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: 65536KTotal Submissions: 13621 Accepted: 9658

Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, and F_n = F_{n − 1} + F_{n − 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 F_n.

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 F_n. If the last four digits of F_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print F_n mod 10000).

Sample Input

099999999991000000000-1

Sample Output

0346266875

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