user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r15
  mov    $0x32,%ecx
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x4f8,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x4e8(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%r12
  lea    0x1b0(%rsp),%rdx
  mov    %r12,%rdi
  lea    0x350(%rsp),%rbx
  lea    0xc(%rsp),%rsi
  rep stos %rax,%es:(%rdi)
  mov    $0x32,%ecx
  movl   $0x0,(%rdi)
  mov    %rdx,%rdi
  rep stos %rax,%es:(%rdi)
  mov    $0x32,%ecx
  movl   $0x0,(%rdi)
  mov    %rbx,%rdi
  rep stos %rax,%es:(%rdi)
  movl   $0x0,(%rdi)
  lea    D25(%rip),%rdi
  call   <scanf@plt>
  mov    0xc(%rsp),%edx
  test   %edx,%edx
  jle    L37
  lea    0x1b4(%rsp),%r15
  mov    $0x1,%r14d
  lea    0x14(%rsp),%r13
  lea    D27(%rip),%rbp
  nopw   0x0(%rax,%rax,1)
L116:
  lea    (%rbx,%r14,4),%rdx
  mov    %r15,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    (%r15),%ecx
  mov    (%rbx,%r14,4),%edx
  cmp    %edx,%ecx
  jg     L34
  movslq %ecx,%rsi
  sub    %ecx,%edx
  add    %rsi,%rdx
  lea    (%r12,%rsi,4),%rax
  lea    0x0(%r13,%rdx,4),%rdx
  nopl   (%rax)
L64:
  addl   $0x1,(%rax)
  add    $0x4,%rax
  cmp    %rdx,%rax
  jne    L64
L34:
  mov    0xc(%rsp),%eax
  add    $0x1,%r14
  add    $0x4,%r15
  cmp    %r14d,%eax
  jge    L116
  test   %eax,%eax
  jle    L37
  lea    0x1(%rax),%esi
  xor    %ebp,%ebp
  mov    $0x1,%eax
  nop
L5:
  movslq %ebp,%rcx
  mov    (%r12,%rax,4),%edx
  cmp    0x10(%rsp,%rcx,4),%edx
  jl     L144
  cmp    %eax,%edx
  cmove  %eax,%ebp
L144:
  add    $0x1,%rax
  cmp    %rax,%rsi
  jne    L5
L60:
  mov    %ebp,%esi
  lea    D8(%rip),%rdi
  lea    D11(%rip),%r12
  call   <printf@plt>
  mov    0xc(%rsp),%eax
  mov    $0x1,%esi
  test   %eax,%eax
  jle    L81
  nopl   (%rax)
L80:
  movslq %esi,%rax
  mov    0x1b0(%rsp,%rax,4),%ebx
  cmp    0x350(%rsp,%rax,4),%ebx
  jle    L119
  jmp    L22
  nopl   (%rax)
L85:
  add    $0x1,%ebx
  cmp    %ebx,0x350(%rsp,%rax,4)
  jl     L22
L119:
  cmp    %ebp,%ebx
  jne    L85
  mov    %r12,%rdi
  add    $0x1,%ebx
  call   <printf@plt>
  movslq 0xc(%rsp),%rax
  mov    %rax,%rsi
  cmp    %ebx,0x350(%rsp,%rax,4)
  jge    L119
  nopl   0x0(%rax)
L22:
  add    $0x1,%esi
  cmp    %esi,0xc(%rsp)
  jge    L80
L81:
  mov    0x4e8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L18
  add    $0x4f8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L37:
  xor    %ebp,%ebp
  jmp    L60
L18:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: data_label: D25
type_guessed: string
name: disassemble_data
arguments: data_label: D27
type_guessed: string
name: disassemble_data
arguments: data_label: D8
type_guessed: string
name: disassemble_data
arguments: data_label: D11
type_guessed: string

tool: D25: %d

tool: D27: %d %d

tool: D8: Numero de guardioes sinceros = %d


tool: D11: Guardiao %d


assistant: ```c
int main() {
  int n, g[101], i, j, min[101], max[101], nSinc = 0;
  for (i = 0; i < 101; i++) {
    g[i] = 0;
    min[i] = 0;
    max[i] = 0;
  }
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d %d", &min[i], &max[i]);
    for (j = min[i]; j <= max[i]; j++)
      g[j]++;
  }
  for (i = 1; i <= n; i++) {
    if (g[i] >= g[nSinc] && i == g[i])
      nSinc = i;
  }
  printf("Numero de guardioes sinceros = %d\n", nSinc);
  for (i = 1; i <= n; i++) {
    for (j = min[i]; j <= max[i]; j++)
      if (j == nSinc) {
        printf("Guardiao %d\n", i);
        i = n;
      }
  }
  return 0;
}
```