ylfeng/ReF-Decompile-dataset huggingface dataset:
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    D49(%rip),%rbp
L28:
  mov    %rbp,%rdi
  call   <printf@plt>
  lea    0xe(%rsp),%rbx
  mov    %rbx,%rdi
  call   <gets@plt>
  mov    %rbx,%rdi
  call   <strlen@plt>
  cmp    $0x4,%eax
  ja     L28
  mov    $0x0,%ebx
  mov    $0x0,%r12d
  lea    0xe(%rsp),%rbp
L41:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jbe    L5
  movsbl (%rbx,%rbp,1),%edi
  call   <isalpha@plt>
  add    $0x1,%rbx
  mov    $0x1,%r12d
  test   %rax,%rax
  je     L41
L69:
  lea    D25(%rip),%r12
  lea    0xe(%rsp),%rbp
L37:
  mov    %r12,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <gets@plt>
  mov    $0x0,%ebx
L85:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jbe    L37
  movsbl (%rbx,%rbp,1),%edi
  call   <isalpha@plt>
  add    $0x1,%rbx
  test   %rax,%rax
  je     L85
L14:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L100
  mov    $0x0,%eax
  add    $0x20,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L5:
  test   %r12d,%r12d
  je     L69
  lea    0xe(%rsp),%rdi
  call   <atoi@plt>
  mov    %eax,%ecx
  mov    $0x190,%esi
  cltd
  idiv   %esi
  mov    %ecx,%eax
  and    $0x3,%eax
  or     %eax,%edx
  je     L108
  lea    D43(%rip),%rdi
  call   <printf@plt>
L0:
  mov    $0x0,%eax
  call   <getchar@plt>
  jmp    L14
L108:
  mov    $0x64,%esi
  mov    %ecx,%eax
  cltd
  idiv   %esi
  test   %edx,%edx
  je     L88
  lea    D24(%rip),%rdi
  call   <printf@plt>
  jmp    L0
L88:
  lea    D24(%rip),%rdi
  call   <printf@plt>
  jmp    L0
L100:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D49
type_guessed: string
name: read_data
arguments: data_tag: D25
type_guessed: string
name: read_data
arguments: data_tag: D43
type_guessed: string
name: read_data
arguments: data_tag: D24
type_guessed: string

tool: D49: Enter 4 digit year: 

tool: D25: Enter only digits: 

tool: D43: It's not a leap year

tool: D24: It's a leap year

assistant: ```c
int main() {
  char s[10];
  int year;
  do {
    printf("Enter 4 digit year: ");
    gets(s);
  } while ((unsigned)strlen(s) > 4);
  int isNumber = 0;
  for (int i = 0; i < (unsigned)strlen(s); i++) {
    if (isalpha(s[i]) == 0) {
      isNumber = 1;
    } else {
      isNumber = 0;
      break;
    }
  }
  while (!isNumber) {
    printf("Enter only digits: ");
    gets(s);
    int isNumber = 0;
    for (int i = 0; i < (unsigned)strlen(s); i++) {
      if (isalpha(s[i]) == 0) {
        isNumber = 1;
      } else {
        return isNumber = 0;
      }
    }
  }
  year = atoi(s);
  if ((year % 4 != 0) || (year % 400 != 0)) {
    printf("It's not a leap year");
  } else if (year % 100 != 0) {
    printf("It's a leap year");
  } else {
    printf("It's a leap year");
  }
  getchar();
  return 0;
}
```
user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    D27(%rip),%rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xe(%rsp),%rbp
  nopl   0x0(%rax,%rax,1)
L87:
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <gets@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    $0x4,%rax
  jg     L87
  xor    %ebx,%ebx
  xor    %edx,%edx
  jmp    L42
  nopl   0x0(%rax)
L31:
  movsbl 0x0(%rbp,%rbx,1),%edi
  add    $0x1,%rbx
  call   <isalpha@plt>
  test   %rax,%rax
  jne    L32
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    $0x1,%edx
L42:
  cmp    %ebx,%eax
  ja     L31
  test   %edx,%edx
  jne    L48
L32:
  lea    D33(%rip),%r12
L109:
  mov    %r12,%rdi
  xor    %ebx,%ebx
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <gets@plt>
  nopl   0x0(%rax)
L4:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jbe    L109
  movsbl 0x0(%rbp,%rbx,1),%edi
  add    $0x1,%rbx
  call   <isalpha@plt>
  test   %rax,%rax
  je     L4
L16:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L39
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L48:
  mov    %rbp,%rdi
  call   <atoi@plt>
  mov    $0x190,%esi
  cltd
  mov    %eax,%ecx
  idiv   %esi
  and    $0x3,%ecx
  or     %ecx,%edx
  jne    L44
  lea    D38(%rip),%rdi
  call   <printf@plt>
L63:
  xor    %eax,%eax
  call   <getchar@plt>
  jmp    L16
L44:
  lea    D35(%rip),%rdi
  call   <printf@plt>
  jmp    L63
L39:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: label: D27
type_guessed: string
name: unpack_data
arguments: label: D33
type_guessed: string
name: unpack_data
arguments: label: D38
type_guessed: string
name: unpack_data
arguments: label: D35
type_guessed: string

tool: D27: Enter 4 digit year: 

tool: D33: Enter only digits: 

tool: D38: It's a leap year

tool: D35: It's not a leap year

assistant: ```c
int main() {
  char s[10];
  int year;
  do {
    printf("Enter 4 digit year: ");
    gets(s);
  } while ((unsigned)strlen(s) > 4);
  int isNumber = 0;
  for (int i = 0; i < (unsigned)strlen(s); i++) {
    if (isalpha(s[i]) == 0) {
      isNumber = 1;
    } else {
      isNumber = 0;
      break;
    }
  }
  while (!isNumber) {
    printf("Enter only digits: ");
    gets(s);
    int isNumber = 0;
    for (int i = 0; i < (unsigned)strlen(s); i++) {
      if (isalpha(s[i]) == 0) {
        isNumber = 1;
      } else {
        return isNumber = 0;
      }
    }
  }
  year = atoi(s);
  if ((year % 4 != 0) || (year % 400 != 0)) {
    printf("It's not a leap year");
  } else if (year % 100 != 0) {
    printf("It's a leap year");
  } else {
    printf("It's a leap year");
  }
  getchar();
  return 0;
}
```
user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<main>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  lea    D14(%rip),%rbx
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xe(%rsp),%rbp
  nopl   0x0(%rax,%rax,1)
L48:
  mov    %rbx,%rdi
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <gets@plt>
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    $0x4,%rax
  jg     L48
  xor    %ebx,%ebx
  xor    %edx,%edx
  jmp    L84
  nopl   0x0(%rax)
L2:
  movsbl 0x0(%rbp,%rbx,1),%edi
  add    $0x1,%rbx
  call   <isalpha@plt>
  test   %rax,%rax
  jne    L94
  mov    %rbp,%rdi
  call   <strlen@plt>
  mov    $0x1,%edx
L84:
  cmp    %ebx,%eax
  ja     L2
  test   %edx,%edx
  jne    L31
L94:
  lea    D3(%rip),%r12
L49:
  mov    %r12,%rdi
  xor    %ebx,%ebx
  call   <printf@plt>
  mov    %rbp,%rdi
  call   <gets@plt>
  nopl   0x0(%rax)
L95:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jbe    L49
  movsbl 0x0(%rbp,%rbx,1),%edi
  add    $0x1,%rbx
  call   <isalpha@plt>
  test   %rax,%rax
  je     L95
L75:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L1
  add    $0x20,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L31:
  mov    %rbp,%rdi
  call   <atoi@plt>
  mov    $0x190,%esi
  cltd
  mov    %eax,%ecx
  idiv   %esi
  and    $0x3,%ecx
  or     %ecx,%edx
  jne    L83
  lea    D31(%rip),%rdi
  call   <printf@plt>
L105:
  xor    %eax,%eax
  call   <getchar@plt>
  jmp    L75
L83:
  lea    D13(%rip),%rdi
  call   <printf@plt>
  jmp    L105
L1:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D14
type_infered: string
name: parse_data
arguments: tag: D3
type_infered: string
name: parse_data
arguments: tag: D31
type_infered: string
name: parse_data
arguments: tag: D13
type_infered: string

tool: D14: Enter 4 digit year: 

tool: D3: Enter only digits: 

tool: D31: It's a leap year

tool: D13: It's not a leap year

assistant: ```c
int main() {
  char s[10];
  int year;
  do {
    printf("Enter 4 digit year: ");
    gets(s);
  } while ((unsigned)strlen(s) > 4);
  int isNumber = 0;
  for (int i = 0; i < (unsigned)strlen(s); i++) {
    if (isalpha(s[i]) == 0) {
      isNumber = 1;
    } else {
      isNumber = 0;
      break;
    }
  }
  while (!isNumber) {
    printf("Enter only digits: ");
    gets(s);
    int isNumber = 0;
    for (int i = 0; i < (unsigned)strlen(s); i++) {
      if (isalpha(s[i]) == 0) {
        isNumber = 1;
      } else {
        return isNumber = 0;
      }
    }
  }
  year = atoi(s);
  if ((year % 4 != 0) || (year % 400 != 0)) {
    printf("It's not a leap year");
  } else if (year % 100 != 0) {
    printf("It's a leap year");
  } else {
    printf("It's a leap year");
  }
  getchar();
  return 0;
}
```
user: What is the c source code of the assembly code below?

```asm
<sqrtf>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  movss  %xmm0,-0x14(%rbp)
  flds   -0x14(%rbp)
  fsqrt
  fstpt  -0x10(%rbp)
  fldt   -0x10(%rbp)
  fstps  -0x18(%rbp)
  movss  -0x18(%rbp),%xmm0
  pop    %rbp
  ret
```

assistant: ```c
float sqrtf(float x) {
  long double t;
  __asm__("fsqrt" : "=t"(t) : "0"(x));
  return (float)t;
}
```
user: 下面的汇编指令对应的C源代码是什么：

```asm
<sqrtf>:
  endbr64
  movss  %xmm0,-0xc(%rsp)
  flds   -0xc(%rsp)
  fsqrt
  fstps  -0xc(%rsp)
  movss  -0xc(%rsp),%xmm0
  ret
```

assistant: ```c
float sqrtf(float x) {
  long double t;
  __asm__("fsqrt" : "=t"(t) : "0"(x));
  return (float)t;
}
```
user: 请将以下的汇编代码转换成C语言代码：

```asm
<sqrtf>:
  endbr64
  movss  %xmm0,-0xc(%rsp)
  flds   -0xc(%rsp)
  fsqrt
  fstps  -0xc(%rsp)
  movss  -0xc(%rsp),%xmm0
  ret
```

assistant: ```c
float sqrtf(float x) {
  long double t;
  __asm__("fsqrt" : "=t"(t) : "0"(x));
  return (float)t;
}
```
user: Please show me the C implementation that matches the assembly code listed below:

```asm
<sqrtf>:
  endbr64
  movss  %xmm0,-0xc(%rsp)
  flds   -0xc(%rsp)
  fsqrt
  fstps  -0xc(%rsp)
  movss  -0xc(%rsp),%xmm0
  ret
```

assistant: ```c
float sqrtf(float x) {
  long double t;
  __asm__("fsqrt" : "=t"(t) : "0"(x));
  return (float)t;
}
```
user: What is the c source code of the assembly code below:

```asm
<partition>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r12
  push   %rbx
  sub    $0x10,%rsp
  mov    %edi,%r10d
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  lea    0x1(%rdi),%eax
  cltq
  lea    0xf(,%rax,4),%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L97:
  cmp    %rdx,%rsp
  je     L112
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L97
L112:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L38
  orq    $0x0,-0x8(%rsp,%rax,1)
L38:
  lea    0x3(%rsp),%r8
  mov    %r8,%rax
  shr    $0x2,%rax
  and    $0xfffffffffffffffc,%r8
  movl   $0x1,0x0(,%rax,4)
  test   %r10d,%r10d
  jle    L18
  lea    0x4(%r8),%rdi
  mov    $0x0,%r9d
  mov    $0x3,%r11d
  jmp    L71
L17:
  add    $0x1,%edx
  neg    %esi
  mov    %edx,%eax
  imul   %edx,%eax
  lea    (%rax,%rax,2),%ebx
  mov    %ebx,%r12d
  sub    %edx,%ebx
  mov    %ebx,%eax
  shr    $0x1f,%eax
  add    %ebx,%eax
  sar    %eax
  mov    %ecx,%ebx
  sub    %eax,%ebx
  js     L42
L76:
  movslq %ebx,%rbx
  mov    %esi,%eax
  imul   (%r8,%rbx,4),%eax
  mov    %eax,%ebx
  add    (%rdi),%ebx
  mov    %ebx,(%rdi)
  add    %edx,%r12d
  mov    %r12d,%eax
  shr    $0x1f,%eax
  add    %r12d,%eax
  sar    %eax
  mov    %ecx,%r12d
  sub    %eax,%r12d
  js     L17
  movslq %r12d,%r12
  mov    %esi,%eax
  imul   (%r8,%r12,4),%eax
  add    %ebx,%eax
  mov    %eax,(%rdi)
  jmp    L17
L42:
  add    $0x4,%rdi
  cmp    %r9d,%r10d
  je     L18
L71:
  mov    %r9d,%eax
  add    $0x1,%r9d
  mov    %r9d,%ecx
  mov    %eax,%ebx
  mov    $0x1,%esi
  mov    $0x1,%edx
  mov    %r11d,%r12d
  test   %eax,%eax
  jns    L76
  jmp    L42
L18:
  movslq %r10d,%r10
  mov    (%r8,%r10,4),%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L77
  lea    -0x10(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %rbp
  ret
L77:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int partition(int n) {
  int dp[n + 1];
  dp[0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1, r = 1; i - (3 * j * j - j) / 2 >= 0; j++, r *= -1) {
      dp[i] += dp[i - (3 * j * j - j) / 2] * r;
      if (i - (3 * j * j + j) / 2 >= 0) {
        dp[i] += dp[i - (3 * j * j + j) / 2] * r;
      }
    }
  }
  return dp[n];
}
```
user: 请从给定的汇编指令推导出C源代码：

```asm
<partition>:
  endbr64
  push   %rbp
  movslq %edi,%r10
  mov    %rsp,%rbp
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    0x1(%r10),%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     L92
L31:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L31
L92:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L57
L25:
  lea    0x3(%rsp),%r9
  mov    %r9,%rax
  and    $0xfffffffffffffffc,%r9
  shr    $0x2,%rax
  movl   $0x1,0x0(,%rax,4)
  test   %r10d,%r10d
  jle    L39
  lea    0x4(%r9),%r8
  mov    $0x1,%edx
  mov    $0x1,%esi
  nopl   (%rax)
L2:
  mov    $0x1,%edi
  mov    $0x1,%ecx
  mov    $0x3,%eax
  jmp    L82
  nopl   0x0(%rax)
L56:
  movslq %r11d,%r11
  mov    (%r9,%r11,4),%edx
L82:
  imul   %edi,%edx
  lea    (%rax,%rcx,1),%r11d
  add    (%r8),%edx
  mov    %r11d,%eax
  mov    %edx,(%r8)
  shr    $0x1f,%eax
  add    %r11d,%eax
  mov    %esi,%r11d
  sar    %eax
  sub    %eax,%r11d
  js     L60
  movslq %r11d,%r11
  mov    (%r9,%r11,4),%eax
  imul   %edi,%eax
  add    %edx,%eax
  mov    %eax,(%r8)
L60:
  add    $0x1,%ecx
  neg    %edi
  mov    %ecx,%eax
  imul   %ecx,%eax
  lea    (%rax,%rax,2),%eax
  mov    %eax,%r11d
  sub    %ecx,%r11d
  mov    %r11d,%edx
  shr    $0x1f,%edx
  add    %r11d,%edx
  mov    %esi,%r11d
  sar    %edx
  sub    %edx,%r11d
  jns    L56
  lea    0x1(%rsi),%eax
  cmp    %esi,%r10d
  je     L39
  mov    (%r8),%edx
  mov    %eax,%esi
  add    $0x4,%r8
  jmp    L2
  nopw   0x0(%rax,%rax,1)
L39:
  mov    (%r9,%r10,4),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L36
  leave
  ret
L57:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L25
L36:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int partition(int n) {
  int dp[n + 1];
  dp[0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1, r = 1; i - (3 * j * j - j) / 2 >= 0; j++, r *= -1) {
      dp[i] += dp[i - (3 * j * j - j) / 2] * r;
      if (i - (3 * j * j + j) / 2 >= 0) {
        dp[i] += dp[i - (3 * j * j + j) / 2] * r;
      }
    }
  }
  return dp[n];
}
```
user: What would the C source look like for the assembly code provided here?

```asm
<partition>:
  endbr64
  push   %rbp
  movslq %edi,%r11
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  lea    0x1(%r11),%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     L108
L44:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L44
L108:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L30
L66:
  lea    0x3(%rsp),%r10
  mov    %r10,%rax
  and    $0xfffffffffffffffc,%r10
  shr    $0x2,%rax
  movl   $0x1,0x0(,%rax,4)
  test   %r11d,%r11d
  jle    L77
  lea    0x4(%r10),%r9
  mov    $0x1,%edx
  mov    $0x1,%edi
  xchg   %ax,%ax
L67:
  mov    (%r9),%esi
  mov    $0x1,%r8d
  mov    $0x1,%ecx
  mov    $0x3,%eax
  jmp    L2
  nopl   (%rax)
L25:
  movslq %ebx,%rbx
  mov    (%r10,%rbx,4),%edx
L2:
  imul   %r8d,%edx
  add    %edx,%esi
  lea    (%rax,%rcx,1),%edx
  mov    %edx,%eax
  mov    %esi,(%r9)
  shr    $0x1f,%eax
  add    %edx,%eax
  mov    %edi,%edx
  sar    %eax
  sub    %eax,%edx
  js     L52
  movslq %edx,%rdx
  mov    (%r10,%rdx,4),%eax
  imul   %r8d,%eax
  add    %eax,%esi
  mov    %esi,(%r9)
L52:
  add    $0x1,%ecx
  neg    %r8d
  mov    %ecx,%eax
  imul   %ecx,%eax
  lea    (%rax,%rax,2),%eax
  mov    %eax,%ebx
  sub    %ecx,%ebx
  mov    %ebx,%edx
  shr    $0x1f,%edx
  add    %ebx,%edx
  mov    %edi,%ebx
  sar    %edx
  sub    %edx,%ebx
  jns    L25
  lea    0x1(%rdi),%eax
  cmp    %edi,%r11d
  je     L77
  mov    (%r9),%edx
  mov    %eax,%edi
  add    $0x4,%r9
  jmp    L67
  xchg   %ax,%ax
L77:
  mov    (%r10,%r11,4),%eax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  jne    L64
  mov    -0x8(%rbp),%rbx
  leave
  ret
L30:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L66
L64:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
int partition(int n) {
  int dp[n + 1];
  dp[0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1, r = 1; i - (3 * j * j - j) / 2 >= 0; j++, r *= -1) {
      dp[i] += dp[i - (3 * j * j - j) / 2] * r;
      if (i - (3 * j * j + j) / 2 >= 0) {
        dp[i] += dp[i - (3 * j * j + j) / 2] * r;
      }
    }
  }
  return dp[n];
}
```
user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<my_put_nbr>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %edi,-0x14(%rbp)
  mov    -0x14(%rbp),%edx
  movslq %edx,%rax
  imul   $0x66666667,%rax,%rax
  shr    $0x20,%rax
  sar    $0x2,%eax
  mov    %edx,%esi
  sar    $0x1f,%esi
  sub    %esi,%eax
  mov    %eax,%ecx
  mov    %ecx,%eax
  shl    $0x2,%eax
  add    %ecx,%eax
  add    %eax,%eax
  sub    %eax,%edx
  mov    %edx,-0x8(%rbp)
  mov    -0x14(%rbp),%eax
  movslq %eax,%rdx
  imul   $0x66666667,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x2,%edx
  sar    $0x1f,%eax
  mov    %eax,%ecx
  mov    %edx,%eax
  sub    %ecx,%eax
  mov    %eax,-0x4(%rbp)
  cmpl   $0xfffffff7,-0x4(%rbp)
  jl     L27
  cmpl   $0x0,-0x4(%rbp)
  jns    L27
  mov    $0x2d,%edi
  call   <my_putchar@plt>
L27:
  cmpl   $0x9,-0x14(%rbp)
  jg     L24
  cmpl   $0xfffffff7,-0x14(%rbp)
  jge    L23
L24:
  mov    -0x4(%rbp),%eax
  mov    %eax,%edi
  call   <my_put_nbr>
L23:
  cmpl   $0x0,-0x8(%rbp)
  jns    L42
  negl   -0x8(%rbp)
L42:
  mov    -0x8(%rbp),%eax
  add    $0x30,%eax
  mov    %eax,%edi
  call   <my_putchar@plt>
  nop
  leave
  ret
```

assistant: ```c
void my_put_nbr(int nb) {
  int div;
  int rest;
  rest = nb % 10;
  div = nb / 10;
  if (div > -10 && div < 0)
    my_putchar('-');
  if (nb >= 10 || nb <= -10)
    my_put_nbr(div);
  if (rest < 0)
    rest = -rest;
  my_putchar(rest + 48);
}
```
user: 请将以下的汇编代码转换成C语言代码:

```asm
<my_put_nbr>:
  endbr64
  push   %r12
  push   %rbp
  push   %rbx
  mov    %edi,%ebp
  movslq %edi,%rbx
  imul   $0x66666667,%rbx,%rbx
  sar    $0x22,%rbx
  mov    %edi,%eax
  sar    $0x1f,%eax
  sub    %eax,%ebx
  mov    %ebx,%r12d
  lea    (%rbx,%rbx,4),%eax
  add    %eax,%eax
  mov    %edi,%ebx
  sub    %eax,%ebx
  cmp    $0xfffffff7,%r12d
  jae    L26
L23:
  lea    0x9(%rbp),%eax
  cmp    $0x12,%eax
  ja     L31
L28:
  mov    %ebx,%edi
  neg    %edi
  cmovs  %ebx,%edi
  add    $0x30,%edi
  call   <my_putchar@plt>
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
L26:
  mov    $0x2d,%edi
  call   <my_putchar@plt>
  jmp    L23
L31:
  mov    %r12d,%edi
  call   <my_put_nbr>
  jmp    L28
```

assistant: ```c
void my_put_nbr(int nb) {
  int div;
  int rest;
  rest = nb % 10;
  div = nb / 10;
  if (div > -10 && div < 0)
    my_putchar('-');
  if (nb >= 10 || nb <= -10)
    my_put_nbr(div);
  if (rest < 0)
    rest = -rest;
  my_putchar(rest + 48);
}
```
user: What is the c source code of the assembly code below?

```asm
<my_put_nbr>:
  endbr64
  push   %r15
  mov    %edi,%eax
  push   %r14
  sar    $0x1f,%eax
  push   %r13
  push   %r12
  push   %rbp
  lea    0x9(%rdi),%ebp
  push   %rbx
  movslq %edi,%rbx
  imul   $0x66666667,%rbx,%rbx
  sub    $0x8,%rsp
  sar    $0x22,%rbx
  sub    %eax,%ebx
  lea    (%rbx,%rbx,4),%eax
  mov    %ebx,%r12d
  mov    %edi,%ebx
  add    %eax,%eax
  sub    %eax,%ebx
  cmp    $0xfffffff7,%r12d
  jae    L136
  cmp    $0x12,%ebp
  ja     L91
L7:
  mov    %ebx,%edi
  neg    %edi
  cmovs  %ebx,%edi
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  add    $0x30,%edi
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1050 <my_putchar@plt>
  nopw   0x0(%rax,%rax,1)
L91:
  movslq %r12d,%rbp
  mov    %r12d,%eax
  lea    0x9(%r12),%r14d
  imul   $0x66666667,%rbp,%rbp
  sar    $0x1f,%eax
  sar    $0x22,%rbp
  sub    %eax,%ebp
  lea    0x0(%rbp,%rbp,4),%eax
  mov    %ebp,%r13d
  mov    %r12d,%ebp
  add    %eax,%eax
  lea    0x9(%r13),%r15d
  sub    %eax,%ebp
  cmp    $0xfffffff7,%r13d
  jae    L139
  cmp    $0x12,%r14d
  jbe    L9
  movslq %r13d,%r12
  mov    %r13d,%eax
  imul   $0x66666667,%r12,%r12
  sar    $0x1f,%eax
  sar    $0x22,%r12
  sub    %eax,%r12d
  lea    (%r12,%r12,4),%eax
  mov    %r12d,%r14d
  add    %eax,%eax
  sub    %eax,%r13d
  mov    %r13d,%r12d
  cmp    $0xfffffff7,%r14d
  jae    L112
L78:
  cmp    $0x12,%r15d
  ja     L90
L98:
  mov    %r12d,%edi
  neg    %edi
  cmovs  %r12d,%edi
  add    $0x30,%edi
  call   <my_putchar@plt>
  jmp    L9
  cs nopw 0x0(%rax,%rax,1)
L136:
  mov    $0x2d,%edi
  call   <my_putchar@plt>
  cmp    $0x12,%ebp
  jbe    L7
  mov    %r12d,%ebp
  jmp    L9
  nopl   0x0(%rax,%rax,1)
L139:
  mov    $0x2d,%edi
  call   <my_putchar@plt>
  cmp    $0x12,%r14d
  ja     L125
L9:
  mov    %ebp,%edi
  neg    %edi
  cmovs  %ebp,%edi
  add    $0x30,%edi
  call   <my_putchar@plt>
  jmp    L7
  nopl   0x0(%rax)
L90:
  mov    %r14d,%edi
  call   <my_put_nbr>
  jmp    L98
  nopw   0x0(%rax,%rax,1)
L112:
  mov    $0x2d,%edi
  call   <my_putchar@plt>
  jmp    L78
  nopw   0x0(%rax,%rax,1)
L125:
  mov    %r13d,%r12d
  jmp    L98
```

assistant: ```c
void my_put_nbr(int nb) {
  int div;
  int rest;
  rest = nb % 10;
  div = nb / 10;
  if (div > -10 && div < 0)
    my_putchar('-');
  if (nb >= 10 || nb <= -10)
    my_put_nbr(div);
  if (rest < 0)
    rest = -rest;
  my_putchar(rest + 48);
}
```
user: 这段汇编代码用C语言表示是什么样的:

```asm
<my_put_nbr>:
  endbr64
  push   %r15
  mov    %edi,%eax
  push   %r14
  sar    $0x1f,%eax
  push   %r13
  push   %r12
  push   %rbp
  lea    0x9(%rdi),%ebp
  push   %rbx
  movslq %edi,%rbx
  imul   $0x66666667,%rbx,%rbx
  sub    $0x8,%rsp
  sar    $0x22,%rbx
  sub    %eax,%ebx
  lea    (%rbx,%rbx,4),%eax
  mov    %ebx,%r12d
  mov    %edi,%ebx
  add    %eax,%eax
  sub    %eax,%ebx
  cmp    $0xfffffff7,%r12d
  jae    L82
  cmp    $0x12,%ebp
  ja     L42
L0:
  mov    %ebx,%edi
  neg    %edi
  cmovs  %ebx,%edi
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  add    $0x30,%edi
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  jmp    1050 <my_putchar@plt>
  nopw   0x0(%rax,%rax,1)
L42:
  movslq %r12d,%rbp
  mov    %r12d,%eax
  lea    0x9(%r12),%r14d
  imul   $0x66666667,%rbp,%rbp
  sar    $0x1f,%eax
  sar    $0x22,%rbp
  sub    %eax,%ebp
  lea    0x0(%rbp,%rbp,4),%eax
  mov    %ebp,%r13d
  mov    %r12d,%ebp
  add    %eax,%eax
  lea    0x9(%r13),%r15d
  sub    %eax,%ebp
  cmp    $0xfffffff7,%r13d
  jae    L24
  cmp    $0x12,%r14d
  jbe    L99
  movslq %r13d,%r12
  mov    %r13d,%eax
  imul   $0x66666667,%r12,%r12
  sar    $0x1f,%eax
  sar    $0x22,%r12
  sub    %eax,%r12d
  lea    (%r12,%r12,4),%eax
  mov    %r12d,%r14d
  add    %eax,%eax
  sub    %eax,%r13d
  mov    %r13d,%r12d
  cmp    $0xfffffff7,%r14d
  jae    L34
L124:
  cmp    $0x12,%r15d
  ja     L17
L125:
  mov    %r12d,%edi
  neg    %edi
  cmovs  %r12d,%edi
  add    $0x30,%edi
  call   <my_putchar@plt>
  jmp    L99
  cs nopw 0x0(%rax,%rax,1)
L82:
  mov    $0x2d,%edi
  call   <my_putchar@plt>
  cmp    $0x12,%ebp
  jbe    L0
  mov    %r12d,%ebp
  jmp    L99
  nopl   0x0(%rax,%rax,1)
L24:
  mov    $0x2d,%edi
  call   <my_putchar@plt>
  cmp    $0x12,%r14d
  ja     L28
L99:
  mov    %ebp,%edi
  neg    %edi
  cmovs  %ebp,%edi
  add    $0x30,%edi
  call   <my_putchar@plt>
  jmp    L0
  nopl   0x0(%rax)
L17:
  mov    %r14d,%edi
  call   <my_put_nbr>
  jmp    L125
  nopw   0x0(%rax,%rax,1)
L34:
  mov    $0x2d,%edi
  call   <my_putchar@plt>
  jmp    L124
  nopw   0x0(%rax,%rax,1)
L28:
  mov    %r13d,%r12d
  jmp    L125
```

assistant: ```c
void my_put_nbr(int nb) {
  int div;
  int rest;
  rest = nb % 10;
  div = nb / 10;
  if (div > -10 && div < 0)
    my_putchar('-');
  if (nb >= 10 || nb <= -10)
    my_put_nbr(div);
  if (rest < 0)
    rest = -rest;
  my_putchar(rest + 48);
}
```