poj 3070 Fibonacci（矩阵快速幂模板题）

裸的矩阵快速幂，题目里连矩阵都告诉你了。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int mod = 1e4;
struct node
{
int s[2][2];
node() {}
node(int a, int b, int c, int d)
{
s[0][0] = a;
s[0][1] = b;
s[1][0] = c;
s[1][1] = d;
}
};

node mul(node a, node b)
{
node t = node(0, 0, 0, 0);
for(int i = 0; i < 2; i++)
for(int j = 0; j < 2; j++)
for(int k = 0; k < 2; k++)
t.s[i][j] = (t.s[i][j]+a.s[i][k]*b.s[k][j])%mod;
return t;
}

node mt_pow(node p, int n)
{
node q = node(1, 0, 0, 1);
while(n)
{
if(n&1) q = mul(p,q);
p = mul(p, p);
n /= 2;
}
return q;
}

int main(void)
{
int n;
node p;
while(scanf("%d", &n), n+1)
{
p = node(1, 1, 1, 0);
p = mt_pow(p, n);
printf("%d\n", p.s[0][1]);
}
return 0;
}

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13621		Accepted: 9658

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:


.

Source

Stanford Local 2006