PROBLEM LINK:
Author: hareesh29
DIFFICULTY:
EASYMEDIUM
PREREQUISITES:
PROBLEM:
Given n cities(1 to n) and an integer k.find the number of combinations of cities the magician can visit using k teleportation given a restriction that he can’t teleport between two cities x and y.
EXPLANATION:
The information we know from the problem statement is that there are n cities and teleportation is restricted between two of these cites. Let’s split these cities into two different states A and B. State A contains n2 cities where there are no travel restrictions and State B contains 2 cities(x,y) where teleportation is not allowed between these two cities.
Let A[k] denote the number of ways the magician ends in state A after k teleports and let B[k] denote the number of ways the magician ends in state B after k teleports.now the total number of combination can be calculated by finding A[k] + B[k].
we know that A[1]=n2 and B[1]=2
now we can find A[2] and B[2] by using this relation
A[2] = A[1]*(n2)+ B[1]*(n2)
B[2] = A[1]*(2)+B[1]*(1)
Explanation of the above relation:

To reach state A from state A total ways are n2 as the magician can teleport to any of the n2 cities from state A

To reach state A from state B the total ways are n2 as the magician can teleport to any of the n2 cities from state B

To reach state B from state A the total ways are 2 as the magician can teleport to any of the 2 cities in state B from State A

To reach state B from state B the total ways are 1 as the magician can teleport only teleport to the same city(if he is in city x he can only travel to x similarly for y)
In general
A[k]=A[k1]*(n2)+B[k1]*(n2)
B[k]=A[K1]*(2) +B[k1]*(1)
It can be represented in matrix form as
\begin{bmatrix}A[k]& B[k] \end{bmatrix} = \begin{bmatrix}A[k1] & B[k1] \end{bmatrix} *\begin{bmatrix}n2 & 2 \\n2 & 1 \end{bmatrix}
Using this reccurence relation we can compute the below relation
\begin{bmatrix}A[k]& B[k] \end{bmatrix} = \begin{bmatrix}A[1] & B[1] \end{bmatrix} *\left[ \begin{array}{}n2 & 2 \\n2 & 1\end{array}\right]^{k1}
now the solution can be found by calculating A[k]+B[k]
Time Complexity:
The time Complexity is O(8*log(k)) where 8 is for matrix multiplication log(k) for exponentiation
SOLUTIONS:
Setter's Solution
mod=int(1e9)+7
def muL1(l,b):
mux=[0,0]
for i in range(2):
mux[i]=(l[0]*b[0][i])%mod
mux[i]=(mux[i]+(l[1]*b[1][i]))%mod
return mux
def mul(b):
mul=[[0,0],[0,0]]
for i in range(2):
for j in range(2):
for k in range(2):
mul[i][j]=(mul[i][j]+b[i][k]*b[k][j])%mod
return mul
def matrixexpo(p,b,l):
while(p>0):
if(p%2==1):
l=muL1(l,b)
b=mul(b)
p//=2
return ((l[0]+l[1]))%mod
for _ in range(int(input())):
n,k,x,y=map(int,input().split())
l=[n2,2]
b=[[n2,2],[n2,1]]
print(matrixexpo(k1,b,l))