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poj 3070 Fibonacci（矩阵快速幂）

2024-07-05 09:33:41   229人阅读

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
代码：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

struct mat
{
    long long  t[2][2];
    void set()
    {
        memset(t,0,sizeof(t));
    }
}a,b;

mat multiple(mat a,mat b,int n,int p)
{
    int i,j,k;
    mat temp;
    temp.set();
    for(i=0;i<n;i++)
    for(j=0;j<n;j++)
    {
        if(a.t[i][j])
        for(k=0;k<n;k++)
            temp.t[i][k]=(temp.t[i][k]+a.t[i][j]*b.t[j][k])%p;
    }
    return temp;
}

mat quick_mod(mat a,mat b,int n,int p)
{
    while(n)
    {
        if(n&1)
        {
           a=multiple(a,b,2,p);
        }
        n>>=1;
        b=multiple(b,b,2,p);
    }
    return a;
}
void init()
{
    a.t[0][0]=2;
    a.t[0][1]=1;
    a.t[1][0]=1;
    a.t[1][1]=0;
    b.t[0][0]=1;
    b.t[0][1]=1;
    b.t[1][0]=1;
    b.t[1][1]=0;
}
int main()
{
    int n;
    while(cin>>n)
    {
        if(n<0) break;
        init();
        a=quick_mod(a,b,n,10000);
        cout<<a.t[1][1]<<endl;
    }
    return 0;
}

































        

            
                
                  
                
                

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