From c0ddbfc3c429db0d8ca03f066a6539d83a6b00f1 Mon Sep 17 00:00:00 2001
From: memset0 <34177126+memset0@users.noreply.github.com>
Date: Tue, 24 Nov 2020 13:43:44 +0800
Subject: [PATCH] =?utf8?q?=E9=AB=98=E6=96=AF=E6=B6=88=E5=85=83=E4=B8=AD?=
 =?utf8?q?=E6=B1=82=E9=80=86=E5=85=83=E5=8F=AA=E9=9C=80=E8=A6=81O(k)?=
 =?utf8?q?=E6=AC=A1?=
MIME-Version: 1.0
Content-Type: text/plain; charset=utf8
Content-Transfer-Encoding: 8bit

---
 docs/graph/lgv.md | 5 +++--
 1 file changed, 3 insertions(+), 2 deletions(-)

diff --git a/docs/graph/lgv.md b/docs/graph/lgv.md
index d22ab497..e403add7 100644
--- a/docs/graph/lgv.md
+++ b/docs/graph/lgv.md
@@ -49,7 +49,7 @@ $$
 
 行列式可以使用高斯消元求。
 
-复杂度为 $O(n+k^2(k + \log p))$ ，其中 $\log p$ 是求逆元复杂度。
+复杂度为 $O(n+k(k^2 + \log p))$ ，其中 $\log p$ 是求逆元复杂度。
 
 ??? note "参考代码"
     ```cpp
@@ -108,9 +108,10 @@ $$
             }
           }
           if (!m[i][i]) continue;
+          int inv = qpow(m[i][i], mod - 2);
           for (int j = i + 1; j <= k; ++j) {
             if (!m[j][i]) continue;
-            int mul = (ll)m[j][i] * qpow(m[i][i], mod - 2) % mod;
+            int mul = (ll)m[j][i] * inv % mod;
             for (int p = i; p <= k; ++p) {
               m[j][p] = (m[j][p] - (ll)m[i][p] * mul % mod + mod) % mod;
             }
-- 
2.11.0