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HDU 3117 Fibonacci Numbers(Fibonacci矩阵加速递推+公式)

2024-07-25 00:14:37   219人阅读

题目意思很简单：求第n个Fibonacci数，如果超过八位输出前四位和后四位中间输出...，否则直接输出Fibonacci数是多少。

后四位很好求，直接矩阵加速递推对10000取余的结果就是。

前四位搜了一下：http://blog.csdn.net/xieqinghuang/article/details/7789908

Fibonacci的通项公式，对，fibonacci数是有通项公式的——

f(n)=1/sqrt(5)(((1+sqrt(5))/2)^n+((1-sqrt(5))/2)^n)

假设F[n]可以表示成 t * 10^k（t是一个小数），那么对于F[n]取对数log10，答案就为log10 t + K，此时很明显log10 t<1，于是我们去除整数部分，就得到了log10 t ，
再用pow（10，log10 t)我们就还原回了t。将t×1000就得到了F[n]的前四位。 具体实现的时候Log10 F[n]约等于((1+sqrt(5))/2)^n/sqrt(5)，这里我们把((1-sqrt(5))/2)^n这一项忽略了，
因为当N>=40时，这个数已经小的可以忽略。于是log10 F[n]就可以化简成log10 1/sqrt(5) + n*log10 (1+sqrt(5))/2

通过求对数可以得到fibonacci数。

还有就是其实后四位存在一个循环节没15000个会重复一次。

Fibonacci Numbers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1882    Accepted Submission(s): 738


Problem Description
The Fibonacci sequence is the sequence of numbers such that every element is equal to the sum of the two previous elements, except for the first two elements f0 and f1 which are respectively zero and one.

What is the numerical value of the nth Fibonacci number?
 

Input
For each test case, a line will contain an integer i between 0 and 108 inclusively, for which you must compute the ith Fibonacci number fi. Fibonacci numbers get large pretty quickly, so whenever the answer has more than 8 digits, output only the first and last 4 digits of the answer, separating the two parts with an ellipsis (“...”). 

There is no special way to denote the end of the of the input, simply stop when the standard input terminates (after the EOF).
 

Sample Input
0
1
2
3
4
5
35
36
37
38
39
40
64
65
 

Sample Output
0
1
1
2
3
5
9227465
14930352
24157817
39088169
63245986
1023...4155
1061...7723
1716...7565
 

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-10
///#define M 1000100
#define LL __int64
///#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)

///#define mod 9973

int mod;
const int maxn = 2010;

using namespace std;

struct matrix
{
    int f[110][110];
};

matrix mul(matrix a, matrix b, int n)///矩阵乘法
{
    matrix c;
    memset(c.f, 0, sizeof(c.f));
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < n; j++)
        {
            for(int k = 0; k < n; k++) c.f[i][j] += a.f[i][k]*b.f[k][j];
            c.f[i][j] %= mod;
        }
    }
    return c;
}

matrix pow_mod(matrix a, int b, int n)///矩阵快速幂
{
    matrix s;
    memset(s.f, 0 , sizeof(s.f));
    for(int i = 0; i < n; i++) s.f[i][i] = 1;
    while(b)
    {
        if(b&1) s = mul(s, a, n);
        a = mul(a, a, n);
        b >>= 1;
    }
    return s;
}

matrix Add(matrix a,matrix b, int n)  ///矩阵加法
{
    matrix c;
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < n; j++)
        {
            c.f[i][j] = a.f[i][j]+b.f[i][j];
            c.f[i][j] %= mod;
        }
    }
    return c;
}

LL num[maxn];

int main()
{
    num[0] = 0LL;
    num[1] = 1LL;
    for(int i = 2; i <= 40; i++) num[i] = num[i-1]+num[i-2];
    mod = 10000;
    int n;
    while(cin >>n)
    {
        if(n < 40)
        {
            cout<<num[n]<<endl;
            continue;
        }
        double ans;
        ans = -0.5*(log10(5.0))+n*log10((sqrt(5.0)+1.0)/2);
        ans -= (int)ans;
        ans = pow(10, ans);
        while(ans < 1000) ans *= 10;
        printf("%d",(int)ans);
        cout<<"...";
        matrix c;
        memset(c.f, 0, sizeof(c.f));
        c.f[0][0] = 1;
        c.f[0][1] = 1;
        c.f[1][0] = 1;
        matrix d;
        d = pow_mod(c, n-2, 2);
        int sum = 0;
        sum += d.f[0][0] + d.f[0][1];
        sum %= mod;
        if(sum < 10) cout<<"000";
        else if(sum < 100) cout<<"00";
        else if(sum < 1000) cout<<"0";
        cout<<sum<<endl;
    }
    return 0;
}


HDU 3117 Fibonacci Numbers(Fibonacci矩阵加速递推+公式)


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