user: Demonstrate the C code that generates the assembly output below:

```asm
<main>:
  endbr64
  push   %r14
  mov    $0x198,%edx
  xor    %esi,%esi
  lea    D10(%rip),%r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x1e0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x1d8(%rsp)
  xor    %eax,%eax
  lea    0x40(%rsp),%rbx
  mov    %rbx,%rdi
  call   <memset@plt>
  lea    0xc(%rsp),%rsi
  mov    %r14,%rdi
  movl   $0x1,0x40(%rsp)
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L52
  lea    0x14(%rsp),%rbp
  mov    $0x1,%r12d
  mov    %rbp,%r13
L7:
  mov    %r13,%rsi
  mov    %r14,%rdi
  add    $0x1,%r12d
  add    $0x4,%r13
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  cmp    %r12d,%eax
  jge    L7
  test   %eax,%eax
  jle    L52
  sub    $0x1,%eax
  lea    0x18(%rsp,%rax,4),%rsi
L54:
  xor    %eax,%eax
  cs nopw 0x0(%rax,%rax,1)
L45:
  mov    (%rbx,%rax,4),%ecx
  test   %ecx,%ecx
  je     L37
  mov    0x0(%rbp),%edx
  add    %eax,%edx
  movslq %edx,%rdx
  add    %ecx,0x40(%rsp,%rdx,4)
L37:
  add    $0x1,%rax
  cmp    $0x65,%rax
  jne    L45
  add    $0x4,%rbp
  cmp    %rsi,%rbp
  jne    L54
L52:
  xor    %ebp,%ebp
  lea    D5(%rip),%r12
  nop
L38:
  mov    (%rbx,%rbp,4),%edx
  mov    %ebp,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbp
  call   <printf@plt>
  cmp    $0x65,%rbp
  jne    L38
  lea    0x1d0(%rsp),%rax
  pxor   %xmm2,%xmm2
  nopl   (%rax)
L81:
  movdqa %xmm2,%xmm0
  add    $0x10,%rbx
  pcmpgtd -0x10(%rbx),%xmm0
  pand   %xmm0,%xmm2
  pandn  -0x10(%rbx),%xmm0
  por    %xmm0,%xmm2
  cmp    %rax,%rbx
  jne    L81
  movdqa %xmm2,%xmm0
  mov    0x1d0(%rsp),%eax
  lea    D8(%rip),%rdi
  psrldq $0x8,%xmm0
  movdqa %xmm0,%xmm1
  pcmpgtd %xmm2,%xmm1
  pand   %xmm1,%xmm0
  pandn  %xmm2,%xmm1
  por    %xmm0,%xmm1
  movdqa %xmm1,%xmm2
  psrldq $0x4,%xmm2
  movdqa %xmm2,%xmm0
  pcmpgtd %xmm1,%xmm0
  pand   %xmm0,%xmm2
  pandn  %xmm1,%xmm0
  por    %xmm2,%xmm0
  movd   %xmm0,%esi
  cmp    %eax,%esi
  cmovl  %eax,%esi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0x1d8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L12
  add    $0x1e0,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L12:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D10
data_type: string
name: read_data
arguments: data_tag: D5
data_type: string
name: read_data
arguments: data_tag: D8
data_type: string

tool: D10: %d

tool: D5: coinChange[%d] = %d


tool: D8: Max Way to Give Money = %d


assistant: ```c
int main() {
  int i, j;
  int coinNumber;
  int coinNumberStore[10];
  int coinChange[102];
  memset(coinChange, 0, sizeof(coinChange));
  coinChange[0] = 1;
  scanf("%d", &coinNumber);
  for (i = 1; i <= coinNumber; i++) {
    scanf("%d", &coinNumberStore[i]);
  }
  for (i = 1; i <= coinNumber; i++) {
    for (j = 0; j <= 100; j++) {
      if (coinChange[j] != 0) {
        coinChange[j + coinNumberStore[i]] = coinChange[j] + coinChange[j + coinNumberStore[i]];
      }
    }
  }
  for (j = 0; j <= 100; j++) {
    printf("coinChange[%d] = %d\n", j, coinChange[j]);
  }
  int MaxWayToGiveMoney = 0;
  for (j = 0; j <= 100; j++) {
    if (coinChange[j] >= MaxWayToGiveMoney)
      MaxWayToGiveMoney = coinChange[j];
  }
  printf("Max Way to Give Money = %d\n", MaxWayToGiveMoney);
  return 0;
}
```