POJ 3070 Fibonacci（矩阵快速幂模板）

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：

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：就是要求第n个Fibonacci数对10000的取余，由于n很大，简单模拟肯定不行，那么就引出了Fibonacci的另一种表示方法（题中已给），根据描述，我们只要求出2*2的矩阵{{1，1}，{1，0}}^n就可以了。

由此引出新算法：矩阵快速幂（根据快速幂改编而来，快速幂计算的是数的n次方，而这个求的是矩阵的n次方）。

#include<stdio.h>
#include<string.h>
#include<queue>
#include<math.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;

const int N=1e6+10;
const int INF=0x3f3f3f3f;
const int MOD=1e4;

typedef long long LL;

struct node
{
    LL m[2][2];
}ans, tmp, cnt;

node Multiply(node a, node b) ///计算两个矩阵相乘之后的矩阵
{
    int i, j, k;

    for (i = 0; i < 2; i++)
    {
        for (j = 0; j < 2; j++)
        {
            cnt.m[i][j] = 0;
            for (k = 0; k < 2; k++)
                cnt.m[i][j] = (cnt.m[i][j]+a.m[i][k]*b.m[k][j])%MOD;
        }
    }

    return cnt;
}

LL Matrix_power(LL n) 
{
    ans.m[0][0] = ans.m[1][1] = 1;
    ans.m[0][1] = ans.m[1][0] = 0;
    tmp.m[0][0] = tmp.m[0][1] = tmp.m[1][0] = 1;
    tmp.m[1][1] = 0;

    while (n) ///和快速幂求法一致
    {
        if (n % 2 != 0)
            ans = Multiply(ans, tmp);

        tmp = Multiply(tmp, tmp);

        n /= 2;
    }

    return ans.m[0][1]; ///由于第n个Fibonacci数存在于{Fn+1，Fn，Fn，Fn-1}中，所以我们返回（0，1）位置上的元素即可
}

int main ()
{
    LL n, num;

    while (scanf("%lld", &n), n != -1)
    {
        num = Matrix_power(n);
        printf("%lld
", num);
    }

    return 0;
}