user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<dynamic_programming>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  mov    %esi,%r15d
  push   %r14
  mov    %rdx,%r14
  push   %r13
  push   %r12
  mov    %edi,%r12d
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  lea    0x1(%rdi),%eax
  mov    %rsp,%rcx
  cltq
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L23
L180:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L180
L23:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L56
L42:
  lea    0x3(%rsp),%rbx
  movslq %r12d,%r13
  mov    %rbx,%rax
  and    $0xfffffffffffffffc,%rbx
  shr    $0x2,%rax
  movl   $0x1,0x0(,%rax,4)
  test   %r12d,%r12d
  jle    L70
  lea    0x0(,%r13,4),%rdx
  lea    0x4(%rbx),%rdi
  xor    %esi,%esi
  call   <memset@plt>
L70:
  test   %r15d,%r15d
  jle    L1
  lea    -0x1(%r15),%eax
  mov    %r14,%rsi
  lea    0x4(%rbx),%r8
  lea    0x4(%r14,%rax,4),%rdi
  nopl   0x0(%rax,%rax,1)
L14:
  mov    (%rsi),%edx
  cmp    %edx,%r12d
  jl     L16
  lea    -0x1(%rdx),%eax
  cmp    $0x2,%eax
  jbe    L18
  cmp    %edx,%r12d
  je     L18
  mov    %r12d,%r11d
  sub    %edx,%r11d
  lea    0x1(%r11),%r10d
  cmp    $0x2,%r11d
  jbe    L45
  mov    %r10d,%r9d
  movslq %edx,%rax
  shr    $0x2,%r9d
  lea    (%rbx,%rax,4),%rcx
  xor    %eax,%eax
  shl    $0x4,%r9
  nopl   0x0(%rax,%rax,1)
L117:
  movdqu (%rcx,%rax,1),%xmm0
  movdqu (%rbx,%rax,1),%xmm2
  paddd  %xmm2,%xmm0
  movups %xmm0,(%rcx,%rax,1)
  add    $0x10,%rax
  cmp    %r9,%rax
  jne    L117
  mov    %r10d,%ecx
  and    $0xfffffffc,%ecx
  lea    (%rcx,%rdx,1),%eax
  cmp    %ecx,%r10d
  je     L16
L177:
  sub    %ecx,%r10d
  cmp    %ecx,%r11d
  je     L157
  movslq %edx,%r9
  movq   (%rbx,%rcx,4),%xmm0
  add    %rcx,%r9
  mov    %r10d,%ecx
  lea    (%rbx,%r9,4),%r9
  and    $0xfffffffe,%ecx
  movq   (%r9),%xmm1
  add    %ecx,%eax
  paddd  %xmm1,%xmm0
  movq   %xmm0,(%r9)
  cmp    %ecx,%r10d
  je     L16
L157:
  movslq %eax,%rcx
  sub    %edx,%eax
  cltq
  mov    (%rbx,%rax,4),%eax
  add    %eax,(%rbx,%rcx,4)
L16:
  add    $0x4,%rsi
  cmp    %rsi,%rdi
  jne    L14
L1:
  mov    (%rbx,%r13,4),%eax
  mov    -0x38(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L115
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L18:
  mov    %r12d,%r9d
  movslq %edx,%r10
  sub    %edx,%r9d
  lea    0x0(,%r10,4),%rcx
  lea    (%r9,%r10,1),%rdx
  lea    (%rbx,%rcx,1),%rax
  neg    %rcx
  lea    (%r8,%rdx,4),%r9
L71:
  mov    (%rax,%rcx,1),%edx
  add    %edx,(%rax)
  add    $0x4,%rax
  cmp    %r9,%rax
  jne    L71
  jmp    L16
L45:
  mov    %edx,%eax
  xor    %ecx,%ecx
  jmp    L177
L56:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L42
L115:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int dynamic_programming(int target, int size, int coins[size]) {
  int ways[target + 1];
  int i, j;
  ways[0] = 1;
  for (i = 1; i <= target; i++)
    ways[i] = 0;
  for (i = 0; i < size; i++) {
    for (j = coins[i]; j <= target; j++) {
      ways[j] += ways[j - coins[i]];
    }
  }
  return ways[target];
}
```