diff --git a/others/matrix_exponentiation.cpp b/others/matrix_exponentiation.cpp index f9b4997e9..d44d22593 100644 --- a/others/matrix_exponentiation.cpp +++ b/others/matrix_exponentiation.cpp @@ -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. +
\f$a_i = b_i\f$ (for \f$i <= k\f$) +
\f$a_i = c_1 a_{i-1} + c_2 a_{i-2} + ... + c_k a_{i-k}\f$ (for \f$i > k\f$) +
Taking the example of Fibonacci series, \f$k=2\f$ +
\f$b_1 = 1,\; b_2=1\f$ +
\f$c_1 = 1,\; c_2=1\f$ +
\f$a = \begin{bmatrix}0& 1& 1& 2& \ldots\end{bmatrix}\f$ +
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 a, b, c; -// To multiply 2 matrix -vector> multiply(vector> A, vector> 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> multiply(const vector> &A, + const vector> &B) { vector> C(k + 1, vector(k + 1)); for (ll i = 1; i <= k; i++) { for (ll j = 1; j <= k; j++) { @@ -46,9 +70,15 @@ vector> multiply(vector> A, vector> B) { return C; } -// computing power of a matrix -vector> power(vector> 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> power(const vector> &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> power(vector> 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 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);