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);