[po] auto sync bot

zhsj · zhsj · commit 9ed6cb6d18b3 · 2019-08-02T13:41:25.000+08:00
diff --git a/library/numbers.po b/library/numbers.po
@@ -181,7 +181,7 @@ msgstr ""
 
 #: ../../library/numbers.rst:159
 msgid "If ``A`` defines an :meth:`__add__` which accepts ``b``, all is well."
-msgstr "如果 ``A`` 被定义成一个承认``b`` :meth:`__add__`，一切都没有问题。"
+msgstr "如果 ``A`` 被定义成一个承认``b`` 的 :meth:`__add__`，一切都没有问题。"
 
 #: ../../library/numbers.rst:161
 msgid ""
@@ -199,20 +199,22 @@ msgstr ""
 msgid ""
 "Then ``B``'s :meth:`__radd__` gets a chance. If it accepts ``a``, all is "
 "well."
-msgstr ""
+msgstr "接着看 ``B`` 的 :meth:`__radd__` 。如果它承认 ``a`` ，一切都没有问题。"
 
 #: ../../library/numbers.rst:169
 msgid ""
 "If it falls back to the boilerplate, there are no more possible methods to "
 "try, so this is where the default implementation should live."
-msgstr ""
+msgstr "如果没有成功回退到模板，就没有更多的方法可以去尝试，因此这里将使用默认的实现。"
 
 #: ../../library/numbers.rst:172
 msgid ""
 "If ``B <: A``, Python tries ``B.__radd__`` before ``A.__add__``. This is ok,"
 " because it was implemented with knowledge of ``A``, so it can handle those "
 "instances before delegating to :class:`Complex`."
 msgstr ""
+"如果 ``B <: A`` ， Python 在 ``A.__add__`` 之前尝试 ``B.__radd__`` 。 这是可行的，是通过对 "
+"``A`` 的认识实现的，因此这可以在交给 :class:`Complex` 处理之前处理这些实例。"
 
 #: ../../library/numbers.rst:177
 msgid ""
