)
L112:
  mov    0x8(%rsp),%rdi
  lea    0x0(%rbp,%r14,1),%eax
  add    $0x1,%r14d
  mov    %eax,(%rdi,%r13,1)
  mov    0x34(%rsp),%eax
  add    $0x4,%r13
  cmp    %r14d,%eax
  jg     L81
  addl   $0x1,0x10(%rsp)
  mov    0x10(%rsp),%ebx
  addl   $0xa,0x18(%rsp)
  addq   $0x190,0x20(%rsp)
  addq   $0x190,0x28(%rsp)
  cmp    %ebx,%eax
  jg     L177
  test   %eax,%eax
  jle    L190
  mov    %eax,%edx
  sub    $0x1,%eax
  mov    %r15,0x20(%rsp)
  xor    %r14d,%r14d
  lea    0x4(%r15,%rax,4),%rax
  mov    %rdx,%rcx
  mov    $0xcccccccd,%r9d
  mov    %rax,0x10(%rsp)
  imul   $0xfffffffffffffe70,%rdx,%rax
  neg    %rcx
  lea    0x0(,%rcx,4),%rbx
  mov    %rbx,0x18(%rsp)
  xor    %ebx,%ebx
  mov    %rax,%r15
L132:
  mov    0x8(%rsp),%r13
  mov    0x20(%rsp),%r11
  xor    %r10d,%r10d
  mov    0x10(%rsp),%rbp
  sub    %r14,%r13
  sub    %r14,%r11
  nopl   (%rax)
L184:
  lea    (%r11,%r14,1),%rsi
  lea    0x0(%r13,%r14,1),%r8
  jmp    L207
  nopl   0x0(%rax,%rax,1)
L180:
  add    $0x4,%rsi
  add    $0x4,%r8
  cmp    %rbp,%rsi
  je     L149
L207:
  lea    (%rsi,%r10,1),%rdx
  mov    (%r11),%eax
  add    (%rdx,%rbx,1),%eax
  cmp    %eax,(%rsi)
  jle    L180
  mov    %eax,(%rsi)
  lea    (%r8,%r10,1),%rax
  mov    (%rax,%rbx,1),%r12d
  test   %r12d,%r12d
  jle    L63
  mov    %r12d,%eax
  xor    %edi,%edi
  nop
L90:
  mov    %eax,%eax
  mov    %edi,%edx
  lea    0x1(%rdi),%edi
  mov    %rax,%rcx
  imul   %r9,%rax
  shr    $0x23,%rax
  cmp    $0x9,%ecx
  jg     L90
  mov    $0x1,%eax
  nopl   0x0(%rax)
L167:
  lea    (%rax,%rax,4),%eax
  add    %eax,%eax
  sub    $0x1,%edx
  jae    L167
L87:
  movslq 0x0(%r13),%rdx
  add    $0x4,%rsi
  add    $0x4,%r8
  mov    %rdx,%rcx
  imul   $0x66666667,%rdx,%rdx
  sar    $0x1f,%ecx
  sar    $0x22,%rdx
  sub    %ecx,%edx
  imul   %edx,%eax
  add    %r12d,%eax
  mov    %eax,-0x4(%r8)
  cmp    %rbp,%rsi
  jne    L207
L149:
  sub    $0x190,%r10
  lea    0x190(%rsi),%rbp
  add    $0x190,%r13
  add    $0x190,%r11
  cmp    %r15,%r10
  jne    L184
  add    $0x190,%rbx
  sub    $0x4,%r14
  cmp    0x18(%rsp),%r14
  jne    L132
L190:
  lea    D70(%rip),%rdi
  xor    %eax,%eax
  lea    0x40(%rsp),%rbp
  xor    %r13d,%r13d
  call   <printf@plt>
  mov    0x34(%rsp),%edx
  lea    D47(%rip),%r12
  test   %edx,%edx
  jle    L36
L22:
  mov    $0xa,%edi
  call   <putchar@plt>
  mov    0x34(%rsp),%eax
  test   %eax,%eax
  jle    L36
  xor    %ebx,%ebx
  nopl   0x0(%rax)
L208:
  mov    0x0(%rbp,%rbx,4),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  mov    0x34(%rsp),%eax
  cmp    %ebx,%eax
  jg     L208
  add    $0x1,%r13d
  add    $0x190,%rbp
  cmp    %r13d,%eax
  jg     L22
L36:
  lea    D42(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x3c(%rsp),%rdx
  lea    0x38(%rsp),%rsi
  xor    %eax,%eax
  lea    D32(%rip),%rdi
  call   <scanf@plt>
  mov    0x38(%rsp),%eax
  mov    0x3c(%rsp),%ebx
  lea    D4(%rip),%rdi
  sub    $0x1,%eax
  lea    -0x1(%rbx),%edx
  mov    %eax,0x38(%rsp)
  cltq
  lea    (%rax,%rax,4),%rax
  mov    %edx,0x3c(%rsp)
  movslq %edx,%rdx
  lea    (%rax,%rax,4),%rax
  lea    (%rdx,%rax,4),%rax
  mov    0x40(%rsp,%rax,4),%edx
  mov    0x9c80(%rsp,%rax,4),%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x138c8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L194
  add    $0x138d8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L63:
  mov    $0x1,%eax
  jmp    L87
L194:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D21
type_infered: string
name: retrieve_data
arguments: tag: D60
type_infered: string
name: retrieve_data
arguments: tag: D6
type_infered: string
name: retrieve_data
arguments: tag: D70
type_infered: string
name: retrieve_data
arguments: tag: D47
type_infered: string
name: retrieve_data
arguments: tag: D42
type_infered: string
name: retrieve_data
arguments: tag: D32
type_infered: string
name: retrieve_data
arguments: tag: D4
type_infered: string

tool: D21: 
Enter the no of vertices 

tool: D60: %d

tool: D6: 
Enter the adjacency matrix 

tool: D70: 
The shortest path adjacency matrix is 

tool: D47: %d 

tool: D42: 
Enter the vertices whose shortest path is required 

tool: D32: %d%d

tool: D4: 
The shortest path is %d of length %d 

assistant: ```c
int main() {
  int A[100][100], i, j, n, B[100][100], k, x, lbnd, ubnd;
  printf("\nEnter the no of vertices ");
  scanf("%d", &n);
  printf("\nEnter the adjacency matrix \n");
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++) {
      scanf("%d", &A[i][j]);
      if (A[i][j] == 0)
        A[i][j] = 9999;
      B[i][j] = (i + 1) * 10 + j + 1;
    }
  for (k = 0; k < n; k++)
    for (i = 0; i < n; i++)
      for (j = 0; j < n; j++) {
        if (A[i][j] > A[i][k] + A[k][j]) {
          int dig = 0, ans = 1;
          A[i][j] = A[i][k] + A[k][j];
          x = B[k][j];
          while (x > 0) {
            dig++;
            x = x / 10;
          }
          while (dig--)
            ans = ans * 10;
          x = B[i][k] / 10;
          B[i][j] = (x)*ans + B[k][j];
        }
      }
  printf("\nThe shortest path adjacency matrix is ");
  for (i = 0; i < n; i++) {
    printf("\n");
    for (j = 0; j < n; j++) {
      printf("%d ", A[i][j]);
    }
  }
  printf("\nEnter the vertices whose shortest path is required ");
  scanf("%d%d", &lbnd, &ubnd);
  lbnd--;
  ubnd--;
  printf("\nThe shortest path is %d of length %d ", B[lbnd][ubnd], A[lbnd][ubnd]);
}
```