document matrix exponentiation

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Krishna Vedala
2020-05-28 11:24:22 -04:00
parent cc5ce16b58
commit 84d33ba5d9

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@@ -1,20 +1,24 @@
/*
Matrix Exponentiation.
/**
@file
@brief Matrix Exponentiation.
The problem can be solved with DP but constraints are high.
ai = bi (for i <= k)
ai = c1*ai-1 + c2*ai-2 + ... + ck*ai-k (for i > k)
Taking the example of Fibonacci series, K=2
b1 = 1, b2=1
c1 = 1, c2=1
a = 0 1 1 2 ....
This way you can find the 10^18 fibonacci number%MOD.
<br/>\f$a_i = b_i\f$ (for \f$i <= k\f$)
<br/>\f$a_i = c_1 a_{i-1} + c_2 a_{i-2} + ... + c_k a_{i-k}\f$ (for \f$i > k\f$)
<br/>Taking the example of Fibonacci series, \f$k=2\f$
<br/>\f$b_1 = 1,\; b_2=1\f$
<br/>\f$c_1 = 1,\; c_2=1\f$
<br/>\f$a = \begin{bmatrix}0& 1& 1& 2& \ldots\end{bmatrix}\f$
<br/>This way you can find the \f$10^{18}\f$ fibonacci number%MOD.
I have given a general way to use it. The program takes the input of B and C
matrix.
Steps for Matrix Expo
1. Create vector F1 : which is the copy of B.
2. Create transpose matrix (Learn more about it on the internet)
3. Perform T^(n-1) [transpose matrix to the power n-1]
4. Multiply with F to get the last matrix of size (1xk).
3. Perform \f$T^{n-1}\f$ [transpose matrix to the power n-1]
4. Multiply with F to get the last matrix of size (1\f$\times\f$k).
The first element of this matrix is the required result.
*/
@@ -25,16 +29,36 @@ using std::cin;
using std::cout;
using std::vector;
/*! shorthand definition for `int64_t` */
#define ll int64_t
#define endl '\n'
/*! shorthand definition for `std::endl` */
#define endl std::endl
/*! shorthand definition for `int64_t` */
#define pb push_back
#define MOD 1000000007
ll ab(ll x) { return x > 0LL ? x : -x; }
/** returns absolute value */
inline ll ab(ll x) { return x > 0LL ? x : -x; }
/** global variable k
* @todo @stepfencurryxiao add documetnation
*/
ll k;
/** global vector variables
* @todo @stepfencurryxiao add documetnation
*/
vector<ll> a, b, c;
// To multiply 2 matrix
vector<vector<ll>> multiply(vector<vector<ll>> A, vector<vector<ll>> B) {
/** To multiply 2 matrices
* \param [in] A matrix 1 of size (m\f$\times\f$n)
* \param [in] B \p matrix 2 of size (p\f$\times\f$q)\n\note \f$p=n\f$
* \result matrix of dimension (m\f$\times\f$q)
*/
vector<vector<ll>> multiply(const vector<vector<ll>> &A,
const vector<vector<ll>> &B) {
vector<vector<ll>> C(k + 1, vector<ll>(k + 1));
for (ll i = 1; i <= k; i++) {
for (ll j = 1; j <= k; j++) {
@@ -46,9 +70,15 @@ vector<vector<ll>> multiply(vector<vector<ll>> A, vector<vector<ll>> B) {
return C;
}
// computing power of a matrix
vector<vector<ll>> power(vector<vector<ll>> A, ll p) {
if (p == 1) return A;
/** computing integer power of a matrix using recursive multiplication.
* @note A must be a square matrix for this algorithm.
* \param [in] A base matrix
* \param [in] p exponent
* \return matrix of same dimension as A
*/
vector<vector<ll>> power(const vector<vector<ll>> &A, ll p) {
if (p == 1)
return A;
if (p % 2 == 1) {
return multiply(A, power(A, p - 1));
} else {
@@ -57,10 +87,15 @@ vector<vector<ll>> power(vector<vector<ll>> A, ll p) {
}
}
// main function
/*! Wrapper for Fibonacci
* \param[in] n \f$n^\text{th}\f$ Fibonacci number
* \return \f$n^\text{th}\f$ Fibonacci number
*/
ll ans(ll n) {
if (n == 0) return 0;
if (n <= k) return b[n - 1];
if (n == 0)
return 0;
if (n <= k)
return b[n - 1];
// F1
vector<ll> F1(k + 1);
for (ll i = 1; i <= k; i++) F1[i] = b[i - 1];
@@ -90,8 +125,7 @@ ll ans(ll n) {
return res;
}
// 1 1 2 3 5
/** Main function */
int main() {
cin.tie(0);
cout.tie(0);