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