Another kind of Fibonacci
Problem Description
As we all known , the Fibonacci series : F(0) = 1, F(1) = 1, F(N) = F(N - 1) + F(N - 2) (N >= 2).Now we define another kind of Fibonacci : A(0) = 1 , A(1) = 1 , A(N) = X * A(N - 1) + Y * A(N - 2) (N >= 2).And we want to Calculate S(N) , S(N) = A(0)2 +A(1)2+……+A(n)2.
Input
There are several test cases.
Each test case will contain three integers , N, X , Y .
N : 2<= N <= 231 – 1
X : 2<= X <= 231 – 1
Y : 2<= Y <= 231 – 1
Output
For each test case , output the answer of S(n).If the answer is too big , divide it by 10007 and give me the reminder.
Sample Input
2 1 1
3 2 3
Sample Output
6
196
思路见下图:
#include<iostream>
#include<cstring>
using namespace std;
const int maxn = 5;
typedef long long LL;
struct Matrix{
int matrix[maxn][maxn];
}ori, ans;
int n = 4, N, X, Y, m = 10007;
void init()
{
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
ori.matrix[i][j] = ans.matrix[i][j] = 0;
ori.matrix[0][0] = ori.matrix[2][1] = 1;
ori.matrix[0][1] = ori.matrix[1][1] = X * X % m;
ori.matrix[0][2] = ori.matrix[1][2] = Y * Y % m;
ori.matrix[0][3] = ori.matrix[1][3] = ((2 * X) % m) * Y % m;
ori.matrix[3][1] = X;
ori.matrix[3][3] = Y;
ans.matrix[0][0] = 2;
ans.matrix[1][0] = ans.matrix[2][0] = ans.matrix[3][0] = 1;
}
Matrix multiply(Matrix a, Matrix b)
{
Matrix temp;
memset(temp.matrix, 0, sizeof(temp.matrix));
for(int i=0;i<n;i++)
for(int j=0;j<n;j++)
for(int k=0;k<n;k++)
temp.matrix[i][j] = (temp.matrix[i][j] + ((a.matrix[i][k] * b.matrix[k][j]) % m)) % m;
return temp;
}
//矩阵的b次幂
Matrix binaryPow(int b)
{
Matrix temp;
memset(temp.matrix, 0, sizeof(temp.matrix));
for(int i=0;i<n;i++)
temp.matrix[i][i] = 1;
while(b > 0)
{
if(b & 1)
temp = multiply(ori, temp);
ori = multiply(ori, ori);
b >>= 1;
}
return temp;
}
int main()
{
//freopen("in.txt","r", stdin);
while(cin>>N>>X>>Y)
{
X %= m;
Y %= m;
init();
Matrix temp = binaryPow(N - 1);
ans = multiply(temp, ans);
cout<<ans.matrix[0][0]<<endl;
}
return 0;
}