POJ3070Fibonacci(矩阵快速幂+高效）

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11587		Accepted: 8229

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：就是求斐波那契数列，只是有点大，递推也会超时，矩阵快速幂0ms,搞定；

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 2;
const int mod = 10000;
struct Mat
{
    int mat[2][2];
};
Mat operator * (Mat a, Mat b)
{
    Mat c;
    memset(c.mat, 0, sizeof(c.mat));
    for(int k = 0; k < N; k++)
    {
        for(int i = 0; i < N; i++)
        {
            for(int j = 0; j < N; j++)
            {
                c.mat[i][j] = (c.mat[i][j] + a.mat[i][k] * b.mat[k][j] % mod) % mod;
            }
        }
    }
    return c;
}
Mat operator ^ (Mat a, int k)
{
    Mat c;
    for(int i = 0; i < N; i++)
        for(int j = 0; j < N; j++)
            c.mat[i][j] = (i == j);
    while(k)
    {
        if(k & 1)
            c = c * a;
        a = a * a;
        k >>= 1;
    }
    return c;
}
int main()
{
    int n;
    while(scanf("%d", &n) != EOF)
    {
        if(n == -1)
            break;
        if(n == 0)
            printf("0
");
        else if(n == 1)
            printf("1
");
        else
        {
            Mat a;
            a.mat[0][0] = 1;
            a.mat[0][1] = 1;
            a.mat[1][0] = 1;
            a.mat[1][1] = 0;
            a = a ^ (n - 1);
            printf("%d
",a.mat[0][0]);
        }
    }

    return 0;
}