In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …
An alternative formula for the Fibonacci sequence is
.
Given an integer n, your goal is to compute the last 4 digits of Fn.
Input
The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1.
Output
For each test case, print the last four digits of Fn. If the last four digits of Fnare all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn mod 10000).
Sample Input
0 9 999999999 1000000000 -1
Sample Output
0 34 626 6875
Hint
As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by
.
Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:
.
题意：求第几项的菲波那切数。
题解：因为数据大所以需要矩阵快速幂. 递推矩阵请看代码：
import java.util.*;
public class Main {
static Scanner cin = new Scanner(System.in);
static class hh{
long [][] ma = new long [2][2];
public hh() {
for (int i = 0; i < 2;i++) {
for (int j = 0; j < 2;j++) {
ma[i][j]=0;
}
}
}
}
static hh mul(hh a, hh b,long x) {
hh c = new hh();
for (int i = 0; i < 2;i++) {
for (int j = 0; j < 2;j++) {
for (int k = 0;k < 2;k++) {
c.ma[i][j]=(c.ma[i][j]+a.ma[i][k]*b.ma[k][j]%x)%x;
}
}
}
return c;
}
static hh quick(hh a,long n,long c) {//模仿数的矩阵快速幂
hh tmp = new hh();
for (int i = 0; i < 2;i++) tmp.ma[i][i]=1;
while(n!=0) {
if(n%2==1) tmp=mul(tmp,a,c);
n/=2;
a=mul(a,a,c);
}
return tmp;
}
public static void main(String[] args){
long t;
while(cin.hasNext()) {
t=cin.nextLong();
if(t==-1) break;
hh a = new hh();
hh b = new hh();
a.ma[0][0]=1;//递推矩阵构建
a.ma[0][1]=1;
a.ma[1][0]=1;
a.ma[1][1]=0;
b=quick(a,t,10000);
System.out.println(b.ma[1][0]%10000);
}
}
}