gh-160050: Use asymptotically faster divmod in l_mod for large integers

Lib/test/test_int.py

@@ -778,6 +778,13 @@ def test_pylong_int_divmod(self):
         a, b = divmod(n*3 + 1, n)
         assert a == 3 and b == 1
 
+    def test_pylong_int_mod(self):
+        n = (1 << 100_000)
+        self.assertEqual((n * 3 + 1) % n, 1)
+        self.assertEqual((-n * 3 - 1) % n, n - 1)
+        self.assertEqual((n * 3 + 1) % -n, 1 - n)
+        self.assertEqual((-n * 3 - 1) % -n, -1)
+
     @support.cpython_only  # tests implementation details of CPython.
     @unittest.skipUnless(_pylong, "_pylong module required")
     def test_pylong_int_divmod_crash(self):
@@ -789,6 +796,7 @@ def test_pylong_int_divmod_crash(self):
         with mock.patch.object(_pylong, "int_divmod", wraps=bad_int_divmod):
             msg = r"tuple of length 2 is required from int_divmod\(\)"
             self.assertRaisesRegex(ValueError, msg, divmod, a, b)
+            self.assertRaisesRegex(ValueError, msg, lambda: a % b)
 
     def test_pylong_str_to_int(self):
         v1 = 1 << 100_000

Misc/NEWS.d/next/Core_and_Builtins/2026-10-23-19-37-00.gh-issue-160050.aB3x9Q.rst

@@ -0,0 +1 @@
+Use asymptotically faster :func:`divmod` implementation for integer modulo of large integers.

Objects/longobject.c

@@ -4564,6 +4564,14 @@ l_mod(PyLongObject *v, PyLongObject *w, PyLongObject **pmod)
         *pmod = (PyLongObject *)fast_mod(v, w);
         return -(*pmod == NULL);
     }
+#if WITH_PYLONG_MODULE
+    Py_ssize_t size_v = _PyLong_DigitCount(v); /* digits in numerator */
+    Py_ssize_t size_w = _PyLong_DigitCount(w); /* digits in denominator */
+    if (size_w > 300 && (size_v - size_w) > 150) {
+        /* Switch to _pylong.int_divmod(). See comments in l_divmod(). */
+        return pylong_int_divmod(v, w, NULL, pmod);
+    }
+#endif
     if (long_rem(v, w, &mod) < 0)
         return -1;
     if ((_PyLong_IsNegative(mod) && _PyLong_IsPositive(w)) ||