已知递推公式:
F(n, 1)=F(n-1, 2) + 2F(n-3, 1) + 5,
F(n, 2)=F(n-1, 1) + 3F(n-3, 1) + 2F(n-3, 2) + 3.
初始值为:F(1, 1)=2, F(1, 2)=3, F(2, 1)=1, F(2, 2)=4,F(3, 1)=6, F(3,2)=5。
输入n,输出F(n, 1)和F(n, 2),由于答案可能很大,你只需要输出答案除以99999999的余数。输入格式
输入第一行包含一个整数n。
输出格式
输出两行,第一行为F(n, 1)除以99999999的余数,第二行为F(n, 2)除以99999999的余数。
样例输入
4
样例输出
14
21
数据规模和约定
1<=n<=10^18
写一个实现矩阵相乘的方法和实现矩阵快速幂的方法,用矩阵表示出递推关系式,注意初始值。
import java.util.Scanner;
public class Main {
final static int mod = 99999999;
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
long n = sc.nextLong();
if (n == 1) {
System.out.println(2);
System.out.println(3);
}else if(n == 2){
System.out.println(1);
System.out.println(4);
}else if(n == 3){
System.out.println(6);
System.out.println(5);
}else{
long[][]a = {{0,1,0,0,2,0,5},
{1,0,0,0,3,2,3},
{1,0,0,0,0,0,0},
{0,1,0,0,0,0,0},
{0,0,1,0,0,0,0},
{0,0,0,1,0,0,0},
{0,0,0,0,0,0,1}};
long[][]b = {{6},{5},{1},{4},{2},{3},{1}};
long[][]x = Multiply_Matrix(Multiply_ksm(a, n-3), b);
long result1 = x[0][0]%mod;
long result2 = x[1][0]%mod;
System.out.println(result1);
System.out.println(result2);
}
sc.close();
}
public static long[][] Multiply_Matrix(long[][] a, long[][] b) {
long[][] c = new long[a.length][b[0].length];
for (int i = 0; i < a.length; i++) {
for (int j = 0; j < b[0].length; j++) {
long temp = 0;
for (int k = 0; k < b.length; k++) {
temp = (temp + ((a[i][k] % mod) * (b[k][j] % mod)) % mod) % mod;
}
c[i][j] = temp;
}
}
return c;
}
public static long[][] Multiply_ksm(long[][] a, long k) {
long[][] d = new long[a.length][a[0].length];
if (k == 1) {
return a;
} else if (k == 2) {
return Multiply_Matrix(a, a);
} else if (k % 2 == 0) {
d = Multiply_ksm(Multiply_Matrix(a, a), k / 2);
return d;
} else {
d = Multiply_ksm(Multiply_Matrix(a, a), k / 2);
return Multiply_Matrix(d, a);
}
}
}