OSDN Git Service
(root)
/
oi-wiki
/
main.git
/ commitdiff
commit
grep
author
committer
pickaxe
?
search:
re
summary
|
shortlog
|
log
|
commit
| commitdiff |
tree
raw
|
patch
| inline |
side by side
(parent:
25b1cac
)
Update inverse.md
author
雷蒻
<34390285+hsfzLZH1@users.noreply.github.com>
Tue, 30 Jul 2019 00:52:33 +0000
(08:52 +0800)
committer
GitHub
<noreply@github.com>
Tue, 30 Jul 2019 00:52:33 +0000
(08:52 +0800)
docs/math/inverse.md
patch
|
blob
|
history
diff --git
a/docs/math/inverse.md
b/docs/math/inverse.md
index
1b95251
..
311a01f
100644
(file)
--- a/
docs/math/inverse.md
+++ b/
docs/math/inverse.md
@@
-61,7
+61,7
@@
inline int qpow(long long a, int b) {
$kj^{-1}+i^{-1} \equiv 0 \pmod p$ ;
- $i^{-1} \equiv -kj^{-1}
+
\pmod p$ ;
+ $i^{-1} \equiv -kj^{-1} \pmod p$ ;
$i^{-1} \equiv -(\frac{p}{i}) (p \mod i)^{-1}$ ;
@@
-81,7
+81,7
@@
for (int i = 2; i <= n; ++i) inv[i] = (long long)(p - p / i) * inv[p % i] % p;
这就是线性求逆元。
-另外,根据线性求逆元方法的式子: $i^{-1} \equiv -kj^{-1}
+
\pmod p$
+另外,根据线性求逆元方法的式子: $i^{-1} \equiv -kj^{-1} \pmod p$
递归求解 $j^-1$ , 直到 $j=1$ 返回 1。