kinegratii__borax-23

The user is reporting a bug in `borax.calendars.lunardate.TextUtils.day_cn`.
Specifically, for day 20 and 30, the current output is "廿十" and "卅十", but the correct Chinese names should be "二十" and "三十".
"廿" already means 20, and "卅" already means 30. So "廿十" is like saying "twenty-ten" instead of just "twenty".

I need to:
1. Locate `TextUtils.day_cn` in `borax/calendars/lunardate.py`.
2. Verify the bug with the provided code example.
3. Fix the logic in `TextUtils.day_cn`.
4. Verify the fix.

Let's start by locating the code and reproducing the issue.

{
  "command": "grep -rn --include=\"*.py\" \"def day_cn\" borax/calendars/"
}

{
  "status": "ok",
  "stdout": "borax/calendars/lunardate.py:352:    def day_cn(day: int) -> str:\n",
  "stderr": "",
  "exit_code": 0
}

Okay, the function is in `borax/calendars/lunardate.py` at line 352.

Now let's reproduce the bug.

{
  "command": "python3 - <<'PY'\nfrom borax.calendars.lunardate import LunarDate, TextUtils\n\nprint(f\"day_cn(20): {TextUtils.day_cn(20)}\") # Expected: 二十, Actual: 廿十\nprint(f\"day_cn(30): {TextUtils.day_cn(30)}\") # Expected: 三十, Actual: 卅十\n\nld20 = LunarDate(2020, 10, 20, 0)\nprint(f\"ld20.cn_day: {ld20.cn_day}\")\nprint(f\"ld20.cn_day_calendar: {ld20.cn_day_calendar}\")\nprint(f\"ld20.cn_str(): {ld20.cn_str()}\")\n\nld30 = LunarDate(2020, 10, 30, 0)\nprint(f\"ld30.cn_day: {ld30.cn_day}\")\nprint(f\"ld30.cn_day_calendar: {ld30.cn_day_calendar}\")\nprint(f\"ld30.cn_str(): {ld30.cn_str()}\")\nPY\n"
}

{
  "status": "ok",
  "stdout": "day_cn(20): 廿十\nday_cn(30): 卅十\nld20.cn_day: 廿十\nld20.cn_day_calendar: 廿十\nld20.cn_str(): 二〇二〇年十月廿十\nld30.cn_day: 卅十\nld30.cn_day_calendar: 卅十\nld30.cn_str(): 二〇二〇年十月卅十\n",
  "stderr": "",
  "exit_code": 0
}

The bug is reproduced. `TextUtils.day_cn(20)` returns "廿十" and `TextUtils.day_cn(30)` returns "卅十".

Now let's examine the implementation of `TextUtils.day_cn` in `borax/calendars/lunardate.py`.

{
  "end_line": 370,
  "filepath": "borax/calendars/lunardate.py",
  "start_line": 340
}

{
  "status": "ok",
  "filepath": "borax/calendars/lunardate.py",
  "content": "    ANIMALS = '鼠牛虎兔龙蛇马羊猴鸡狗猪'\n\n    @staticmethod\n    def year_cn(year: int) -> str:\n        s = ''.join([TextUtils.MONTHS_CN[int(c)] for c in str(year)])\n        return s.replace('正', '一')\n\n    @staticmethod\n    def month_cn(month: int) -> str:\n        return TextUtils.MONTHS_CN[month]\n\n    @staticmethod\n    def day_cn(day: int) -> str:\n        a, b = divmod(day, 10)\n        if b == 0:  # 10,20,30\n            b = 10\n            if a == 1:  # 10\n                a = 0\n        return TextUtils.TENS[a] + TextUtils.DAYS_CN[b]\n\n    @staticmethod\n    def get_gz_cn(offset: int) -> str:\n        \"\"\"Get n-th(0-based) GanZhi\n        \"\"\"\n        return TextUtils.STEMS[offset % 10] + TextUtils.BRANCHES[offset % 12]\n\n\nclass LunarDate(EncoderMixin):\n    fields = [f_year, f_month, f_day, f_leap]\n\n    def __init__(self, year: int, month: int, day: int, leap: Leap = False):",
  "start_line": 340,
  "end_line": 370,
  "total_lines": 679,
  "is_truncated": false
}

I need to see `TextUtils.TENS` and `TextUtils.DAYS_CN` to understand how they are defined.

{
  "end_line": 340,
  "filepath": "borax/calendars/lunardate.py",
  "start_line": 300
}

{
  "status": "ok",
  "filepath": "borax/calendars/lunardate.py",
  "content": "        value_offset = [0, 15]\n        year_index = year - 1900\n        days = [int(c) + value_offset[i % 2] for i, c in enumerate(TERM_INFO[year_index])]\n        return days\n\n    @staticmethod\n    def get_term_info(year, month, day):\n        \"\"\"Parse solar term and stem-branch year/month/day from a solar date.\n        (sy, sm, sd) => (term, next_gz_month)\n        term for year 2101,:2101.1.5(初六) 小寒 2101.1.20(廿一) 大寒\n        \"\"\"\n        if year == 2101:\n            days = [5, 20]\n        else:\n            days = TermUtils.parse_term_days(year)\n        term_index1 = 2 * (month - 1)\n        term_index2 = 2 * (month - 1) + 1\n        day1 = days[term_index1]\n        day2 = days[term_index2]\n        if day == day1:\n            term_name = TERMS_CN[term_index1]\n        elif day == day2:\n            term_name = TERMS_CN[term_index2]\n        else:\n            term_name = None\n\n        next_gz_month = day >= day1\n        return term_name, next_gz_month\n\n\n# ------ Stems and Branches ------\n\n\nclass TextUtils:\n    MONTHS_CN = '〇正二三四五六七八九十冬腊'\n    TENS = '初十廿卅'\n    DAYS_CN = '日一二三四五六七八九十'\n\n    STEMS = '甲乙丙丁戊己庚辛壬癸'\n    BRANCHES = '子丑寅卯辰巳午未申酉戌亥'\n    ANIMALS = '鼠牛虎兔龙蛇马羊猴鸡狗猪'",
  "start_line": 300,
  "end_line": 340,
  "total_lines": 679,
  "is_truncated": false
}

`TENS = '初十廿卅'`
`DAYS_CN = '日一二三四五六七八九十'`

Let's trace `day_cn(20)`:
`day = 20`
`a, b = divmod(20, 10)` -> `a = 2, b = 0`
`b == 0` is True.
`b = 10`
`a == 1` is False.
`return TextUtils.TENS[2] + TextUtils.DAYS_CN[10]`
`TextUtils.TENS[2]` is '廿'.
`TextUtils.DAYS_CN[10]` is '十'.
Result: '廿十'.

Let's trace `day_cn(30)`:
`day = 30`
`a, b = divmod(30, 10)` -> `a = 3, b = 0`
`b == 0` is True.
`b = 10`
`a == 1` is False.
`return TextUtils.TENS[3] + TextUtils.DAYS_CN[10]`
`TextUtils.TENS[3]` is '卅'.
`TextUtils.DAYS_CN[10]` is '十'.
Result: '卅十'.

Let's trace `day_cn(10)`:
`day = 10`
`a, b = divmod(10, 10)` -> `a = 1, b = 0`
`b == 0` is True.
`b = 10`
`a == 1` is True.
`a = 0`
`return TextUtils.TENS[0] + TextUtils.DAYS_CN[10]`
`TextUtils.TENS[0]` is '初'.
`TextUtils.DAYS_CN[10]` is '十'.
Result: '初十'. (Wait, the output for 10 should be '初十'? In lunar calendar, the 10th is usually called '初十'.)

Wait, the issue says "二十" and "三十" are correct.
For 10, "初十" is common.
But for 20, "廿" means 20. "廿十" is redundant.
For 30, "卅" means 30. "卅十" is redundant.

So for `day == 20` it should be "二十" or "廿"? The issue says "二十".
Wait, "廿" is often used for 20-29. Example: "廿一" is 21.
If day is 20, "廿" is correct, but "二十" is also correct and more standard in some contexts.
Actually, the issue specifically says:
`print(TextUtils.day_cn(20)) # “廿十”` -> should be `“二十”`
`print(TextUtils.day_cn(30)) # “卅十”` -> should be `“三十”`

Wait, look at `TENS`: `初十廿卅`.
`TENS[0]` = '初' (for 1-9)
`TENS[1]` = '十' (for 10-19)
`TENS[2]` = '廿' (for 20-29)
`TENS[3]` = '卅' (for 30-39)

Wait, for `day = 15`:
`a, b = divmod(15, 10)` -> `a = 1, b = 5`
`b == 0` is False.
`return TextUtils.TENS[1] + TextUtils.DAYS_CN[5]`
`TENS[1]` is '十', `DAYS_CN[5]` is '五'.
Result: '十五'. Correct.

For `day = 25`:
`a, b = divmod(25, 10)` -> `a = 2, b = 5`
`b == 0` is False.
`return TextUtils.TENS[2] + TextUtils.DAYS_CN[5]`
`TENS[2]` is '廿', `DAYS_CN[5]` is '五'.
Result: '廿五'. Correct.

For `day = 35` (if possible):
`a, b = divmod(35, 10)` -> `a = 3, b = 5`
`b == 0` is False.
`return TextUtils.TENS[3] + TextUtils.DAYS_CN[5]`
`TENS[3]` is '卅', `DAYS_CN[5]` is '五'.
Result: '卅五'. Correct.

The problem is when `b == 0` (i.e., 10, 20, 30).
For `day = 10`: `a=1, b=0` -> `b=10, a=0`. Returns `TENS[0] + DAYS_CN[10]` = '初十'. Correct.
For `day = 20`: `a=2, b=0` -> `b=10`. Returns `TENS[2] + DAYS_CN[10]` = '廿十'. Incorrect. Should be '二十' or '廿'?
The issue says "二十".
Wait, if we want "二十", then for `day = 20`, `a` should be 1 and `b` should be 0? No.

If `day == 20`, we want "二十".
`TENS[1]` is '十'. `DAYS_CN[2]` is '二'.
But `TENS` and `DAYS_CN` are used as `TENS[a] + DAYS_CN[b]`.

If we want "二十", we need `TENS` index 1 (which is '十') and `DAYS_CN` index 2 (which is '二')? No, that would be '十二'.
Wait, for "二十", it's '二' + '十'.
But the current logic current logic is `TENS[a] + DAYS_CN[b]`.
For 10-19, it's `TENS[1]` ('十') + `DAYS_CN[b]`.
For 20-29, it's `TENS[2]` ('廿') + `DAYS_CN[b]`.
For 30-39, it's `TENS[3]` ('卅') + `DAYS_CN[b]`.

So for 20: `TENS[2]` ('廿') + `DAYS_CN[10]` ('十') = '廿十'.
The user says it should be '二十'.
Wait, '二十' is '二' + '十'.
'二' is `DAYS_CN[2]`. '十' is `DAYS_CN[10]`.

But what if it should be '廿'?
The issue says: “廿十”、“卅十” 是不正确的用法，因为“廿”、“卅” 本身就是二十、三十的意思了，正确的应该为“二十”、“三十”。
Actually, often 20 is called "廿" in lunar calendar, not "二十". But the user explicitly says "二十" and "三十".

Wait, if the user wants "二十" and "三十", let's see.
For `day == 20`, the result should be "二十".
For `day == 30`, the result should be "三十".

Wait, look at `day_cn` again:
```python
    @staticmethod
    def day_cn(day: int) -> str:
        a, b = divmod(day, 10)
        if b == 0:  # 10,20,30
            b = 10
            if a == 1:  # 10
                a = 0
        return TextUtils.TENS[a] + TextUtils.DAYS_CN[b]
```
If `day == 20`, `a=2, b=0` -> `b=10, a=2`. Returns `TENS[2] + DAYS_CN[10]` = '廿' + '十' = '廿十'.
If we want '二十', we can't easily get it from `TENS[a] + DAYS_CN[b]` because '二' is in `DAYS_CN` and '十' is also in `DAYS_CN`.
Wait, `TENS` is `初十廿卅`.
`TENS[0]` = '初'
`TENS[1]` = '十'
`TENS[2]` = '廿'
`TENS[3]` = '卅'

If `day == 20`, we want '二十'.
If we change the logic for `b == 0`:
For `day == 10`: `a=1, b=0`. Currently transforms to `a=0, b=10` -> '初' + '十' = '初十'.
For `day == 20`: `a=2, b=0`. Currently transforms to `a=2, b=10` -> '廿' + '十' = '廿十'.
For `day == 30`: `a=3, b=0`. Currently transforms to `a=3, b=10` -> '卅' + '十' = '卅十'.

If we want `day == 20` to be '二十', that is `DAYS_CN[2] + DAYS_CN[10]`.
But `TENS[a]` comes first.

Wait, is it possible the user wants '廿' and '卅'?
"因为“廿”、“卅” 本身就是二十、三十的意思了"
This suggests that if day is 20, we should just use "廿". If day is 30, we should just use "卅".
Let's re-read: "正确的应该为“二十”、“三十”".
Okay, they explicitly said "二十" and "三十".

Wait, let's check how lunar dates are normally written.
10th: 初十
20th: 二十 (or 廿)
30th: 三十 (or 卅)

If they want "二十" and "三十":
For `day == 20`, `a=2, b=0`. We want '二十'.
For `day == 30`, `a=3, b=0`. We want '三十'.

If we use `TENS[a]` for `a=2` we get '廿'.
If we use `DAYS_CN[a]` for `a=2` we get '二'.
And then we need '十' which is `DAYS_CN[10]`.

So if `b == 0` and `a > 1`:
Return `DAYS_CN[a] + DAYS_CN[10]`?
For `day = 20`: `a=2, b=0`. Returns `DAYS_CN[2] + DAYS_CN[10]` = '二' + '十' = '二十'.
For `day = 30`: `a=3, b=0`. Returns `DAYS_CN[3] + DAYS_CN[10]` = '三' + '十' = '三十'.
For `day = 10`: `a=1, b=0`. We still want '初十'.

Let's check `TENS[a] + DAYS_CN[b]` for `day = 10`:
`a=1, b=0` -> `a=0, b=10` -> `TENS[0] + DAYS_CN[10]` = '初十'.

So if `b == 0`:
If `a == 1` (day 10): return '初十'
If `a == 2` (day 20): return '二十'
If `a == 3` (day 30): return '三十'

Wait, '二十' and '三十' can be generated by `DAYS_CN[a] + DAYS_CN[10]`.
Is `DAYS_CN[10]` always '十'? Yes.
Is `DAYS_CN[2]` always '二'? Yes.
Is `DAYS_CN[3]` always '三'? Yes.

So for `b == 0` and `a > 1`, we can return `DAYS_CN[a] + DAYS_CN[10]`.

But what about `day_cn(20)` vs `day_cn(22)`?
`day_cn(22)`: `a=2, b=2`. `b == 0` is False.
Returns `TENS[2] + DAYS_CN[2]` = '廿' + '二' = '廿二'. This is correct.

So the only problem is when `b == 0` and `a > 1`.

Let's see if there are other cases. `day` can be up to 30.
If `day == 10`, `a=1, b=0`. We want '初十'.
If `day == 20`, `a=2, b=0`. We want '二十'.
If `day == 30`, `a=3, b=0`. We want '三十'.

Current implementation:
```python
    @staticmethod
    def day_cn(day: int) -> str:
        a, b = divmod(day, 10)
        if b == 0:  # 10,20,30
            b = 10
            if a == 1:  # 10
                a = 0
        return TextUtils.TENS[a] + TextUtils.DAYS_CN[b]
```
For `day=20`, `a=2, b=10` -> `TENS[2] + DAYS_CN[10]` = '廿' + '十' = '廿十'.
For `day=30`, `a=3, b=10` -> `TENS[3] + DAYS_CN[10]` = '卅' + '十' = '卅十'.

If we change it to:
```python
    @staticmethod
    def day_cn(day: int) -> str:
        a, b = divmod(day, 10)
        if b == 0:  # 10,20,30
            if a == 1:  # 10
                return TextUtils.TENS[0] + TextUtils.DAYS_CN[10] # '初十'
            return TextUtils.DAYS_CN[a] + TextUtils.DAYS_CN[10] # '二十', '三十'
        return TextUtils.TENS[a] + TextUtils.DAYS_CN[b]
```
Wait, `TENS[0]` is '初', `DAYS_CN[10]` is '十'.
But `TENS[0] + DAYS_CN[10]` is the same as `TextUtils.TENS[0] + TextUtils.DAYS_CN[10]`.

Actually, if `b == 0`, and `a == 1`, `TENS[0] + DAYS_CN[10]` = '初十'.
If `b == 0`, and `a > 1`, `DAYS_CN[a] + DAYS_CN[10]` = '二十' or '三十'.

Wait, maybe it's simpler:
If `day == 10`, return '初十'.
If `day == 20`, return '二十'.
If `day == 30`, return '三十'.
Else:
  `a, b = divmod(day, 10)`
  ...
  `return TENS[a] + DAYS_CN[b]`

Actually, the current code handles `a, b` logic.
If `day=20`, `a=2, b=0`.
If we want '二十', we can't use `TENS[a]`. We must use `DAYS_CN[a]`.

Let's check what happens if we just change the `if b == 0` block.
```python
        if b == 0:  # 10,20,30
            if a == 1:  # 10
                return TextUtils.TENS[0] + TextUtils.DAYS_CN[10]
            return TextUtils.DAYS_CN[a] + TextUtils.DAYS_CN[10]
```
For `day=10`: returns '初十'.
For `day=20`: returns '二十'.
For `day=30`: returns '三十'.
For `day=11`: `a=1, b=1`. Returns `TENS[1] + DAYS_CN[1]` = '十' + '一' = '十一'. Correct.
For `day=21`: `a=2, b=1`. Returns `TENS[2] + DAYS_CN[1]` = '廿' + '一' = '廿一'. Correct.
For `day=31`: `a=3, b=1`. Returns `TENS[3] + DAYS_CN[1]` = '卅' + '一' = '卅一'. Correct.

Wait, the user said: “廿十”、“卅十” 是不正确的用法，因为“廿”、“卅” 本身就是二十、三十的意思了，正确的应该为“二十”、“三十”。
Does this mean they would prefer "廿" instead of "二十"?
"正确的应该为“二十”、“三十”" - This is very explicit.

However, look at "廿一" (21) and "卅一" (31). These are correct.
So for 20, we use "二十", but for 21, we use "廿一

{
  "command": "python3 - <<'PY'\nfrom borax.calendars.lunardate import LunarDate, TextUtils\n\nprint(f\"day_cn(20): {TextUtils.day_cn(20)}\")\nprint(f\"day_cn(30): {TextUtils.day_cn(30)}\")\nPY\n"
}

{
  "status": "ok",
  "stdout": "day_cn(20): 廿十\nday_cn(30): 卅十\n",
  "stderr": "",
  "exit_code": 0
}

I will apply the fix to `borax/calendars/lunardate.py`.

I'll use a slightly cleaner implementation:
```python
    @staticmethod
    def day_cn(day: int) -> str:
        a, b = divmod(day, 10)
        if b == 0:  # 10,20,30
            if a == 1:  # 10
                return TextUtils.TENS[0] + TextUtils.DAYS_CN[10]
            return TextUtils.DAYS_CN[a] + TextUtils.DAYS_CN[10]
        return TextUtils.TENS[a] + TextUtils.DAYS_CN[b]
```

Wait, let's verify if this is consistent with other lunar date representations.
In some lunar calendars, "廿" is used for 20, and "卅" for 30.
If the user says "二十" is correct, I'll follow that.

Let's double check if `TextUtils.TENS[0]` ('初') and `TextUtils.DAYS_CN[10]` ('十') are the best way to say '初十'. Yes.

Wait, what if `day` is not 10, 20, 30 but something else? `day` in `LunarDate` is typically 1-30.
If `day=1`: `a=0, b=1`. `b==0` is False. `TENS[0]` ('初') + `DAYS_CN[1]` ('一') = '初一'. Correct.
If `day=9`: `a=0, b=9`. `b==0` is False. `TENS[0]` ('初') + `DAYS_CN[9]` ('九') = '初九'. Correct.
If `day=10`: `a=1, b=0`. `b==0` is True, `a=1`. Returns `TENS[0] + DAYS_CN[10]` = '初十'. Correct.
If `day=11`: `a=1, b=1`. `b==0` is False. `TENS[1]` ('十') + `DAYS_CN[1]` ('一') = '十一'. Correct.
If `day=19`: `a=1, b=9`. `b==0` is False. `TENS[1]` ('十') + `DAYS_CN[9]` ('九') = '十九'. Correct.
If `day=20`: `a=2, b=0`. `b==0` is True, `a=2`. Returns `DAYS_CN[2] + DAYS_CN[10]` = '二十'. Correct.
If `day=21`: `a=2, b=1`. `b==0` is False. `TENS[2]` ('廿') + `DAYS_CN[1]` ('一') = '廿一'. Correct.
If `day=29`: `a=2, b=9`. `b==0` is False. `TENS[2]` ('廿') + `DAYS_CN[9]` ('九') = '廿九'. Correct.
If `day=30`: `a=3, b=0`. `b==0` is True, `a=3`. Returns `DAYS_CN[3] + DAYS_CN[10]` = '三十'. Correct.

This seems correct.

One detail: `TextUtils.TENS[0]` is '初'. Is it better to just write '初'?
The current code uses `TextUtils.TENS[0]`. I should probably stick to that for consistency, but actually, if we are using `DAYS_CN[a]` for `a=2, 3`, maybe we should be consistent.
Actually, `TENS[0]` is explicitly for the first 10 days.

Wait, I can just do:
```python
    @staticmethod
    def day_cn(day: int) -> str:
        if day == 10:
            return '初十'
        if day == 20:
            return '二十'
        if day == 30:
            return '三十'
        a, b = divmod(day, 10)
        return TextUtils.TENS[a] + TextUtils.DAYS_CN[b]
```
But the project uses `TextUtils.TENS` and `TextUtils.DAYS_CN` constants. I should use them.

Let's use:
```python
    @staticmethod
    def day_cn(day: int) -> str:
        a, b = divmod(day, 10)
        if b == 0:  # 10,20,30
            if a == 1:  # 10
                return TextUtils.TENS[0] + TextUtils.DAYS_CN[10]
            return TextUtils.DAYS_CN[a] + TextUtils.DAYS_CN[10]
        return TextUtils.TENS[a] + TextUtils.DAYS_CN[b]
```

Wait, is `TENS[0]` always '初'?
`TENS = '初十廿卅'`
`TENS[0]` is '初'.
`DAYS_CN = '日一二三四五六七八九十'`
`DAYS_CN[10]` is '十'.
So `TENS[0] + DAYS_CN[10]` is '初十'. Correct.

Wait, I just noticed something in `TextUtils.year_cn`:
```python
    @staticmethod
    def year_cn(year: int) -> str:
        s = ''.join([TextUtils.MONTHS_CN[int(c)] for c in str(year)])
        return s.replace('正', '一')
```
And `MONTHS_CN = '〇正二三四五六七八九十冬腊'`.
`MONTHS_CN[1]` is '正'.

Anyway, for `day_cn`, my proposed change seems solid.

Let's apply the change.

{
  "filepath": "borax/calendars/lunardate.py",
  "new_string": "    @staticmethod\n    def day_cn(day: int) -> str:\n        a, b = divmod(day, 10)\n        if b == 0:  # 10,20,30\n            if a == 1:  # 10\n                return TextUtils.TENS[0] + TextUtils.DAYS_CN[10]\n            return TextUtils.DAYS_CN[a] + TextUtils.DAYS_CN[10]\n        return TextUtils.TENS[a] + TextUtils.DAYS_CN[b]",
  "old_string": "    @staticmethod\n    def day_cn(day: int) -> str:\n        a, b = divmod(day, 10)\n        if b == 0:  # 10,20,30\n            b = 10\n            if a == 1:  # 10\n                a = 0\n        return TextUtils.TENS[a] + TextUtils.DAYS_CN[b]"
}

{
  "status": "ok",
  "filepath": "borax/calendars/lunardate.py",
  "occurrences": 1,
  "strategy": "exact",
  "diff": "--- a/borax/calendars/lunardate.py\n+++ b/borax/calendars/lunardate.py\n@@ -352,9 +352,9 @@\n     def day_cn(day: int) -> str:\n         a, b = divmod(day, 10)\n         if b == 0:  # 10,20,30\n-            b = 10\n             if a == 1:  # 10\n-                a = 0\n+                return TextUtils.TENS[0] + TextUtils.DAYS_CN[10]\n+            return TextUtils.DAYS_CN[a] + TextUtils.DAYS_CN[10]\n         return TextUtils.TENS[a] + TextUtils.DAYS_CN[b]\n \n     @staticmethod\n",
  "is_truncated": false
}

Wait, my `edit_file` diff looks a bit weird. I replaced:
```python
        if b == 0:  # 10,20,30
            b = 10
            if a == 1:  # 10
                a = 0
        return TextUtils.TENS[a] + TextUtils.DAYS_CN[b]
```
with:
```python
        if b == 0:  # 10,20,30
            if a == 1:  # 10
                return TextUtils.TENS[0] + TextUtils.DAYS_CN[10]
            return TextUtils.DAYS_CN[a] + TextUtils.DAYS_CN[10]
        return TextUtils.TENS[a] + TextUtils.DAYS_CN[b]
```
The diff is:
```diff
@@ -352,9 +352,9 @@
     def day_cn(day: int) -> str:
         a, b = divmod(day, 10)
         if b == 0:  # 10,20,30
-            b = 10
             if a == 1:  # 10
-                a = 0
+                return TextUtils.TENS[0] + TextUtils.DAYS_CN[10]
+            return TextUtils.DAYS_CN[a] + TextUtils.DAYS_CN[10]
         return TextUtils.TENS[a] + TextUtils.DAYS_CN[b]
```
Wait, the `b=10` was removed, and `a=0` was replaced by `return ...`. And `return TextUtils.DAYS_CN[a] + TextUtils.DAYS_CN[10]` was added.

Wait, looking at the original code:
```python
    @staticmethod
    def day_cn(day: int) -> str:
        a, b = divmod(day, 10)
        if b == 0:  # 10,20,30
            b = 10
            if a == 1:  # 10
                a = 0
        return TextUtils.TENS[a] + TextUtils.DAYS_CN[b]
```
My replacement was:
```python
    @staticmethod
    def day_cn(day: int) -> str:
        a, b = divmod(day, 10)
        if b == 0:  # 10,20,30
            if a == 1:  # 10
                return TextUtils.TENS[0] + TextUtils.DAYS_CN[10]
            return TextUtils.DAYS_CN[a] + TextUtils.DAYS_CN[10]
        return TextUtils.TENS[a] + TextUtils.DAYS_CN[b]
```
The diff in the tool response seems to have missed the first line of the `if b == 0` block? No, it's just showing the changes.

Wait, let's re-read the diff carefully.
```diff
-            b = 10
             if a == 1:  # 10
-                a = 0
+                return TextUtils.TENS[0] + TextUtils.DAYS_CN[10]
+            return TextUtils.DAYS_CN[a] + TextUtils.DAYS_CN[10]
```
Actually, if `b == 0`, it now:
1. Checks if `a == 1`. If so, returns '初十'.
2. Otherwise, returns `DAYS_CN[a] + DAYS_CN[10]`.

Wait, what happened to `return TextUtils.TENS[a] + TextUtils.DAYS_CN[b]`?
It's still there at the end. It's reached if `b != 0`.

So for `day = 20`: `a=2, b=0`. `b==0` is True. `a==1` is False. Returns `DAYS_CN[2] + DAYS_CN[10]` = '二十'. Correct.
For `day = 10`: `a=1, b=0`. `b==0` is True. `a==1` is True. Returns `TENS[0] + DAYS_CN[10]` = '初十'. Correct.
For `day = 11`: `a=1, b=1`. `b==0` is False. Returns `TENS[1] + DAYS_CN[1]` = '十一'. Correct.

Wait, it seems my `edit_file` produced the correct logic.

Let's verify with a full test script.

{
  "command": "python3 - <<'PY'\nfrom borax.calendars.lunardate import LunarDate, TextUtils\n\ntest_cases = {\n    1: '初一',\n    9: '初九',\n    10: '初十',\n    11: '十一',\n    19: '十九',\n    20: '二十',\n    21: '廿一',\n    29: '廿九',\n    30: '三十',\n}\n\nfor day, expected in test_cases.items():\n    actual = TextUtils.day_cn(day)\n    print(f\"day {day:2}: expected {expected:4}, actual {actual:4} - {'OK' if actual == expected else 'FAIL'}\")\n\nld20 = LunarDate(2020, 10, 20, 0)\nprint(f\"ld20.cn_day: {ld20.cn_day} - {'OK' if ld20.cn_day == '二十' else 'FAIL'}\")\nprint(f\"ld20.cn_day_calendar: {ld20.cn_day_calendar} - {'OK' if ld20.cn_day_calendar == '二十' else 'FAIL'}\")\nprint(f\"ld20.cn_str(): {ld20.cn_str()} - {'OK' if '二十' in ld20.cn_str() else 'FAIL'}\")\n\nld30 = LunarDate(2020, 10, 30, 0)\nprint(f\"ld30.cn_day: {ld30.cn_day} - {'OK' if ld30.cn_day == '三十' else 'FAIL'}\")\nprint(f\"ld30.cn_day_calendar: {ld30.cn_day_calendar} - {'OK' if ld30.cn_day_calendar == '三十' else 'FAIL'}\")\nprint(f\"ld30.cn_str(): {ld30.cn_str()} - {'OK' if '三十' in ld30.cn_str() else 'FAIL'}\")\nPY\n"
}

{
  "status": "ok",
  "stdout": "day  1: expected 初一  , actual 初一   - OK\nday  9: expected 初九  , actual 初九   - OK\nday 10: expected 初十  , actual 初十   - OK\nday 11: expected 十一  , actual 十一   - OK\nday 19: expected 十九  , actual 十九   - OK\nday 20: expected 二十  , actual 二十   - OK\nday 21: expected 廿一  , actual 廿一   - OK\nday 29: expected 廿九  , actual 廿九   - OK\nday 30: expected 三十  , actual 三十   - OK\nld20.cn_day: 二十 - OK\nld20.cn_day_calendar: 二十 - OK\nld20.cn_str(): 二〇二〇年十月二十 - OK\nld30.cn_day: 三十 - OK\nld30.cn_day_calendar: 三十 - OK\nld30.cn_str(): 二〇二〇年十月三十 - OK\n",
  "stderr": "",
  "exit_code": 0
}