POJ 3070 Fibonacci

模算法+矩阵快速幂
                                                                                      Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7138		Accepted: 5044

Description

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

099999999991000000000-1

Sample Output

0346266875

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

 

 

#include <iostream>
#include <cstring>

#define MOD 10000

typedef long long LL[2][2];

using namespace std;

void mul(LL a,LL b,LL c)
{
    c[0][0]=((a[0][0]%MOD*b[0][0]%MOD)%MOD+(a[0][1]%MOD*b[1][0]%MOD)%MOD)%MOD;
    c[0][1]=((a[0][0]%MOD*b[0][1]%MOD)%MOD+(a[0][1]%MOD*b[1][1]%MOD)%MOD)%MOD;
    c[1][0]=((a[1][0]%MOD*b[0][0]%MOD)%MOD+(a[1][1]%MOD*b[1][0]%MOD)%MOD)%MOD;
    c[1][1]=((a[1][0]%MOD*b[0][1]%MOD)%MOD+(a[1][1]%MOD*b[1][1]%MOD)%MOD)%MOD;
}



int main()
{
    int n;
    while(cin>>n&&n!=-1)
    {
        LL e={1,0,0,1};
        LL c={1,1,1,0};
        LL t;
        while(n)
        {
            for(; n; n >>= 1)
            {
            if(n&1)
            {
               mul(e,c,t);
               memcpy(e,t,sizeof(t));
            }
               mul(c,c,t);
               memcpy(c,t,(sizeof(t)));
            }
        }

        cout<<e[0][1]<<endl;
    }


    return 0;
}