user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<main>:
  endbr64
  push   %r15
  lea    D76(%rip),%rdi
  xor    %r15d,%r15d
  push   %r14
  lea    D29(%rip),%r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x78,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  call   <puts@plt>
  lea    0x60(%rsp),%rax
  mov    %rax,0x8(%rsp)
  mov    %rax,%r12
L99:
  lea    -0x50(%r12),%rbx
L70:
  mov    %rbx,%rsi
  mov    %r14,%rdi
  add    $0x14,%rbx
  call   <scanf@plt>
  cmp    %r12,%rbx
  jne    L70
  add    $0x1,%r15
  lea    0x4(%rbx),%r12
  cmp    $0x5,%r15
  jne    L99
  mov    $0xa,%edi
  mov    %rbp,%rbx
  xor    %r14d,%r14d
  call   <putchar@plt>
  lea    D16(%rip),%r15
  lea    D18(%rip),%r12
L128:
  add    $0x1,%r14d
  mov    %r15,%rdi
  xor    %eax,%eax
  xor    %r13d,%r13d
  mov    %r14d,%esi
  call   <printf@plt>
L80:
  mov    (%rbx,%r13,4),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%r13
  call   <printf@plt>
  cmp    $0x5,%r13
  jne    L80
  mov    $0xa,%edi
  add    $0x14,%rbx
  call   <putchar@plt>
  cmp    $0x4,%r14d
  jne    L128
  lea    D75(%rip),%rdi
  mov    %rbp,%r12
  lea    D28(%rip),%r14
  call   <puts@plt>
L23:
  xor    %ebx,%ebx
L28:
  mov    (%r12,%rbx,4),%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x5,%rbx
  jne    L28
  mov    $0xa,%edi
  add    $0x14,%r12
  call   <putchar@plt>
  cmp    %r12,0x8(%rsp)
  jne    L23
  mov    0x10(%rsp),%r10d
  xor    %esi,%esi
  xor    %r12d,%r12d
  xor    %r13d,%r13d
  xor    %r14d,%r14d
  xor    %r11d,%r11d
  xor    %ebx,%ebx
  xor    %r9d,%r9d
  mov    %r10d,%eax
  mov    %r10d,%r8d
L120:
  mov    %esi,%edi
  lea    0x0(%rbp,%rsi,4),%rcx
  xor    %edx,%edx
  cmp    %r10d,%eax
  jg     L105
L62:
  cmp    %r8d,%eax
  jge    L82
  mov    %edi,%ebx
  mov    %edx,%r9d
  mov    %eax,%r8d
L82:
  cmp    $0x80000000,%eax
  sbb    $0xffffffff,%r13d
  add    %r12d,%eax
  cmp    $0x2,%rsi
  cmove  %eax,%r12d
  add    $0x1,%edx
  add    $0x14,%rcx
  cmp    $0x4,%edx
  je     L44
  mov    (%rcx),%eax
  cmp    %r10d,%eax
  jle    L62
L105:
  mov    %edi,%r14d
  mov    %edx,%r11d
  mov    %eax,%r10d
  jmp    L82
  nop
L44:
  add    $0x1,%rsi
  cmp    $0x5,%rsi
  je     L5
  mov    0x0(%rbp,%rsi,4),%eax
  jmp    L120
L5:
  sub    $0x8,%rsp
  mov    %r11d,%edx
  mov    %r10d,%esi
  mov    %r14d,%ecx
  push   %rbx
  lea    D45(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12d,%esi
  lea    D59(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x14,%edx
  mov    %r13d,%esi
  xor    %eax,%eax
  sub    %r13d,%edx
  lea    D8(%rip),%rdi
  call   <printf@plt>
  pop    %rax
  pop    %rdx
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L127
  add    $0x78,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L127:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D76
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D29
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D16
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D18
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D75
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D28
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D45
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D59
type_guessed: string
name: extract_symbolic_data
arguments: data_tag: D8
type_guessed: string

tool: D76: Vamos analisar uma matriz 4x5!
Forneca os valores dela, ao todo 20.
(A matriz sera orientada por Linhas)

tool: D29: %d

tool: D16: Linha %d:  

tool: D18: %d 

tool: D75: 

Matriz dos Dados:

tool: D28: [%d]	

tool: D45: 

O maior numero digitado e o: %d [Pos: %d %d]
 e o menor numero digitado e o: %d [Pos: %d %d]



tool: D59: A soma dos elementos da terceira coluna e: %d



tool: D8: Foram inseridos %d numeros negativos e %d numeros positivos.


assistant: ```c
int main() {
  int numero[4][5];
  int i, j, maior, menor, menori, menorj, maiori, maiorj, negativos = 0, col3 = 0;
  printf("Vamos analisar uma matriz 4x5!\nForneca os valores dela, ao todo 20.\n(A matriz sera orientada por Linhas)\n");
  for (j = 0; j < 5; j++) {
    for (i = 0; i < 4; i++) {
      scanf("%d", &numero[i][j]);
    }
  }
  printf("\n");
  for (i = 0; i < 4; i++) {
    printf("Linha %d:  ", (i + 1));
    for (j = 0; j < 5; j++) {
      printf("%d ", numero[i][j]);
    }
    printf("\n");
  }
  printf("\n\nMatriz dos Dados:\n");
  for (i = 0; i < 4; i++) {
    for (j = 0; j < 5; j++) {
      printf("[%d]\t", numero[i][j]);
    }
    printf("\n");
  }
  maior = numero[0][0];
  menor = numero[0][0];
  maiori = 0;
  maiorj = 0;
  menori = 0;
  menorj = 0;
  for (j = 0; j < 5; j++) {
    for (i = 0; i < 4; i++) {
      if (numero[i][j] > maior) {
        maior = numero[i][j];
        maiori = i;
        maiorj = j;
      } else if (numero[i][j] < menor) {
        menor = numero[i][j];
        menori = i;
        menorj = j;
      }
      if (numero[i][j] < 0) {
        negativos++;
      }
      if (j == 2) {
        col3 = col3 + numero[i][j];
      }
    }
  }
  printf("\n\nO maior numero digitado e o: %d [Pos: %d %d]\n e o menor numero digitado e o: %d [Pos: %d %d]\n\n", maior, maiori, maiorj, menor, menori, menorj);
  printf("A soma dos elementos da terceira coluna e: %d\n\n", col3);
  printf("Foram inseridos %d numeros negativos e %d numeros positivos.\n", negativos, (20 - negativos));
  return 0;
}
```