1. Chinese Remainder Theorem (CRT)
1.1 problems introduced
- Given congruence equation, the form
\[ \begin{cases} x\equiv a_1 \pmod {p_1} \\ x\equiv a_2 \pmod {p_2} \\ \dots \\ x\equiv a_n \pmod {p_n} \\ \end{cases} \]
- Wherein \ (P_i \) pairwise quality.
Solutions of the Equations 1.2
- We remember \ (M = \ prod_ = {I}. 1 np_i ^ \) , \ (m_i = \ {P_i FRAC {m}} \) .
- We assume've got a solution \ (x_0 \) , then \ (x_0 + kM (k \ in \ mathbb Z) \) are clearly the solution of equations.
- I.e. all solutions constituting about \ (M \) a residual category.
- We therefore only requires a \ ([0, M) \ ) integer solution to the.
- Chinese remainder theorem gives conclusion, this equation \ ([0, M) \ ) is present in a unique solution .
Solution 1.3 Equations
- Let's consider how to find the solution.
- We consider \ (n \) th equations, the \ (i \) equations group
\[ \begin{cases} x_i\equiv 0 \pmod {p_1} \\ x_i\equiv 0 \pmod {p_2} \\ ~~~~ ~ ~~ ~\dots \\ x_i \equiv a_i \pmod {p_i}\\ ~~~~~ ~~ ~ \dots \\ x_i\equiv 0 \pmod {p_{n-1}} \\ x_i\equiv 0 \pmod {p_n} \\ \end{cases} \]
- Then \ (\ sum_ {i = 1 } ^ nx_i \) to a solution.
- Only needs to \ (M \) modulo obtain \ ([0, n) \ ) Special Solution therein.
- Consider \ (x_i \) how demand.
- Let us consider a system of equations
\[ \begin{cases} y_i\equiv 0 \pmod {p_1} \\ y_i\equiv 0 \pmod {p_2} \\ ~~~~ ~ ~~ ~\dots \\ y_i \equiv 1 \pmod {p_i}\\ ~~~~~ ~~ ~ \dots \\ y_i\equiv 0 \pmod {p_{n-1}} \\ y_i\equiv 0 \pmod {p_n} \\ \end{cases} \]
- Then we can make \ (x_i = a_iy_i \) .
- Since \ (p_j | y_i (J \ neq i) \) , and \ (p_i \) between any two relatively prime, so the \ (m_i \) in the mold \ (p_i \) there is an inverse element in the sense, so we can direct order \ (y_i m_i = \ Times m_i ^ {-}. 1 \) , \ (m_i ^ {-}. 1 \) represents the inverse element.
- Inverse, as the case may be with \ (exgcd \) or Fermat's little theorem requirements.
- 所以 \(x_i=m_i\times m_i^{-1}\times a_i\)。
- So \ (ANS = \ sum_ = {I}. 1 ^ nm_i \ Times m_i ^ {-}. 1 \ Times a_i \) , in order to prevent overflow, we only need to die \ (M \) made to the sense operation.
- Note that \ (CRT \) and \ (exCRT \) sometimes multiplication could overflow on how to deal with this problem see [tips] O (1) rapid multiplication - changle_cyx .
1.4 solution in the mold \ (M \) unique in the sense
- Next we prove that the solution in the mold \ (M \) unique in the sense.
- I.e., two solutions prove absence \ (x_0, x_1 \) , so \ (M \ x_1 the nmid-x_0 \) .
- Reductio ad absurdum:
- Order \ (x_0 \) represents a particular solution of the original equations, there is another solution \ (D x_0 + \) , so \ (M \ the nmid D \) .
- 则 \(\exists i\in [1,n]，p_i\nmid d\)，即 \(d\not \equiv 0\pmod {p_i}\)。
- By the \ (x_0 \) is the solution of the original equations available \ (ans_0 \ equiv a_i \ PMOD {P_i} \) .
- So \ (x_0 + D \ \ equiv a_i PMOD {P_i} \ \ Not) , (D x_0 + \) \ Solution not the original equations, and contradicts the assumption.
- QED.
- So we get the Chinese remainder theorem.
- This unique wide applications in the Chinese remainder theorem, two examples are described in section 3.
2. Expand the Chinese remainder theorem (exCRT)
2.1 problems introduced
- Chinese remainder theorem to solve the equations modulo twenty-two relatively prime.
- So for modulus does not meet twenty-two equations relatively prime, how to solve it?
- 即求解
\[ \begin{cases} x\equiv a_1 \pmod {p_1} \\ x\equiv a_2 \pmod {p_2} \\ \dots \\ x\equiv a_n \pmod {p_n} \\ \end{cases} \] - Where \ (p_i \) no special constraints.
Solutions of the Equations 2.2
- Not necessarily solvable equations, we can determine the way in the solution process.
- We remember \ (m_i = \ {text} LCM (P_1, P_2, \ DOTS, P_i) \) , \ (M = \ {text} LCM (P_1, P_2, \ DOTS, P_n) \) .
- Suppose we get a particular solution \ (x_0 \) , so \ (x_0 + kM (k \ in \ mathbb Z) \) solution are equations.
- Solution out \ (X \) is also one of the remaining classes.
- 1.4 similar evidence, we can still get if the equation has a solution, the solution of the equations in the mold \ (M \) unique in the sense.
Solution of Equations 2.3
- The reason we can not be like prime to do that is the case, because they do not relatively prime, so we need the multiplicative inverse may not exist.
- We consider front to back, one by one equation to merge.
- We assume the solution into the first \ (i \) equations, we make \ (x_i \) denotes a modulo \ (m_i \) a unique solution in the sense.
- For the first \ (1 \) equations, we make direct \ (x_1 = A_1 \) .
- For the first \ (i (i> 1) \) equations, we know the first \ (i-1 \) equations of the general solution for the \ (x_ {i-1} + km_ {i-1} (k \ in \ the Z-mathbb) \) .
- So now we add the first \ (i \) limit equations of \ (the X-\ equiv a_i \ PMOD P_i} {\) .
- Then the first \ (I \) after the simultaneous equations is equivalent to the equation
\[x_{i-1}+km_{i-1}\equiv a_i(mod~p_i)\]
- Indeterminate equation that is written in the form
\[m_{i-1}k+p_iy=a_i-x_{i-1}\]
- Then Extended Euclidean Algorithm ( \ (exgcd \) solution at this equation).
- If there is a \ (I \) , so that this equation has no solution, i.e. \ ((I-m_Low {}. 1, P_i) \ a_i the nmid. 1-I-X_ {} \) , also means that the entire set of equations has no solution .
- Otherwise letting solve special solution \ (K \) accessible to the mold \ (frac {p_i} {( m_ {i-1}, p_i)} \ \) unique solution in the sense, and even if \ (K \) of \ (\ frac {p_i} {( m_ {i-1}, p_i)} \) modulo.
- Then get \ (I-x_i = X_ {+}. 1. 1-m_Low {I} K \) , and also make \ (x_i \) of \ (m_i \) modulo to.
- Finally \ (x_n \) is the answer.
- Template question: Luo Gu P4777 - Extended Chinese remainder theorem (EXCRT)
#include <cmath>
#include <cstdio>
#include <cctype>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
template <class T>
inline void read(T &x)
{
static char ch;
while (!isdigit(ch = getchar()));
x = ch - '0';
while (isdigit(ch = getchar()))
x = x * 10 + ch - '0';
}
typedef long long s64;
const int MaxN = 1e5 + 5;
int n;
s64 p[MaxN], a[MaxN], pre_lcm[MaxN];
inline s64 qmul(s64 b, s64 p, const s64 &mod)
{
b = (b % mod + mod) % mod;
p = (p % mod + mod) % mod;
s64 res = 0;
for (; p; p >>= 1, (b = b + b) > mod ? (b -= mod) : 0)
if (p & 1)
{
res = res + b;
if (res >= mod)
res -= mod;
}
return res;
}
inline s64 ex_gcd(const s64 &a, const s64 &b, s64 &x, s64 &y)
{
if (!b)
return x = 1, y = 0, a;
s64 res = ex_gcd(b, a % b, y, x);
y -= a / b * x;
return res;
}
inline s64 solve_equ(const s64 &a, const s64 &b, const s64 &c)
{
s64 x, y;
s64 d = ex_gcd(a, b, x, y);
return qmul(x, c / d, b / d);
}
int main()
{
read(n);
for (int i = 1; i <= n; ++i)
read(p[i]), read(a[i]);
pre_lcm[1] = p[1];
for (int i = 2; i <= n; ++i)
pre_lcm[i] = p[i] / std::__gcd(pre_lcm[i - 1], p[i]) * pre_lcm[i - 1];
s64 x = a[1];
for (int i = 2; i <= n; ++i)
{
s64 k = solve_equ(pre_lcm[i - 1], p[i], a[i] - x);
x = x + qmul(pre_lcm[i - 1], k, pre_lcm[i]);
if (x >= pre_lcm[i])
x -= pre_lcm[i];
}
std::cout << x << std::endl;
return 0;
}3. Simple Application Chinese Remainder Theorem
We give two very very very simple example (dalao do not d).
3.1 Euler function is a multiplicative function of proof
- Euler function \ (\ varphi (n-) = \ sum_ {I} ^ n-1 = [I \ n-PERP] \) , i.e. \ (1 \) to \ (\ n-) with \ (n-\) prime the number of numbers.
- Euler function is a multiplicative function.
- If \ (P \ PERP Q \) , then \ (\ varphi (PQ) = \ varphi (P) \ varphi (Q) \) .
- Proof is given below from the perspective of the Chinese Remainder Theorem:
- We set \ (A \) represents \ (1 \) to \ (P \) with (P \) \ prime number set composed, \ (B \) represents \ (1 \) to \ (Q \ ) with \ (Q \) prime number set composed, \ (C \) represents \ (1 \) to \ (PQ \) with (PQ \) \ a set of prime numbers.
- Since \ (P \ PERP Q \) , so \ (i \ perp pq \ Leftrightarrow i \ perp p and I \ PERP Q \) .
- The subtractive removed, we know \ ((n-, m) = (n-, m + KN), K \ in \ mathbb the Z \) .
- Combined with the above conclusion, we can get
- The Chinese remainder theorem, we have for each \ (A \ in A, B \ in B \) , we can obtain congruence equation
\ [\ begin {cases} x \ equiv a (mod ~ p) \\ x \ equiv b (mod ~ q) \ end {cases} \] - The unique solution \ (X \) , and apparently \ ([1, pq] \ ) in the presence of a not \ (X \) , so \ (X \) is the solution of equations of two different congruence .
- Thus we can \ (A \ B Times \) (Cartesian product) and \ (C \) to establish a bijection between.
- Therefore set \ (C \) the number of elements is set equal to \ (A \) and set \ (B \) the product of the number of elements.
- QED.
3.2 Solving the number of special topics modulus
- To give a classic topic: BZOJ1951 [SDOI2010] ancient pig culture
- Title effect: seeking \ (G ^ {\ sum_ {D |} n-n-C_ {D}} ^ {} \ MOD P (P = 999 911 659, n-, G \ ^ LE10. 9) \) .
- We know that \ (p \) is a prime number.
- I am sure you will ask, most number of topics modulus is not prime it, this is what's special? ?
- But you find that \ (G \) index is large, the exact value can not be obtained directly.
- Euler's theorem we know \ (A ^ B \ equiv A ^ B {\% \ varphi (P)} (~ MOD P), A \ PERP P \) .
- When \ | (p G \) , the answer is \ (0 \) .
- Otherwise, \ (G \ the p-PERP \) , we can solve by Euler's theorem.
- Is our index need only take \ (\ sum_ {d | n } C_ {n} ^ {D} \ MOD \ varphi (P) \) , i.e., \ (\ sum_ {d | n } C_ {n} ^ D} {\ MOD 999 911 658 \) .
- Now the question is, if we directly \ (Lucas \) Theorem to solve, does not meet the conditions of the modulus is a prime number.
Some people say: I will expand Lucas! !(But the authors also do not)- But this question is no need to use any extensions \ (Lucas \) .
- Since \ (= 2 999 911 658 \ times3 \ times4679 \ times35617 \) , i.e., the number of prime factors of each number are \ (1 \) .
- So we find that we simply requires \ (\ sum_ {d | n } C_ {n} ^ {d} \) these results after four modulo number, with \ (the CRT \) be merged ( \ ( CRT \) tells us find answers to these four lists congruence equation, you can get mold \ (M \) a unique solution in the sense).
- Then transformed into modulus of issues of prime numbers, the direct use of \ (Lucas \) Theorem fix it.
- \ (Lucas \) Theorem ( \ (P \) is a prime number): \ (\ Binom {n-} {m} = \ Binom {n-/ P} {m / P} \ Binom {n-\% P} {m \ P}% \) .
#include <bits/stdc++.h>
//999911658=2*3*4679*35617
const int mod = 999911659;
const int p[5] = {0, 2, 3, 4679, 35617};
const int MaxN = 4e4 + 5;
int n, G;
int a[5], fac[MaxN], inv[MaxN];
inline void add(int &x, const int &y, const int &mod)
{
x += y;
x >= mod ? x -= mod : 0;
}
inline int qpow(int a, int b, const int &mod)
{
int res = 1;
for (; b; b >>= 1, a = 1LL * a * a % mod)
if (b & 1)
res = 1LL * res * a % mod;
return res;
}
inline int lucas(const int &n, const int &m, const int &mod)
{
if (n < m) return 0;
if (n < mod && m < mod) return 1LL * fac[n] * inv[m] % mod * inv[n - m] % mod;
return 1LL * lucas(n % mod, m % mod, mod) * lucas(n / mod, m / mod, mod) % mod;
}
inline int solve(const int &p)
{
fac[0] = 1;
for (int i = 1; i < p; ++i)
fac[i] = 1LL * fac[i - 1] * i % p;
inv[p - 1] = qpow(fac[p - 1], p - 2, p);
for (int i = p - 2; i >= 0; --i)
inv[i] = 1LL * inv[i + 1] * (i + 1) % p;
int res = 0;
for (int i = 1; i * i <= n; ++i)
if (n % i == 0)
{
add(res, lucas(n, i, p), p);
if (i * i != n)
add(res, lucas(n, n / i, p), p);
}
return res;
}
int main()
{
scanf("%d%d", &n, &G);
if (G == mod)
{
puts("0");
return 0;
}
for (int i = 1; i <= 4; ++i)
a[i] = solve(p[i]);
int lcm = mod - 1, ans = 0;
for (int i = 1; i <= 4; ++i)
{
int x = lcm / p[i];
add(ans, 1LL * x * qpow(x, p[i] - 2, p[i]) % lcm * a[i] % lcm, lcm);
}
printf("%d\n", qpow(G, ans, mod));
return 0;
}