poj-3070-Fibonacci

Fibonacci（ 传送门）

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

矩阵快速幂的入门题，，我也是第一次练

code：

#include
#include
#include
#include

using namespace std;
const int N=2;
const int mod=10000;
int res[N][N];
int temp[N][N];
int a[N][N];
void Mul(int a[][N],int b[][N])///矩阵乘法
{
    memset(temp,0,sizeof(temp));
    for(int i=0;i>=1;
    }
}
int main()
{
    int n;
    while(~scanf("%d",&n)&&n!=-1){
        a[0][0]=1;
        a[1][0]=1;
        a[0][1]=1;
        a[1][1]=0;
        if(n==0||n==1)printf("%d\n",n);
        else{
            fun(a,n);
            printf("%d\n",res[0][1]);
        }
    }
}