user: Translate the below assembly code into C programming language:

```asm
<solve>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  test   %esi,%esi
  jle    L37
  mov    %rdi,%rbx
  mov    %esi,%ebp
  lea    (%rsi,%rsi,1),%r15d
  mov    %esi,%r14d
  lea    -0x1(%rsi),%r12d
  xor    %r13d,%r13d
  nopl   (%rax)
L140:
  mov    %rbx,%rdi
  mov    %ebp,%edx
  mov    %r13d,%esi
  call   <pivoting@plt>
  lea    0x0(,%r13,4),%r8
  xor    %edi,%edi
  nopw   0x0(%rax,%rax,1)
L18:
  cmp    %r13d,%edi
  je     L69
  mov    (%rbx,%rdi,8),%rdx
  mov    (%rbx,%r13,8),%rsi
  xor    %eax,%eax
  movss  (%rdx,%r8,1),%xmm2
  divss  (%rsi,%r8,1),%xmm2
  nopl   0x0(%rax,%rax,1)
L142:
  movss  (%rsi,%rax,4),%xmm1
  movss  (%rdx,%rax,4),%xmm0
  mulss  %xmm2,%xmm1
  subss  %xmm1,%xmm0
  movss  %xmm0,(%rdx,%rax,4)
  add    $0x1,%rax
  cmp    %eax,%r15d
  jg     L142
L69:
  add    $0x1,%rdi
  cmp    %r14,%rdi
  jne    L18
  add    $0x1,%r13
  cmp    %rdi,%r13
  jne    L140
  movslq %ebp,%rax
  shl    $0x2,%r13
  xor    %edi,%edi
  lea    0x0(,%rax,4),%rcx
  mov    %rax,0x28(%rsp)
  lea    0x1(%rax,%r12,1),%r8
  mov    %rcx,0x10(%rsp)
  shl    $0x2,%r8
  cs nopw 0x0(%rax,%rax,1)
L45:
  mov    (%rbx,%rdi,2),%rdx
  mov    0x10(%rsp),%rax
  lea    (%rdx,%rdi,1),%rsi
  add    %rdx,%rax
  add    %r8,%rdx
  nopl   0x0(%rax,%rax,1)
L48:
  movss  (%rax),%xmm0
  add    $0x4,%rax
  divss  (%rsi),%xmm0
  movss  %xmm0,-0x4(%rax)
  cmp    %rax,%rdx
  jne    L48
  add    $0x4,%rdi
  cmp    %rdi,%r13
  jne    L45
  xor    %eax,%eax
  nopl   0x0(%rax)
L120:
  mov    (%rbx,%rax,2),%rdx
  add    %rax,%rdx
  add    $0x4,%rax
  movss  (%rdx),%xmm0
  divss  %xmm0,%xmm0
  movss  %xmm0,(%rdx)
  cmp    %rax,%r13
  jne    L120
  lea    D74(%rip),%rdi
  mov    %rbx,%r14
  lea    D25(%rip),%r13
  call   <puts@plt>
  lea    0x0(%rbp,%rbp,1),%eax
  mov    %eax,0x24(%rsp)
  lea    0x8(%rbx,%r12,8),%rax
  mov    %rax,0x18(%rsp)
  mov    %r15d,%eax
  mov    %rax,0x8(%rsp)
  nopl   0x0(%rax)
L111:
  xor    %r12d,%r12d
  jmp    L96
  nopl   (%rax)
L143:
  mov    %r13,%rdi
  mov    $0x1,%eax
  add    $0x1,%r12
  call   <printf@plt>
  cmp    %r12,0x8(%rsp)
  je     L116
L96:
  mov    (%r14),%rax
  pxor   %xmm0,%xmm0
  cvtss2sd (%rax,%r12,4),%xmm0
  cmp    %r12d,%ebp
  jne    L143
  lea    D4(%rip),%rdi
  mov    $0x1,%eax
  add    $0x1,%r12
  call   <printf@plt>
  cmp    %r12,0x8(%rsp)
  jne    L96
L116:
  mov    $0xa,%edi
  add    $0x8,%r14
  call   <putchar@plt>
  cmp    %r14,0x18(%rsp)
  jne    L111
  lea    D26(%rip),%rdi
  lea    D45(%rip),%r13
  call   <puts@plt>
  lea    D23(%rip),%rdi
  call   <puts@plt>
  mov    %r15d,%ecx
  sub    %ebp,%ecx
  lea    -0x1(%rcx),%eax
  mov    0x28(%rsp),%rcx
  lea    0x1(%rcx,%rax,1),%r12
  shl    $0x2,%r12
  nopl   0x0(%rax,%rax,1)
L127:
  mov    0x10(%rsp),%r15
  cmp    0x24(%rsp),%ebp
  jge    L13
  nopl   0x0(%rax,%rax,1)
L139:
  mov    (%rbx),%rax
  pxor   %xmm0,%xmm0
  mov    %r13,%rdi
  cvtss2sd (%rax,%r15,1),%xmm0
  mov    $0x1,%eax
  add    $0x4,%r15
  call   <printf@plt>
  cmp    %r12,%r15
  jne    L139
L13:
  mov    $0xa,%edi
  add    $0x8,%rbx
  call   <putchar@plt>
  cmp    %rbx,0x18(%rsp)
  jne    L127
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L37:
  lea    D74(%rip),%rdi
  call   <puts@plt>
  lea    D26(%rip),%rdi
  call   <puts@plt>
  add    $0x38,%rsp
  lea    D23(%rip),%rdi
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1090 <puts@plt>
```

tool_calls: name: read_data
arguments: tag: D74
type_infered: string
name: read_data
arguments: tag: D25
type_infered: string
name: read_data
arguments: tag: D4
type_infered: string
name: read_data
arguments: tag: D26
type_infered: string
name: read_data
arguments: tag: D45
type_infered: string
name: read_data
arguments: tag: D23
type_infered: string

tool: D74: 
 The augmented matrix is :

tool: D25: %f 

tool: D4:  |  %f 

tool: D26: 


tool: D45: %f  

tool: D23: The Inverse Matrix is :-


assistant: ```c
void solve(float **arr, int n) {
  int i, j, k, l, p;
  float m, y, max, temp;
  float x[n];
  for (k = 0; k < n; k++) {
    pivoting(arr, k, n);
    for (i = 0; i < n; i++) {
      if (i == k)
        continue;
      m = arr[i][k] / arr[k][k];
      for (j = 0; j < 2 * n; j++)
        arr[i][j] -= m * arr[k][j];
    }
  }
  for (j = 0; j < n; j++) {
    for (i = 0; i < n; i++) {
      arr[j][i + n] = arr[j][i + n] / arr[j][j];
    }
  }
  for (j = 0; j < n; j++)
    arr[j][j] /= arr[j][j];
  printf("\n The augmented matrix is :\n");
  for (j = 0; j < n; j++) {
    for (i = 0; i < 2 * n; i++) {
      if (i == n)
        printf(" |  %f ", arr[j][i]);
      else
        printf("%f ", arr[j][i]);
    }
    printf("\n");
  }
  printf("\n\n");
  printf("The Inverse Matrix is :-\n\n");
  for (i = 0; i < n; i++) {
    for (j = n; j < 2 * n; j++) {
      printf("%f  ", arr[i][j]);
    }
    printf("\n");
  }
}
```