HDU 4549 M斐波那契数列

M斐波那契数列

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 520    Accepted Submission(s): 146

Problem Description

M斐波那契数列F[n]是一种整数数列，它的定义如下：

F[0] = a
F[1] = b
F[n] = F[n-1] * F[n-2] ( n > 1 )

现在给出a, b, n，你能求出F[n]的值吗？

 

Input

输入包含多组测试数据；
每组数据占一行，包含3个整数a, b, n（ 0 <= a, b, n <= 10^9 ）

 

Output

对每组测试数据请输出一个整数F[n]，由于F[n]可能很大，你只需输出F[n]对1000000007取模后的值即可，每组数据输出一行。

 

Sample Input

0 1 06 10 2

 

Sample Output

060

 

Source

2013金山西山居创意游戏程序挑战赛——初赛（2）

 

 

package acm.hpu;import java.util.Scanner;public class Main {private static long mod = 1000000007l;private static Matrix fibn = null;public static void main(String[] args) {Scanner in = new Scanner(System.in);long a,b;long n;while(in.hasNextLong()){fibn = null;a = in.nextLong();b = in.nextLong();n = in.nextLong();if(n == 0){System.out.println(a%mod);}else if(n==1){System.out.println(b%mod);}else{System.out.println(solve(a,b,n));}}}private static long solve(long a, long b, long n){fib(n-1);long an = fibn.data[0][1];long bn = fibn.data[1][1];a = Matrix.pow(a, an, mod);b = Matrix.pow(b,bn, mod);long result = a * b % mod;return result;}private static Matrix fib(long n){if(fibn == null){Matrix m = new Matrix(Matrix.FIB);m = m.pow(n, mod);fibn = m;}return fibn;} }class Matrix{public static final int UNIT = 1;public static final int FIB = 2;public static final int COMMON = 0;long [][] data = null;public Matrix(){data = new long[2][2];}public Matrix(int type){data = new long[2][2];switch(type){case FIB:data[0][1] = data[1][0] = data[1][1] = 1;break;case UNIT:data[0][0] = data[1][1] = 1;data[0][1] = data[1][0] = 0;break;default:}}public Matrix multiply(Matrix matrix,long mod){int k,i,j;Matrix m = new Matrix();for(i = 0; i < data.length; ++i){for(j = 0; j < data[i].length; ++j){for(k = 0; k < data.length; ++k){m.data[i][j] = (m.data[i][j] + this.data[i][k] * matrix.data[k][j])%(mod-1);}}}return m;}public Matrix pow(long n, long mod){Matrix m = copy();Matrix unit = new Matrix(Matrix.UNIT);while(n>0){if((n&1) == 1){unit = unit.multiply(m,mod);}m = m.multiply(m,mod);n >>= 1;}return unit;}public Matrix copy(){Matrix m = new Matrix();for(int i = 0; i < m.data.length; ++i){for(int j = 0; j < m.data[i].length; ++j){m.data[i][j] = this.data[i][j];}}return m;}public static long pow(long x, long n, long mod){long c = 1;while(n > 0){if((n&1) == 1){c = c * x % mod;}x = x * x % mod;n >>= 1;}return c;}}