Added Tutorial for LOJ-1070 (en)

1070/en.md

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+# LOJ 1070 - Algebraic Problem
+
+## Summary
+Given three non-negative integers, a+b = **p**, ab = **q** and **n** we have to find the value of $a^n$ + $b^n$. a and b not necessarily have to be integers. There will be no such input so that we have to find the value of $0^0$
+
+## Prerequisite
+Matrix Exponentiation: https://www.youtube.com/watch?v=QcT5T-46iFA
+
+**unsigned long long int** in C/C++: https://www.geeksforgeeks.org/maximum-value-of-unsigned-long-long-int-in-c/ (language specific). Implementations of **unsigned long long int** have a modulo behavior when performing arithmetic operations. Any arithmetic manipulation with unsigned long long int's will be taken modulo $2^{64}$ automatically.
+
+## Solution
+Let's define:
+
+F(n) = $a^n$ + $b^n$
+
+=> F(0) = $a^0$ + $b^0$ = 2
+
+=> F(1) = a + b = **p**
+
+$a^2$ +  $b^2$ = $(a+b)^2$ - 2ab
+
+=> $a^2$ +  $b^2$ = (a + b) * (a + b) - ab * ( $a^0$ + $b^0$ )
+
+=> F(2) = $a^2$ +  $b^2$ = **p** * (a + b) - **q** * ( $a^0$ + $b^0$ )
+
+$a^3$ +  $b^3$ = (a + b) * ( $a^2$ +  $b^2$ ) - ab * (a + b)
+
+=> F(3) = $a^3$ +  $b^3$ = **p** * ( $a^2$ +  $b^2$ ) - **q** * (a + b)
+
+$a^4$ +  $b^4$ = (a + b) * ( $a^3$ +  $b^3$ ) - ab * ( $a^2$ +  $b^2$ )
+
+=> F(4) = $a^4$ +  $b^4$ = **p** * ( $a^3$ +  $b^3$ ) - **q** * ( $a^2$ +  $b^2$ )
+
+Observing the pattern we can conclude:
+
+F(n) = **p** * F(n-1) - **q** * F(n-2) 
+
+It's a linear recurrence that can be solved using matrix exponentiation technique.
+
+$$
+\begin{pmatrix}
+p & -q \\
+1 & 0
+\end{pmatrix}
+\begin{pmatrix}
+F(1) \\
+F(0)
+\end{pmatrix} =
+\begin{pmatrix}
+F(2) \\
+F(1)
+\end{pmatrix}
+$$
+
+$$
+\begin{pmatrix}
+p & -q \\
+1 & 0
+\end{pmatrix} ^ {2}
+\begin{pmatrix}
+p \\
+2
+\end{pmatrix} =
+\begin{pmatrix}
+F(3) \\
+F(2)
+\end{pmatrix}
+$$
+
+$$
+\begin{pmatrix}
+p & -q \\
+1 & 0
+\end{pmatrix} ^ {n}
+\begin{pmatrix}
+p \\
+2
+\end{pmatrix} =
+\begin{pmatrix}
+F(n+1) \\
+F(n)
+\end{pmatrix}
+$$
+
+## Complexity
+- Time Complexity: O(T * $k^3$ $log{_2}{n}$). Here, k = 2
+- Memory Complexity: O( $k^2$ ).
+
+## Code
+
+### C++
+
+```cpp
+#include <bits/stdc++.h>
+
+using namespace std;
+
+
+typedef unsigned long long ull;
+
+struct Matrix {
+    vector <vector <ull>> mat;
+
+    Matrix(int n) {
+        mat.assign(n, vector <ull> (n, 0));
+    }
+};
+
+Matrix mat_multiply(const Matrix& A, const Matrix& B) {
+    int n = A.mat.size();
+    Matrix res(n);
+
+    for(int i = 0; i < n; ++i) {
+        for(int j = 0; j < n; ++j) {
+            for(int k = 0; k < n; ++k) {
+                res.mat[i][j] += A.mat[i][k] * B.mat[k][j];
+            }
+        }
+    }
+
+    return res;
+}
+
+Matrix binpow(Matrix &A, int power) {
+    int n = A.mat.size();
+    Matrix res(n);
+
+    for(int i = 0; i < n ; ++i) {
+        for(int j = 0; j < n; ++j) {
+            res.mat[i][j] = (i == j? 1:0); // identity matrix
+        }
+    }
+
+    while (power > 0) {
+        if (power & 1) {
+            res = mat_multiply(res, A);
+        }
+        power >>= 1;
+        A = mat_multiply(A, A);
+    }
+
+    return res;
+}
+
+int main(int argc, const char *argv[]) {
+
+    // for fast I/O
+    ios_base::sync_with_stdio(false);
+    cin.tie(nullptr);
+
+    int t;
+    cin >> t;
+
+    for(int ts = 1; ts <= t; ++ts) {
+        ull p, q, n;
+        cin >> p >> q >> n;
+
+        Matrix A(2);
+        A.mat[0][0] = p;
+        A.mat[0][1] = -q;
+        A.mat[1][0] = 1;
+
+        A = binpow(A, n);
+
+        cout << "Case " << ts << ": " << A.mat[1][0]*p + (A.mat[1][1]<<1) << '\n';
+    }
+
+    return 0;
+} 
+```