HDU 3070 Fibonacci

最新推荐文章于 2024-08-26 15:30:00 发布

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最新推荐文章于 2024-08-26 15:30:00 发布

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本文介绍了一种计算斐波那契数列第n项最后四位数字的方法，使用矩阵快速幂来提高效率，适用于n值非常大的情况。

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Fibonacci

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 15 Accepted Submission(s) : 6

Problem 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 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

Sample Output

Source

PKU

快速幂模板：

long long modexp(long long a, long long b, int mod)  
{  
    long long res=1;  
    while(b>0)  
    {  
        //a=a%mod;(有时候n的值太大了会超出long long的储存，所以要先取余)  
        if(b&1)//&位运算：判断二进制最后一位是0还是1，&的运算规则为前后都是1的时候才是1，奇数偶数判断；  
            res=res*a%mod;  
        b=b>>1;//相当于除以2；  
        a=a*a%mod;  
    }  
    return res;  
}

代码：

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
struct fast_mod
{
    int t[2][2];
};
fast_mod matrixmul(fast_mod a,fast_mod b)
{
    int i,j,k;
    fast_mod c;
    for(i=0; i<2; i++)
    {
        for(j=0; j<2; j++)
        {
            c.t[i][j]=0;
            for(k=0; k<2; k++)
            {
                c.t[i][j]+=(a.t[i][k]*b.t[k][j])%10000;
            }
        }
    }
    return c;
}
int main()
{
    int n,i;
    while(scanf("%d",&n),n!=-1)
    {
        if(!n)
        {
            printf("0\n");
            continue;
        }
        fast_mod s;
        s.t[0][0]=s.t[0][1]=s.t[1][0]=1;
        s.t[1][1]=0;
        fast_mod ans;
        memset(ans.t,0,sizeof(ans.t));
        for(i=0; i<2; i++)
        {
            ans.t[i][i]=1;
        }
        while(n)
        {
            if(n&1)
                ans=matrixmul(ans,s);
            n=n>>1;
            s=matrixmul(s,s);
        }
        cout<<ans.t[0][1]%10000<<endl;
    }

}