Pagini recente » Cod sursa (job #53549) | Cod sursa (job #1938624) | Cod sursa (job #362094) | Cod sursa (job #1680627) | Cod sursa (job #2276517)
//
// main.cpp
// iepuri
//
// Created by Tereza Oprea on 04/11/2018.
// Copyright © 2018 Tereza Oprea. All rights reserved.
//
#include <iostream>
#include <fstream>
#define M 666013
using namespace std;
ifstream fin ("iepuri.in");
ofstream fout ("iepuri.out");
int n, x, y, z, a, b, c, t;
struct matrix
{
int m[3][3];
matrix( )
{
memset (m, 0, sizeof(m));
}
matrix operator * (matrix b)
{
matrix out = matrix( );
for (int i=0; i<3; i++)
for (int j=0; j<3; j++)
for (int k=0; k<3; k++)
{
out.m[i][j] = (out.m[i][j] + 1LL * m[ i ][ k ] * b.m[ k ][ j ])%M;
}
return out;
}
};
matrix A, B;
matrix unit = matrix( );
matrix mPow (matrix x, int n)
{
if (n==0)
return unit;
matrix half = mPow (x, n/2);
half = half * half;
if (n%2)
half = half * x;
return half;
}
int main() {
unit.m[0][0] = unit.m[1][1] = unit.m[2][2] = 1;
fin >> t;
for (int i=1; i<=t; i++){
A = matrix( );
B = matrix( );
fin >> x >> y >> z;
fin >> a >> b >> c;
fin >> n;
B.m[0][0] = 0; B.m[0][1] = 0; B.m[0][2] = c;
B.m[1][0] = 1; B.m[1][1] = 0; B.m[1][2] = b;
B.m[2][0] = 0; B.m[2][1] = 1; B.m[2][2] = a;
A.m[0][0] = x; A.m[0][1] = y; A.m[0][2] = z;
B = mPow(B, n-2);
A = A * B;
fout << A.m[0][2] << '\n';
}
return 0;
}