user: Please show me the C implementation that matches the assembly code listed below:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1f0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x1d8(%rbp)
  lea    -0x1a0(%rbp),%rax
  mov    $0x190,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  movl   $0x1,-0x1a0(%rbp)
  lea    -0x1e4(%rbp),%rax
  mov    %rax,%rsi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  movl   $0x1,-0x1e0(%rbp)
  jmp    L112
L88:
  lea    -0x1d0(%rbp),%rdx
  mov    -0x1e0(%rbp),%eax
  cltq
  shl    $0x2,%rax
  add    %rdx,%rax
  mov    %rax,%rsi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  mov    -0x1e0(%rbp),%eax
  cltq
  mov    -0x1d0(%rbp,%rax,4),%eax
  add    %eax,-0x1d8(%rbp)
  addl   $0x1,-0x1e0(%rbp)
L112:
  mov    -0x1e4(%rbp),%eax
  cmp    %eax,-0x1e0(%rbp)
  jle    L88
  movl   $0x1,-0x1e0(%rbp)
  jmp    L91
L77:
  mov    -0x1d8(%rbp),%eax
  mov    %eax,-0x1dc(%rbp)
  jmp    L30
L13:
  mov    -0x1dc(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%eax
  test   %eax,%eax
  je     L3
  mov    -0x1dc(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%ecx
  mov    -0x1e0(%rbp),%eax
  cltq
  mov    -0x1d0(%rbp,%rax,4),%edx
  mov    -0x1dc(%rbp),%eax
  add    %edx,%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%edx
  mov    -0x1e0(%rbp),%eax
  cltq
  mov    -0x1d0(%rbp,%rax,4),%esi
  mov    -0x1dc(%rbp),%eax
  add    %esi,%eax
  add    %ecx,%edx
  cltq
  mov    %edx,-0x1a0(%rbp,%rax,4)
L3:
  subl   $0x1,-0x1dc(%rbp)
L30:
  cmpl   $0x0,-0x1dc(%rbp)
  jns    L13
  addl   $0x1,-0x1e0(%rbp)
L91:
  mov    -0x1e4(%rbp),%eax
  cmp    %eax,-0x1e0(%rbp)
  jle    L77
  movl   $0x0,-0x1dc(%rbp)
  jmp    L53
L39:
  mov    -0x1dc(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%edx
  mov    -0x1dc(%rbp),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x1dc(%rbp)
L53:
  mov    -0x1dc(%rbp),%eax
  cmp    -0x1d8(%rbp),%eax
  jle    L39
  movl   $0x0,-0x1d4(%rbp)
  movl   $0x0,-0x1dc(%rbp)
  jmp    L4
L97:
  mov    -0x1dc(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%eax
  cmp    %eax,-0x1d4(%rbp)
  jg     L110
  mov    -0x1dc(%rbp),%eax
  cltq
  mov    -0x1a0(%rbp,%rax,4),%eax
  mov    %eax,-0x1d4(%rbp)
L110:
  addl   $0x1,-0x1dc(%rbp)
L4:
  mov    -0x1dc(%rbp),%eax
  cmp    -0x1d8(%rbp),%eax
  jle    L97
  mov    -0x1d4(%rbp),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L20
  call   <__stack_chk_fail@plt>
L20:
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D23
type_guessed: string
name: read_data
arguments: label: D5
type_guessed: string
name: read_data
arguments: label: D0
type_guessed: string

tool: D23: %d

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


tool: D0: Max Way to Give Money = %d


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