Fibonacci

题目联接:http://acm.tzc.edu.cn/acmhome/problemdetail.do?&method=showdetail&contestId=282&id=3492

Fibonacci

Time Limit(Common/Java):1000MS/3000MS     Memory Limit:65536KByte
Total Submit: 15            Accepted: 11

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

分析：

用编程之美上提到的分治策略，即有些人讲的快速指数相乘法。

代码：

#include<iostream>
using namespace std;
void multiply(int *a, int *b)    //计算矩阵乘法（模4）
{
    int tmp1[4], tmp2[4];
    int i;
    for (i=0; i<4; i++) tmp1[i]=a[i];
    for (i=0; i<4; i++) tmp2[i]=b[i];
    a[0]=(tmp1[0]*tmp2[0] + tmp1[1]*tmp2[2])%10000;
    a[1]=(tmp1[0]*tmp2[1] + tmp1[1]*tmp2[3])%10000;
    a[2]=(tmp1[2]*tmp2[0] + tmp1[3]*tmp2[2])%10000;
    a[3]=(tmp1[2]*tmp2[1] + tmp1[3]*tmp2[3])%10000;    
}
int main()
{
    long n;
    while (cin>>n, n!=-1)
    {
        int a[4]={1, 0, 0, 1};    //存放矩阵A^n；
        int tmp[4]={1, 1, 1, 0};    //存放矩阵A^i;
        for (; n; n>>=1)    //快速指数相乘法
        {
            if (n&1)
            {
                multiply(a,tmp);
            }
            multiply(tmp,tmp);
        }
        cout<<a[1]<<endl;
    }
    return 0;
}