poj_3070Fibonacci(矩阵快速幂)

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12732		Accepted: 9060

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:

.

Source

Stanford Local 2006

 

//快速幂矩阵
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct Mat{
    int mat[2][2];
};
const int Mod = 10000;
Mat operator *(Mat a, Mat b)
{
    Mat c;
    memset(c.mat,0,sizeof(c.mat));
    for(int i = 0; i < 2; i++)
    {
        for(int j = 0; j < 2; j++)
        {
            for(int k = 0; k < 2; k++)
            {
                c.mat[i][j] = (c.mat[i][j]+(a.mat[i][k]*b.mat[k][j])%Mod)%Mod;
            }
        }
    }
    return c;
}
Mat multi(int n)
{
    Mat c;
    memset(c.mat,0,sizeof(c.mat));
    c.mat[0][0] = c.mat[1][1] = 1;
    Mat a;
    memset(a.mat,0,sizeof(a.mat));
    a.mat[0][0] = a.mat[0][1] = a.mat[1][0] = 1;
    while(n)
    {
        if(n&1) c = c*a;
        a = a*a;
        n>>=1;
    }
    return c;
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        if(n==-1) return 0;
        Mat ans = multi(n);
        printf("%d\n",ans.mat[0][1]);
    }
    return 0;
}