user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %r15
  lea    D35(%rip),%r15
  push   %r14
  push   %r13
  xor    %r13d,%r13d
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x198,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x188(%rsp)
  xor    %eax,%eax
  lea    0x54(%rsp),%rax
  lea    0x40(%rsp),%rbp
  mov    %rax,0x8(%rsp)
L135:
  mov    0x8(%rsp),%rax
  xor    %r14d,%r14d
  lea    -0x14(%rax),%rbx
L121:
  mov    %r14d,(%rbx)
  mov    %r14d,%esi
  mov    %r15,%rdi
  xor    %eax,%eax
  add    %r13d,%r14d
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    0x8(%rsp),%rbx
  jne    L121
  mov    $0xa,%edi
  add    $0x1,%r13d
  call   <putchar@plt>
  lea    0x14(%rbx),%rax
  mov    %rax,0x8(%rsp)
  cmp    $0x5,%r13d
  jne    L135
  movq   $0x0,0x30(%rsp)
  pxor   %xmm0,%xmm0
  lea    0x64(%rbp),%rdi
  mov    %rbp,%rcx
  movaps %xmm0,0x10(%rsp)
  movaps %xmm0,0x20(%rsp)
L76:
  xor    %eax,%eax
L166:
  movslq (%rcx,%rax,4),%rdx
  add    $0x1,%rax
  addl   $0x1,0x10(%rsp,%rdx,4)
  cmp    $0x5,%rax
  jne    L166
  add    $0x14,%rcx
  cmp    %rdi,%rcx
  jne    L76
  lea    0x10(%rsp),%rdx
  xor    %eax,%eax
  xor    %esi,%esi
  nopl   (%rax)
L125:
  test   %eax,%eax
  je     L106
  cmp    %esi,(%rdx)
  jle    L104
L106:
  mov    (%rdx),%esi
  mov    %eax,%r12d
L104:
  add    $0x1,%eax
  add    $0x4,%rdx
  cmp    $0xa,%eax
  jne    L125
  xor    %r13d,%r13d
  xor    %r14d,%r14d
L181:
  xor    %eax,%eax
L184:
  mov    0x0(%rbp,%rax,4),%edx
  cmp    %r12d,%edx
  jge    L52
  movslq %r14d,%rcx
  add    $0x1,%r14d
  mov    %edx,0xb0(%rsp,%rcx,4)
L52:
  cmp    %r12d,%edx
  jle    L175
  movslq %r13d,%rcx
  add    $0x1,%r13d
  mov    %edx,0x120(%rsp,%rcx,4)
L175:
  add    $0x1,%rax
  cmp    $0x5,%rax
  jne    L184
  add    $0x14,%rbp
  cmp    %rdi,%rbp
  jne    L181
  xor    %eax,%eax
  mov    %r12d,%edx
  lea    D11(%rip),%rdi
  call   <printf@plt>
  test   %r14d,%r14d
  je     L126
  lea    -0x1(%r14),%eax
  lea    0xb0(%rsp),%rbx
  lea    0xb4(%rsp,%rax,4),%r12
  lea    D49(%rip),%rbp
  nop
L123:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %rbx,%r12
  jne    L123
L126:
  lea    D25(%rip),%r15
  xor    %eax,%eax
  mov    %r14d,%esi
  mov    %r15,%rdi
  call   <printf@plt>
  test   %r13d,%r13d
  je     L152
  lea    0x120(%rsp),%rax
  lea    D49(%rip),%rbp
  mov    %rax,0x8(%rsp)
  mov    %rax,%rbx
  lea    -0x1(%r13),%eax
  mov    %rax,%r12
  lea    0x124(%rsp,%rax,4),%r14
  nopl   0x0(%rax,%rax,1)
L44:
  mov    (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x4,%rbx
  call   <printf@plt>
  cmp    %r14,%rbx
  jne    L44
L20:
  mov    %r13d,%esi
  mov    %r15,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x8(%rsp),%rax
  xor    %edx,%edx
  jmp    L47
  nopl   0x0(%rax,%rax,1)
L39:
  mov    (%rax),%ecx
  add    $0x4,%rax
  add    $0x1,%edx
  cmp    (%rax),%ecx
  jg     L132
L47:
  cmp    %r12d,%edx
  jl     L39
  lea    D65(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
L77:
  mov    0x188(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L200
  add    $0x198,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L132:
  lea    D34(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L77
L152:
  lea    0x120(%rsp),%rax
  or     $0xffffffff,%r12d
  mov    %rax,0x8(%rsp)
  jmp    L20
L200:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: label: D35
type: string
name: disassemble_data
arguments: label: D11
type: string
name: disassemble_data
arguments: label: D49
type: string
name: disassemble_data
arguments: label: D25
type: string
name: disassemble_data
arguments: label: D65
type: string
name: disassemble_data
arguments: label: D34
type: string

tool: D35: %2d 

tool: D11: fmax = %d, valmax = %d


tool: D49: %d 

tool: D25:  (%d elementi)


tool: D65: Vettore monotono crescente.

tool: D34: Vettore NON monotono crescente.

assistant: ```c
int main() {
  int m[5][5];
  int v1[5 * 5], v2[5 * 5];
  int freq[10];
  int fmax = 0;
  int valmax;
  int u, z;
  int i, j;
  for (i = 0; i < 5; i++) {
    for (j = 0; j < 5; j++) {
      m[i][j] = i * j;
      printf("%2d ", m[i][j]);
    }
    printf("\n");
  }
  for (i = 0; i < 10; i++)
    freq[i] = 0;
  for (i = 0; i < 5; i++) {
    for (j = 0; j < 5; j++) {
      freq[m[i][j]]++;
    }
  }
  for (i = 0; i < 10; i++)
    if (i == 0 || freq[i] > fmax) {
      valmax = i;
      fmax = freq[i];
    }
  u = z = 0;
  for (i = 0; i < 5; i++) {
    for (j = 0; j < 5; j++) {
      if (m[i][j] < valmax) {
        v1[u] = m[i][j];
        u++;
      }
      if (m[i][j] > valmax) {
        v2[z] = m[i][j];
        z++;
      }
    }
  }
  printf("fmax = %d, valmax = %d\n", fmax, valmax);
  for (i = 0; i < u; i++)
    printf("%d ", v1[i]);
  printf(" (%d elementi)\n", u);
  for (i = 0; i < z; i++)
    printf("%d ", v2[i]);
  printf(" (%d elementi)\n", z);
  i = 0;
  while (i < z - 1 && v2[i] <= v2[i + 1])
    i++;
  if (i < z - 1)
    printf("Vettore NON monotono crescente.");
  else
    printf("Vettore monotono crescente.");
}
```