user: What would the C source look like for the assembly code provided here?

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x4000(%rsp),%r11
L185:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L185
  sub    $0x848,%rsp
  lea    D42(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x4838(%rsp)
  xor    %eax,%eax
  lea    0x482d(%rsp),%rbp
  lea    0x20(%rsp),%r13
  mov    %rbp,%rsi
  call   <scanf@plt>
  lea    0x1c(%rsp),%rsi
  lea    D62(%rip),%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  movslq %eax,%rcx
  xor    %eax,%eax
  mov    %rcx,%rbx
  test   %ecx,%ecx
  jle    L132
  nopl   0x0(%rax)
L96:
  movsbl 0x0(%rbp,%rax,1),%edx
  sub    $0x30,%edx
  mov    %edx,0x0(%r13,%rax,4)
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L96
L132:
  lea    -0x1(%rbx),%r13d
  xor    %ebp,%ebp
  xor    %r12d,%r12d
  test   %r13d,%r13d
  jle    L82
  nopl   0x0(%rax)
L175:
  mov    %r12d,%esi
  mov    $0x2,%edi
  add    $0x1,%r12d
  call   <pow@plt>
  add    %eax,%ebp
  cmp    %r13d,%r12d
  jne    L175
  test   %ebp,%ebp
  js     L8
L82:
  lea    0x50(%rsp),%r15
  lea    -0x2(%rbx),%r9d
  xor    %edi,%edi
  mov    %r15,%rsi
  xchg   %ax,%ax
L110:
  movslq %r9d,%rax
  mov    %edi,%edx
  test   %r9d,%r9d
  js     L109
  nopw   0x0(%rax,%rax,1)
L192:
  mov    %edx,%ecx
  sar    %edx
  and    $0x1,%ecx
  mov    %ecx,(%rsi,%rax,4)
  sub    $0x1,%rax
  test   %eax,%eax
  jns    L192
L109:
  add    $0x1,%edi
  add    $0x24,%rsi
  cmp    %edi,%ebp
  jge    L110
  mov    0x20(%rsp),%eax
  movl   $0x0,0xc(%rsp)
  xor    %r12d,%r12d
  mov    %eax,0x8(%rsp)
L164:
  mov    0x8(%rsp),%edi
  mov    $0x1,%r14d
  lea    0x20(%rsp),%r13
  cmp    $0x1,%ebx
  jle    L22
  nopl   0x0(%rax)
L199:
  mov    -0x4(%r15,%r14,4),%edx
  mov    0x0(%r13,%r14,4),%esi
  add    $0x1,%r14
  call   <calculation@plt>
  mov    %eax,%edi
  cmp    %r14d,%ebx
  jg     L199
  cmp    %eax,0x1c(%rsp)
  je     L62
L21:
  add    $0x1,%r12d
  add    $0x24,%r15
  cmp    %r12d,%ebp
  jge    L164
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  je     L8
L193:
  mov    0x4838(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L137
  add    $0x4848,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L22:
  mov    0x8(%rsp),%eax
  cmp    %eax,0x1c(%rsp)
  jne    L21
  mov    0x8(%rsp),%esi
  lea    D62(%rip),%rdi
  xor    %eax,%eax
  addl   $0x1,0xc(%rsp)
  call   <printf@plt>
L50:
  mov    0x1c(%rsp),%esi
  lea    D28(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L21
L62:
  mov    0x8(%rsp),%esi
  lea    D62(%rip),%rdi
  xor    %eax,%eax
  addl   $0x1,0xc(%rsp)
  mov    $0x1,%r14d
  call   <printf@plt>
  nopw   0x0(%rax,%rax,1)
L178:
  xor    %edi,%edi
  cmpl   $0x1,-0x4(%r15,%r14,4)
  setne  %dil
  lea    0x2b(%rdi,%rdi,1),%edi
  call   <putchar@plt>
  mov    0x0(%r13,%r14,4),%esi
  lea    D62(%rip),%rdi
  xor    %eax,%eax
  add    $0x1,%r14
  call   <printf@plt>
  cmp    %r14d,%ebx
  jg     L178
  jmp    L50
L8:
  lea    D26(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L193
L137:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: disassemble_data
arguments: tag: D42
datatype: string
name: disassemble_data
arguments: tag: D62
datatype: string
name: disassemble_data
arguments: tag: D28
datatype: string
name: disassemble_data
arguments: tag: D26
datatype: string

tool: D42: %s

tool: D62: %d

tool: D28: =%d


tool: D26: -1

assistant: ```c
int main() {
  char nums[11];
  int a[11];
  int pp[511][9];
  int n, l = 0, i, sum = 0, j, k = 0, max = 0, current, elem, found = 0;
  char oper;
  scanf("%s", nums);
  scanf("%d", &n);
  l = strlen(nums);
  for (i = 0; i < l; i++) {
    a[i] = nums[i] - '0';
  }
  for (i = 0; i < l - 1; i++) {
    max = max + pow(2, i);
  }
  for (i = 0; i <= max; i++) {
    current = i;
    for (j = l - 2; j >= 0; j--) {
      elem = current % 2;
      current = current / 2;
      pp[i][j] = elem;
    }
  }
  sum = a[0];
  for (i = 0; i <= max; i++) {
    for (k = 1; k < l; k++) {
      sum = calculation(sum, a[k], pp[i][k - 1]);
    }
    if (sum == n) {
      found++;
      printf("%d", a[0]);
      for (k = 1; k < l; k++) {
        if (pp[i][k - 1] == 1) {
          oper = '+';
        } else {
          oper = '-';
        }
        printf("%c", oper);
        printf("%d", a[k]);
      }
      printf("=%d\n", n);
    }
    sum = a[0];
  }
  if (found == 0) {
    printf("-1");
  }
  return 0;
}
```