前言
上一次，我讲到了求解线性同余方程，并挖了一个线性求逆元的大坑，现在补上吧。
逆元的定义：若
a
∗
x
≡
1
(
m
o
d
b
)
a * x\equiv 1 \pmod b
a ∗ x ≡ 1 ( m o d b )
a
a
a
b
b
b
x
x
x
a
a
a
b
b
b
a
−
1
a^{-1}
a − 1
p
p
p
1
−
1
≡
1
(
m
o
d
p
)
1^{-1} \equiv 1 \pmod p
1 − 1 ≡ 1 ( m o d p )
p
=
k
∗
i
+
r
p = k * i + r
p = k ∗ i + r
k
,
i
,
r
k,i,r
k , i , r
r
<
k
r < k
r < k
k
=
⌊
p
i
⌋
k = \lfloor { p \over i}\rfloor
k = ⌊ i p ⌋
r
=
p
%
i
r = p \% i
r = p % i
p
≡
0
(
m
o
d
p
)
p \equiv 0 \pmod p
p ≡ 0 ( m o d p )
k
∗
i
+
r
≡
0
(
m
o
d
p
)
k * i + r \equiv 0 \pmod p
k ∗ i + r ≡ 0 ( m o d p )
i
−
1
∗
r
−
1
i^{-1}*r^{-1}
i − 1 ∗ r − 1
p
p
p
k
∗
r
−
1
+
i
−
1
≡
0
(
m
o
d
p
)
k * r^{-1} + i^{-1} \equiv 0 \pmod p
k ∗ r − 1 + i − 1 ≡ 0 ( m o d p )
i
−
1
≡
−
k
∗
r
−
1
(
m
o
d
p
)
i^{-1} \equiv -k * r^{-1} \pmod p
i − 1 ≡ − k ∗ r − 1 ( m o d p )
i
−
1
≡
−
⌊
p
i
⌋
∗
(
p
%
i
)
−
1
(
m
o
d
p
)
i^{-1} \equiv - \lfloor { p \over i}\rfloor * ( p \% i)^{-1} \pmod p
i − 1 ≡ − ⌊ i p ⌋ ∗ ( p % i ) − 1 ( m o d p )
不难得出代码：
inv[ i] = - ( p/ i) * inv[ p% i] ;
非常简洁，
O
(
n
)
O(n)
O ( n )
求逆元还有用费马小定理的方法，但本蒟蒻不会 ，就不赘述了。
中国剩余定理是用来求线性同余方程组的解的，出自《孙子算经》中的题目：今有物不知其数，三三数之剩二（模3得2），五五数之剩三（模5得3），七七数之剩二（模7得2），问物几何？这个问题就是孙子问题，也就是中国剩余定理的由来了。第一步 ，找出5和7的公倍数中模3得1的最小数为70，找出3和7的公倍数中模5得3的最小数为21，找出3和5的公倍数中模7得1的最小数为15；第二步 ，用70乘以2，21乘以3，15乘以2，得出和为233，不难发现，这个数模3得2，模5得3，模7得2；第三步 ，算出3，5，7三个数的最小公倍数为105，将233除以105，得到余数为23，这个即为符合要求的最小正整数。
{
x
≡
2
(
m
o
d
3
)
x
≡
3
(
m
o
d
5
)
x
≡
2
(
m
o
d
7
)
\begin{cases} x \equiv 2 \pmod {3} \\[1ex] x \equiv 3 \pmod {5} \\[1ex] x \equiv 2 \pmod {7} \end{cases}
⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x ≡ 2 ( m o d 3 ) x ≡ 3 ( m o d 5 ) x ≡ 2 ( m o d 7 )
{
x
≡
1
(
m
o
d
3
)
x
≡
0
(
m
o
d
5
)
(1)
x
≡
0
(
m
o
d
7
)
\begin{cases} x \equiv 1 \pmod {3} \\[1ex] x \equiv 0 \pmod {5} & & & \text{(1)} \\[1ex] x \equiv 0 \pmod {7} \end{cases}
⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x ≡ 1 ( m o d 3 ) x ≡ 0 ( m o d 5 ) x ≡ 0 ( m o d 7 ) (1)
{
x
≡
0
(
m
o
d
3
)
x
≡
1
(
m
o
d
5
)
(2)
x
≡
0
(
m
o
d
7
)
\begin{cases} x \equiv 0 \pmod {3} \\[1ex] x \equiv 1 \pmod {5}&&& \text{(2)} \\[1ex] x \equiv 0 \pmod {7} \end{cases}
⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x ≡ 0 ( m o d 3 ) x ≡ 1 ( m o d 5 ) x ≡ 0 ( m o d 7 ) (2)
{
x
≡
0
(
m
o
d
3
)
x
≡
0
(
m
o
d
5
)
(3)
x
≡
1
(
m
o
d
7
)
\begin{cases} x \equiv 0 \pmod {3} \\[1ex] x \equiv 0 \pmod {5}&&& \text{(3)} \\[1ex] x \equiv 1 \pmod {7} \end{cases}
⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x ≡ 0 ( m o d 3 ) x ≡ 0 ( m o d 5 ) x ≡ 1 ( m o d 7 ) (3)
35
∗
x
≡
1
(
m
o
d
3
)
35*x \equiv 1 \pmod 3
3 5 ∗ x ≡ 1 ( m o d 3 )
1
0
=
70
0
\begin{matrix} & 1 \\ & 0 & = & 70 \\ & 0 \\ \end{matrix}
1 0 0 = 7 0
0
1
=
21
0
\begin{matrix} & 0 \\ & 1 & = & 21 \\ & 0 \\ \end{matrix}
0 1 0 = 2 1
0
0
=
15
1
\begin{matrix} & 0 \\ & 0 & = & 15 \\ & 1 \\ \end{matrix}
0 0 1 = 1 5
2
3
2
\begin{matrix} & 2 \\ & 3 \\ & 2 \\ \end{matrix}
2 3 2
a
%
p
+
b
%
p
=
(
a
+
b
)
%
p
a \% p + b \% p =(a + b) \% p
a % p + b % p = ( a + b ) % p
2
1
0
0
3
=
2
∗
0
+
3
∗
1
+
2
∗
0
=
2
∗
70
+
3
∗
21
+
2
∗
15
≡
23
(
m
o
d
105
)
2
0
0
1
\begin{matrix} & 2 & & & & 1& &&&0&&&&0\\ & 3 & = & 2 & * & 0 & + & 3 & * & 1&+&2&*&0 & = 2 * 70 + 3 * 21 + 2 * 15 \equiv23 \pmod {105} \\ & 2& & & & 0 &&&&0&&&&1\\ \end{matrix}
2 3 2 = 2 ∗ 1 0 0 + 3 ∗ 0 1 0 + 2 ∗ 0 0 1 = 2 ∗ 7 0 + 3 ∗ 2 1 + 2 ∗ 1 5 ≡ 2 3 ( m o d 1 0 5 )
m
1
,
m
2
,
.
.
.
,
m
r
m_1,m_2,...,m_r
m 1 , m 2 , . . . , m r 两两互素 ，并记
N
=
m
1
∗
m
2
∗
.
.
.
∗
m
r
N = m_1*m_2*...*m_r
N = m 1 ∗ m 2 ∗ . . . ∗ m r
{
x
≡
b
1
(
m
o
d
m
1
)
x
≡
b
2
(
m
o
d
m
2
)
.
.
.
.
.
.
x
≡
b
r
(
m
o
d
m
r
)
\begin{cases} x \equiv b_1\pmod {m_1} \\[1ex] x \equiv b_2\pmod {m_2} \\[1ex] ...\\[1ex] ...\\[1ex] x \equiv b_r\pmod {m_r} \end{cases}
⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ x ≡ b 1 ( m o d m 1 ) x ≡ b 2 ( m o d m 2 ) . . . . . . x ≡ b r ( m o d m r )
N
N
N
{
x
≡
0
(
m
o
d
m
1
)
.
.
.
x
≡
0
(
m
o
d
m
i
−
1
)
x
≡
1
(
m
o
d
m
i
)
x
≡
0
(
m
o
d
m
i
+
1
)
.
.
.
x
≡
0
(
m
o
d
m
r
)
\begin{cases} x \equiv 0 \pmod {m_1} \\[1ex] ...\\[1ex] x \equiv 0 \pmod {m_{i - 1}} \\[1ex] x \equiv 1 \pmod {m_i} \\[1ex] x \equiv 0 \pmod {m_{i + 1}} \\[1ex] ...\\[1ex] x \equiv 0 \pmod {m_r} \end{cases}
⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ x ≡ 0 ( m o d m 1 ) . . . x ≡ 0 ( m o d m i − 1 ) x ≡ 1 ( m o d m i ) x ≡ 0 ( m o d m i + 1 ) . . . x ≡ 0 ( m o d m r )
1
≤
i
≤
r
1 \le i \le r
1 ≤ i ≤ r
x
i
=
(
N
m
i
)
x_i = ({N \over m_i} )
x i = ( m i N )
m
m
m
k
i
∗
x
i
≡
1
(
m
o
d
m
i
)
k_i*x_i \equiv 1 \pmod {m_i}
k i ∗ x i ≡ 1 ( m o d m i )
b
i
∗
k
i
∗
x
i
b_i * k_i * x_i
b i ∗ k i ∗ x i
N
N
N 板子 https://www.luogu.org/problem/P3868
{
x
≡
a
1
(
m
o
d
b
1
)
x
≡
a
2
(
m
o
d
b
2
)
.
.
.
.
.
.
x
≡
a
k
(
m
o
d
b
k
)
\begin{cases} x \equiv a_1\pmod {b_1} \\[1ex] x \equiv a_2\pmod {b_2} \\[1ex] ...\\[1ex] ...\\[1ex] x \equiv a_k\pmod {b_k} \end{cases}
⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ x ≡ a 1 ( m o d b 1 ) x ≡ a 2 ( m o d b 2 ) . . . . . . x ≡ a k ( m o d b k )
#include <iostream>
#include <cstdio>
#include <cstring>
#define ll long long
#define N 13
using namespace std;
ll a[ N] , b[ N] , k , n;
inline void read ( ll & res ) {
res = 0 ;
ll pd = 1 ;
char aa = getchar ( ) ;
while ( aa < '0' || aa > '9' ) {
if ( aa == '-' ) {
pd = - pd;
}
aa = getchar ( ) ;
}
while ( aa >= '0' && aa <= '9' ) {
res = ( res << 1 ) + ( res << 3 ) + ( aa - '0' ) ;
aa = getchar ( ) ;
}
res * = pd;
return ;
}
inline ll mul ( ll tem , ll t ) {
ll ans = 0 ;
while ( t ) {
if ( t & 1 ) {
ans = ( ans + tem ) % n;
}
tem = ( tem + tem ) % n;
t >>= 1 ;
}
return ans;
}
inline ll exgcd ( ll a , ll b , ll & x , ll & y ) {
if ( ! b ) {
x = 1 ;
y = 0 ;
return a;
}
ll d = exgcd ( b , a % b , x , y ) ;
ll tem = x;
x = y;
y = tem - a / b * y;
return d;
}
inline ll china ( ) {
ll res = 0 ;
for ( int i = 1 ; i <= k ; ++ i ) {
ll x , y;
exgcd ( n / b[ i] , b[ i] , x , y ) ;
res = ( res + mul ( mul ( x , n/ b[ i] ) % n , a[ i] ) % n ) % n;
}
return ( res % n + n ) % n;
}
int main ( ) {
n = 1 ;
read ( k) ;
for ( int i = 1 ; i <= k ; ++ i ) {
read ( a[ i] ) ;
}
for ( int i = 1 ; i <= k ; ++ i ) {
read ( b[ i] ) ;
n * = b[ i] ;
a[ i] = ( a[ i] % b[ i] + b[ i] ) % b[ i] ;
}
printf ( "%lld" , china ( ) ) ;
return 0 ;
}
求解：
{
x
≡
b
1
(
m
o
d
m
1
)
x
≡
b
2
(
m
o
d
m
2
)
.
.
.
.
.
.
x
≡
b
r
(
m
o
d
m
r
)
\begin{cases} x \equiv b_1\pmod {m_1} \\[1ex] x \equiv b_2\pmod {m_2} \\[1ex] ...\\[1ex] ...\\[1ex] x \equiv b_r\pmod {m_r} \end{cases}
⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ x ≡ b 1 ( m o d m 1 ) x ≡ b 2 ( m o d m 2 ) . . . . . . x ≡ b r ( m o d m r )
m
1
,
m
2
,
.
.
.
,
m
r
m_1,m_2,...,m_r
m 1 , m 2 , . . . , m r 不再两两互素 。
b
1
b_1
b 1
{
x
≡
b
i
(
m
o
d
m
i
)
x
≡
b
j
(
m
o
d
m
j
)
\begin{cases} x \equiv b_i\pmod {m_i} \\[1ex] x \equiv b_j\pmod {m_j} \\[1ex] \end{cases}
⎩ ⎨ ⎧ x ≡ b i ( m o d m i ) x ≡ b j ( m o d m j )
{
x
=
m
i
∗
k
i
+
b
i
x
=
m
j
∗
k
j
+
b
j
\begin{cases} x = m_i * k_i + b_i \\[1ex] x = m_j * k_j + b_j \\[1ex] \end{cases}
⎩ ⎨ ⎧ x = m i ∗ k i + b i x = m j ∗ k j + b j
m
i
∗
k
i
−
m
j
∗
k
j
=
b
j
−
b
i
m_i * k_i - m_j * k_j = b_j -b_i
m i ∗ k i − m j ∗ k j = b j − b i
a
=
m
i
a = m_i
a = m i
b
=
m
j
b = m_j
b = m j
c
=
b
j
−
b
i
c = b_j -b_i
c = b j − b i
a
∗
x
′
≡
c
(
m
o
d
b
)
a * x' \equiv c \pmod {b}
a ∗ x ′ ≡ c ( m o d b )
k
i
k_i
k i
x
′
∗
c
G
C
D
（
a
,
b
）
x' *{ c \over GCD（a,b）}
x ′ ∗ G C D （ a , b ） c
c
%
G
C
D
（
a
,
b
）
!
=
0
c \% GCD（a,b） != 0
c % G C D （ a , b ） ! = 0
z
=
m
j
G
C
D
(
m
i
,
m
j
)
z = {m_j \over GCD(m_i,m_j)}
z = G C D ( m i , m j ) m j
k
i
k_i
k i
(
k
i
%
z
+
z
)
%
z
(k_i \%z+z)\%z
( k i % z + z ) % z
x
=
m
i
∗
k
i
+
b
i
x = m_i * k_i + b_i
x = m i ∗ k i + b i
L
C
M
(
m
i
,
m
j
)
LCM(m_i,m_j)
L C M ( m i , m j )
{
x
≡
N
(
m
o
d
M
)
x
≡
b
i
(
m
o
d
m
i
)
\begin{cases} x \equiv N \pmod {M} \\[1ex] x \equiv b_i\pmod {m_i} \\[1ex] \end{cases}
⎩ ⎨ ⎧ x ≡ N ( m o d M ) x ≡ b i ( m o d m i )
N
N
N
i
−
1
i - 1
i − 1
L
C
M
(
m
1
,
m
2
,
.
.
.
,
m
i
−
1
)
LCM(m_1 , m_2 , ... , m_{i - 1})
L C M ( m 1 , m 2 , . . . , m i − 1 )
N
N
N
i
−
1
i - 1
i − 1
x
≡
N
(
m
o
d
M
)
x \equiv N \pmod {M}
x ≡ N ( m o d M )
i
−
1
i - 1
i − 1
i
i
i
i
i
i
{
x
=
M
∗
k
1
+
N
x
=
m
i
∗
k
2
+
b
i
\begin{cases} x = M * k_1 + N \\[1ex] x = m_i * k_2 + b_i \\[1ex] \end{cases}
⎩ ⎨ ⎧ x = M ∗ k 1 + N x = m i ∗ k 2 + b i
M
∗
k
1
−
m
i
∗
k
2
=
b
i
−
N
M * k_1 - m_i * k_2 = b_i -N
M ∗ k 1 − m i ∗ k 2 = b i − N
r
r
r 板子 https://www.luogu.org/problem/P4777
#include <iostream>
#include <cstdio>
#include <cstring>
#define ll long long
using namespace std;
const int N = 1e5 + 5 ;
ll aa[ N] , bb[ N] ;
ll n;
inline void read ( ll & res ) {
res = 0 ;
ll pd = 1 ;
char a = getchar ( ) ;
while ( a < '0' || a > '9' ) {
if ( a == '-' ) {
pd = - pd;
}
a = getchar ( ) ;
}
while ( a >= '0' && a <= '9' ) {
res = ( res << 1 ) + ( res << 3 ) + ( a - '0' ) ;
a = getchar ( ) ;
}
res * = pd;
return ;
}
inline ll mul ( ll a , ll b , ll mod ) {
ll ans = 0 ;
while ( b ) {
if ( b & 1 ) {
ans = ( ans + a ) % mod;
}
a = ( a + a ) % mod;
b >>= 1 ;
}
return ans;
}
inline ll exgcd ( ll a , ll b , ll & x , ll & y ) {
if ( ! b ) {
x = 1 ;
y = 0 ;
return a;
}
ll d = exgcd ( b , a % b , x , y ) ;
ll t = x;
x = y;
y = t - a / b * y;
return d;
}
inline ll china ( ) {
ll ans = bb[ 1 ] ;
ll m = aa[ 1 ] ;
for ( int i = 2 ; i <= n ; ++ i ) {
ll a = m;
ll b = aa[ i] ;
ll c = ( bb[ i] - ans % b + b ) % b;
ll x , y;
ll d = exgcd ( a , b , x , y ) ;
if ( c % d != 0 ) {
return - 1 ;
}
ll tem = b/ d;
ll res = mul ( x , c/ d , tem ) ;
ans + = res * m;
m * = tem;
ans = ( ans % m + m ) % m;
}
return ( ans % m + m ) % m;
}
int main ( ) {
read ( n) ;
for ( int i = 1 ; i <= n ; ++ i ) {
read ( aa[ i] ) ;
read ( bb[ i] ) ;
}
printf ( "%lld" , china ( ) ) ;
return 0 ;
}