[POJ3070]Fibonacci

Fibonacci

Time Limit: 1000MS Memory Limit: 65536K

Description

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 Fn are 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:

题解

矩阵乘法入门

（不要脸地骗一道AC…..）

var
x:array[0..10000]of longint;
i,j,k:longint;
t,y:array[1..2,1..2]of int64;
n:longint;
function g(b:longint):longint;
var c,d,e,f:int64;
begin
if b<0 then exit(0);
t[1,1]:=1; t[1,2]:=0; t[2,1]:=0; t[2,2]:=1;
y[1,1]:=0; y[1,2]:=1; y[2,1]:=1; y[2,2]:=1;
while b<>0 do
begin
if (b and 1)=1
then begin
c:=(t[1,1]*y[1,1]+t[1,2]*y[2,1])mod k;
d:=(t[1,1]*y[1,2]+t[1,2]*y[2,2])mod k;
e:=(t[2,1]*y[1,1]+t[2,2]*y[2,1])mod k;
f:=(t[2,1]*y[1,2]+t[2,2]*y[2,2])mod k;
t[1,1]:=c; t[1,2]:=d; t[2,1]:=e; t[2,2]:=f;
end;
c:=(y[1,1]*y[1,1]+y[1,2]*y[2,1])mod k;
d:=(y[1,1]*y[1,2]+y[1,2]*y[2,2])mod k;
e:=(y[2,1]*y[1,1]+y[2,2]*y[2,1])mod k;
f:=(y[2,1]*y[1,2]+y[2,2]*y[2,2])mod k;
y[1,1]:=c; y[1,2]:=d; y[2,1]:=e; y[2,2]:=f;
b:=b shr 1;
end;
exit((t[2,1]*x[0]+t[2,2]*x[1])mod k);
end;

begin
k:=10000;
readln(n);
while n<>-1 do
begin
x[0]:=0; x[1]:=1;
writeln(g(n-1));
readln(n);
end;
end.