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* Matrix Exponentiation * Update and rename Others/Matrix_Expo.cpp to others/matrix_exponentiation.cpp * Update matrix_exponentiation.cpp * clang-format -i -style="{BasedOnStyle: Google, IndentWidth: 4}" matrix_exponentiation.cpp * clang-format -i -style="{IndentWidth: 4}" matrix_exponentiation.cpp * Fix cpplint readability/braces issue * using std::cin; using std::cout; using std::vector; * added int_64 instead of long long * Minor changes * Update matrix_exponentiation.cpp
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/* | ||
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. | ||
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). | ||
The first element of this matrix is the required result. | ||
*/ | ||
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#include <bits/stdc++.h> | ||
using std::cin; | ||
using std::cout; | ||
using std::vector; | ||
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#define ll int64_t | ||
#define endl '\n' | ||
#define pb push_back | ||
#define MOD 1000000007 | ||
ll ab(ll x) { return x > 0LL ? x : -x; } | ||
ll k; | ||
vector<ll> a, b, c; | ||
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// To multiply 2 matrix | ||
vector<vector<ll>> multiply(vector<vector<ll>> A, 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++) { | ||
for (ll z = 1; z <= k; z++) { | ||
C[i][j] = (C[i][j] + (A[i][z] * B[z][j]) % MOD) % MOD; | ||
} | ||
} | ||
} | ||
return C; | ||
} | ||
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// computing power of a matrix | ||
vector<vector<ll>> power(vector<vector<ll>> A, ll p) { | ||
if (p == 1) | ||
return A; | ||
if (p % 2 == 1) { | ||
return multiply(A, power(A, p - 1)); | ||
} else { | ||
vector<vector<ll>> X = power(A, p / 2); | ||
return multiply(X, X); | ||
} | ||
} | ||
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// main function | ||
ll ans(ll n) { | ||
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]; | ||
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// Transpose matrix | ||
vector<vector<ll>> T(k + 1, vector<ll>(k + 1)); | ||
for (ll i = 1; i <= k; i++) { | ||
for (ll j = 1; j <= k; j++) { | ||
if (i < k) { | ||
if (j == i + 1) | ||
T[i][j] = 1; | ||
else | ||
T[i][j] = 0; | ||
continue; | ||
} | ||
T[i][j] = c[k - j]; | ||
} | ||
} | ||
// T^n-1 | ||
T = power(T, n - 1); | ||
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// T*F1 | ||
ll res = 0; | ||
for (ll i = 1; i <= k; i++) { | ||
res = (res + (T[1][i] * F1[i]) % MOD) % MOD; | ||
} | ||
return res; | ||
} | ||
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// 1 1 2 3 5 | ||
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int main() { | ||
cin.tie(0); | ||
cout.tie(0); | ||
ll t; | ||
cin >> t; | ||
ll i, j, x; | ||
while (t--) { | ||
cin >> k; | ||
for (i = 0; i < k; i++) { | ||
cin >> x; | ||
b.pb(x); | ||
} | ||
for (i = 0; i < k; i++) { | ||
cin >> x; | ||
c.pb(x); | ||
} | ||
cin >> x; | ||
cout << ans(x) << endl; | ||
b.clear(); | ||
c.clear(); | ||
} | ||
return 0; | ||
} |