矩阵快速幂和几个应用

模板：

#define maxn 1005
struct Matrix 
{
    int m[maxn][maxn];
}a,b,ans,res;
void init()
{
    for(int i = 1; i <= n; i ++)
    {
        for(int j = 1; j <= n; j ++)
        {
            if(i==j)
                a.m[i][j] = 1;
            else
                a.m[i][j] = 0;
        }
    }
}
Matrix mul(Matrix a,Matrix b)
{
    Matrix c;
    for(int i = 1; i <= n; i ++)
    {
        for(int j = 1; j <= n; j ++)
        {
            c.m[0][0] = 0;
            for(int k = 1; k <= n; k ++)
                c.m[i][j] += a.m[i][k]*b.m[k][j];
        }
    }
    return c;
}
Matrix q_pow(Matrix a,int b)
{
    res = a;
    while(b)
    {
        if(b&1)
            res = mul(res,a);
        a = mul(a,a);
        b >>= 1;
    }
    return res;
}

Fibonacci

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

题意：求斐波那契数列的第n项

解析：利用矩阵快速幂，因为题意给出了

即求这个矩阵的n次幂，然后再输出矩阵的第一列第二个就行啦~

#include<iostream>
#include<cstring>
#include<algorithm>