user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r14
  lea    D14(%rip),%r14
  push   %r13
  lea    D57(%rip),%r13
  push   %r12
  xor    %r12d,%r12d
  push   %rbp
  push   %rbx
  sub    $0x70,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x68(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
  mov    %rbp,%rbx
L31:
  add    $0x1,%r12d
  mov    %r14,%rdi
  xor    %eax,%eax
  mov    %r12d,%esi
  call   <printf@plt>
  lea    0x8(%rbx),%rcx
  lea    0x4(%rbx),%rdx
  mov    %rbx,%rsi
  lea    0x10(%rbx),%r9
  lea    0xc(%rbx),%r8
  mov    %r13,%rdi
  add    $0x14,%rbx
  call   <scanf@plt>
  cmp    $0x5,%r12d
  jne    L31
  lea    D32(%rip),%rdi
  mov    %rbp,%r12
  xor    %r13d,%r13d
  call   <puts@plt>
  lea    D1(%rip),%rbx
L28:
  mov    0x4(%r12),%edx
  add    (%r12),%edx
  mov    %r13d,%esi
  mov    %rbx,%rdi
  add    0x8(%r12),%edx
  add    0xc(%r12),%edx
  add    $0x1,%r13d
  add    $0x14,%r12
  add    -0x4(%r12),%edx
  movslq %edx,%rcx
  mov    %edx,%eax
  imul   $0x66666667,%rcx,%rcx
  sar    $0x1f,%eax
  sar    $0x21,%rcx
  sub    %eax,%ecx
  xor    %eax,%eax
  call   <printf@plt>
  cmp    $0x5,%r13d
  jne    L28
  lea    D56(%rip),%rdi
  xor    %r12d,%r12d
  lea    D22(%rip),%rbx
  call   <puts@plt>
L33:
  mov    0x0(%rbp),%eax
  mov    0x14(%rbp),%r11d
  mov    $0x65,%ecx
  mov    0x28(%rbp),%r10d
  mov    0x3c(%rbp),%r9d
  lea    (%rax,%r11,1),%edx
  mov    0x50(%rbp),%edi
  add    %r10d,%edx
  add    %r9d,%edx
  add    %edi,%edx
  movslq %edx,%r8
  mov    %edx,%esi
  mov    %r11d,%edx
  imul   $0x66666667,%r8,%r8
  sar    $0x1f,%esi
  sar    $0x21,%r8
  cmp    %r11d,%eax
  cmovle %eax,%edx
  cmp    %ecx,%edx
  cmovg  %ecx,%edx
  cmp    %r10d,%edx
  cmovg  %r10d,%edx
  cmp    %r9d,%edx
  cmovg  %r9d,%edx
  cmp    %edi,%edx
  mov    %edx,%ecx
  cmovg  %edi,%ecx
  cmp    %r11d,%eax
  cmovl  %r11d,%eax
  xor    %edx,%edx
  test   %eax,%eax
  cmovs  %edx,%eax
  cmp    %r10d,%eax
  cmovl  %r10d,%eax
  cmp    %r9d,%eax
  cmovl  %r9d,%eax
  cmp    %edi,%eax
  cmovge %eax,%edi
  sub    %esi,%r8d
  xor    %eax,%eax
  mov    %r12d,%esi
  add    $0x1,%r12d
  add    $0x4,%rbp
  mov    %edi,%edx
  mov    %rbx,%rdi
  call   <printf@plt>
  cmp    $0x5,%r12d
  jne    L33
  mov    0x68(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L7
  add    $0x70,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L7:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: data_label: D14
data_type: string
name: retrieve_data
arguments: data_label: D57
data_type: string
name: retrieve_data
arguments: data_label: D32
data_type: string
name: retrieve_data
arguments: data_label: D1
data_type: string
name: retrieve_data
arguments: data_label: D56
data_type: string
name: retrieve_data
arguments: data_label: D22
data_type: string

tool: D14: Enter student %d's 5 quiz scores: 

tool: D57: %d %d %d %d %d

tool: D32: Student quiz averages:

tool: D1: Student %d's Total: %d Average: %d 


tool: D56: Quiz stats:

tool: D22: Quiz %d: High: %d Low: %d Avg: %d


assistant: ```c
int main(void) {
  int a[5][5];
  int sum = 0;
  int low;
  int high;
  for (int i = 0; i < 5; i += 1) {
    printf("Enter student %d's 5 quiz scores: ", i + 1);
    scanf("%d %d %d %d %d", &a[i][0], &a[i][1], &a[i][2], &a[i][3], &a[i][4]);
  }
  printf("Student quiz averages:\n");
  for (int i = 0; i < 5; i += 1) {
    sum = 0;
    for (int j = 0; j < 5; j += 1) {
      sum += a[i][j];
    }
    printf("Student %d's Total: %d Average: %d \n", i, sum, sum / 5);
  }
  printf("Quiz stats:\n");
  for (int j = 0; j < 5; j += 1) {
    sum = 0;
    high = 0;
    low = 101;
    for (int i = 0; i < 5; i += 1) {
      sum += a[i][j];
      if (a[i][j] > high) {
        high = a[i][j];
      }
      if (a[i][j] < low) {
        low = a[i][j];
      }
    }
    printf("Quiz %d: High: %d Low: %d Avg: %d\n", j, high, low, sum / 5);
  }
  return 0;
}
```