Fibonacci

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

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
很简单的矩阵快速幂，几乎就是一道模板试手题，原理学过矩阵乘法的都知道，就不再赘述了。
直接上代码：
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<cmath>
#define ll long long
int gi()
{
    int x=0,y=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')
            y=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*y;
}
int n;
struct ju
{
    ll a[145][145];
    inline ju operator *(const ju &b)const
    {
        ju tmp;
        for(int i=1; i<=2; i++)
            for(int j=1; j<=2; j++)
            {
                tmp.a[i][j]=0;
                for(int k=1; k<=2; k++)
                {
                    tmp.a[i][j]+=a[i][k]*b.a[k][j];
                    tmp.a[i][j]%=10000;
                }
            }
        return tmp;
    }
} ans;
ju pow(ju a,int k)
{
    ju tmp=a;
    while(k!=0)
    {
        if(k&1)
            tmp=tmp*a;
        a=a*a;
        k>>=1;
    }
    return tmp;
}
using namespace std;
int main()
{
    while(1)
    {
        n=gi();
        if(n==-1)
            return 0;
        ans.a[1][1]=1;
        ans.a[1][2]=1;
        ans.a[2][1]=1;
        ans.a[2][2]=0;
        ans=pow(ans,n);
        printf("%lld\n",ans.a[2][2]);
    }
}

//        FOR C.H

 

posted @ 2017-07-10 23:37  GSHDYJZ  阅读(152)  评论(0)  编辑  收藏  举报