1. The basic concepts and common code
(1) prime number (质数)
int prime[maxn],tot=0;
bool vis[maxn];
void init(int n)
{
vis[1]=1;
for(int i=2;i<=n;++i)
{
if(!vis[i]) prime[++tot]=i;
for(int j=1;j<=tot&&prime[j]*i<=n;++j)
{
vis[i*prime[j]]=1;
if(i%prime[j]==0) break;//每个数都只会被它最小的素数因子筛掉
}
}
}//线性筛素数(2) Euclidean algorithm (Euclidean)
#define ll long long
ll gcd(ll a,ll b)
{
return b==0?a:gcd(b,a%b);
}Extended Euclidean Algorithm
int exgcd(int a,int b,int &d,int &x,int &y)
{
if(!b)
{
x=1,y=0;
return d=a;
}
else
{
exgcd(b,a%b,d,y,x);
y-=x*(a/b);
}
}//可用于求ax+by=d的一组解，当且仅当d=gcd(a,b)时有解（若要通过此特解推出其他解，每次x减小b/d，y增加a/d即可）Extended Euclid can inverse yuan (inverse perhaps be understood as meaning the countdown in the mold)
If required a inverse element in the mold b sense, first under a die b significance has the inverse element with the proviso that gcd (a, b) = 1, i.e., a and b are relatively prime, we assume that at a die b Significance the inverse is x, then the ax in the mold b congruent to 1, you can get ax + by = 1 (should be ax = by + 1, after transposition becomes the ax-by = 1, we let b = -b, the ax + by = 1)
(3) rapid power
#define LL long long
LL pow_mod(LL a,LL b,int mod)
{
LL ans=1;
while(b)
{
if(b&1)
{
ans=(ans*a)%mod;
}
a=(a*a)%mod;
b>>=1;
}
}(4) the Euler function (not more than the counted number with their own prime)
\[ \varphi(n)=n(1-\frac{1}{p_1})(1-\frac{1}{p_2})\cdots(1-\frac{1}{p_k}) \]
prove:
Positive integer n is given a unique factorization
\ [n = p_1 ^ {a_1
} p_2 ^ {a_2} p_3 ^ {a_3} \ cdots p_k ^ {a_k} \] with inclusion and exclusion, is first subtracted from the total n, p1 number multiple of p2, p3 ... pk of (pi is a prime number, it shall not multiples its prime), i.e. nn / p1-n / p2 ...- n / pk ( pi largest provided n is less than or equal to meet multiple of pi * t, t is determined so that n / pi), and then add "is also the number of multiples of two, and subtracting a multiple of the number is three while the number of .. ., that is, the resulting formula:
\ [\ varphi (n-) = \ sum_ {S \ subseteq P_1, P_2, ... P_K} (-. 1) ^ {| S |} \ {n-FRAC} {\ prod_ {p_i \ epsilon S}} p_i \]
int phi[maxn];
void init(int n)
{
phi[1]=1;
for(int i=2;i<=n;++i)
{
phi[i]=i;
}
for(int i=2;i<=n;++i)
{
if(phi[i]==i)
{
for(int j=i;j<=n;j+=i)
{
phi[j]=phi[j]/i*(i-1);
}
}
}
}//筛表法求欧拉函数(5) the remaining lines, the inverse modulo multiplication
Popular that lines the remainder modulo n is entirely {0,1,2, ..., n-1}, to simplify the remaining lines (also referred to as shrink-based) system that is completely the remaining number prime with n.
The most common system remaining completely written modulus n is Z / nZ, may be written as:
\ [the Z / n or Z_n \]
, referred to as a shrink-based
\ [Z_n ^ * \]
#define LL long long
LL inv(LL a,LL mod)
{
LL d,x,y;
return exgcd(a,mod,d,x,y)==1?(x+mod)%mod:-1;
}//模乘法的逆Another method is to use the inverse Euler's theorem
Given an arbitrary integer n> 1, for any shrinking lines n in an element a,
\ [a ^ {\ varphi (n)} \ equiv1 (MOD \ Quad n) \]
Thus a inverse element is
\ [ a ^ {\ varphi (n)
-1} mod \ quad n \] if n is prime, then
\ [\ varphi (n) =
n-1 \] Therefore, a reverse is pow_mod (a, n-2, n )
2. mode equation
(1) linear model equations
\[ ax\equiv b(mod\quad n) \]
Put it into ax-ny = b, no solution when the number of about d = gcd (a, n) is not b, or both sides while divided by d, to give a'x-n'y = b ', a ' = a / d, n '= n / d, b' = b / d, namely
\ [a'x \ equiv b '(
mod \ quad n') \] At this time, a 'and n' have coprime, so then left by a 'mold n' inverse element in the sense, the solution is:
\ [X \ equiv (a ') ^ {-. 1} B' (MOD \ Quad n ') \]
this solution is the modulus n' the remaining elements of a system, we must put the remaining modulo n represented as element lines. Order (A ') ^ -. 1 B' = P , corresponding to the above solution x = p, x = p + n ', x p + 2n =', x p + 3n '=, .... For mold n , if p + in 'and p + jn' congruence, the (p + in ') - ( p + jn') = (ij) n ' is a multiple of n , and therefore, (ij of) a d multiples (gcd (a, n)) of. In other words, the remaining lines modulo n,
\ [AX \ equiv B (MOD \ Quad n) \]
exactly d a solution of p, p + n ', p + 2n', p + 3n ',. .., p + (d-1 ) n '.
If you have multiple equations, variables, or only one, how should we do it?
Then use the following theorem
(2) Chinese remainder theorem (Chinese Remainder Theorem)
Consider equations
\ [x \ equiv a_i (mod
\ quad m_i) \] and all pairwise relatively prime mi. Let M be the product of all of mi, wi = M / mi, the gcd (wi, mi) = 1 .
Extended Euclidean Algorithm with pi and qi can be found such that wi * pi + mi * qi = 1. Then allowed ei = wi * pi, the equations equivalent to a single equation
\ [X \ equiv e_1a_1 + e_2a_2 + \ cdots + e_na_n (MOD \ Quad M) \]
, i.e. in modulo M remaining lines, primary equation with a unique solution.
Proof: The equation Wi PI mi + Qi = 1 on both sides of the mold can be obtained immediately mi
\ [e_i \ equiv 1 (MOD \ Quad m_i) \]
, whereas for all i not equal to j, wi mj is a multiple of, and therefore
\ [e_i \ equiv 0 (MOD \ Quad M_j) \]
, so that, when x0 mi of modulo addition eiai which a remainder is other than 1 * ai = ai, the remainder remaining terms are zero.
#define LL long long
LL crt(int n,int* a,int* m)
{
LL M=1,d,x,y,x=0;
for(int i=1;i<=n;++i) M*=m[i];
for(int i=1;i<=n;++i)
{
LL w=M/m[i];
exgcd(m[i],w,d,d,y);
x=(x+y*w*a[i])%M;
}
return (x+M)%M;
}(3) discrete logarithm
For simplicity, we only consider the simplest case, i.e., when n is a prime number, MODULUS equation
\ [a ^ x \ equiv b
(mod \ quad n) \] Since n is a prime number, as long as a is not 0 , there must be an inverse a ^ -1. Euler's theorem, simply by checking x = 0,1,2, ..., n- 1 is not a solution to. Since
\ [A ^ {n-1} \ equiv. 1 (MOD \ n-Quad) \]
, n-1 when x exceeds a ^ x when the cycle starts.
We check the front m (m is we make n ^ (1/2)) item, i.e., a ^ 0, a ^ 1, ..., a ^ value modulo n (m-1) if is b, and the a ^ imodn stored in the ei, and calculates the inverse a ^ m a ^ (- m).
Consider a ^ m, a ^ (m + 1), ..., a ^ (2m-1). The eleven do not check, because if one of them has a solution, that corresponds to the presence of i
\ [e_i * A ^ m \ equiv B (MOD \ n-Quad) \]
, with both sides by a ^ (- m) to give
\ [e_i \ equiv b '(mod
\ quad n) \] wherein
\ [b' = a ^ {
- m} b (mod \ quad n) \] Thus simply test whether such ei is equal to b 'to .
If not, then consider a ^ 2m, a ^ (2m + 1), ..., a ^ (3m-1), so that b '' = (a ^ (- 2m) * b) mod n
Therefore we have to enumerate a method according to a ^ (m * m-1)
int pow_mod(int a,int b,int mod)
{
int ans=1;
while(b)
{
if(b&1) ans=(ans*a)%mod;
a=(a*a)%mod;
b>>=1;
}
return ans;
}
int log_mod(int a,int b,int n)
{
int m,v,e=1;
m=(int)sqrt(n+0.5);
v=pow_mod(pow_mod(a,m,n),n-2,n);
map<int,int> x;
x[1]=0;
for(int i=1;i<m;++i)
{
e=(e*a)%n;
if(!x.count(e)) x[e]=i;
}
for(int i=0;i<m;++i)//考虑a^(im),a^(im+1),...,a^(im+m-1)
{
if(x.count(b)) return i*m+x[b];
b=(b*v)%n;
}
return -1;
}//这就是可用于解决离散对数的大步小步算法（Baby_Step_Giant_Step Algorithm），复杂
//度O(n^(1/2)logn)