HDU - 3117 Fibonacci Numbers

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Fibonacci Numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 82 Accepted Submission(s): 47

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

Sample Output

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

Source

IPCP 2005 Northern Preliminary for Northeast North-America

求斐波那契数列的后四十项的前4位跟后4位。前四位可以用通项公式 + log10（科学计数法）来搞定，后4位就可以用构造矩阵 + 矩阵快速幂了。

斐波那契数的通项公式是：f(n)=1/sqrt(5)(((1+sqrt(5))/2)^n+((1-sqrt(5))/2)^n)，对于40个后((1-sqrt(5))/2)^n可以忽略不计。

构造矩阵 | f[n] | * | 1 1| = | f[n-1] | ===== > | f[n] | * | 1 1 | ^ (n-2) = | f[2] |

| f[n-1]| | 1 0| | f[n-2] | | f[n-1] | | 1 0 | | f[1] |

#include <iostream>
#include <cmath>
#include <string>
#include <algorithm>
#include <string.h>
#include <map>
#include <typeinfo>
using namespace std;
const int mod = 10000;
int width;
struct Matrix {
    int a[35][35];
    void init() {
        memset(this->a, 0, sizeof(this->a));
    }
};

Matrix operator * (const Matrix& a, const Matrix& b) {
    Matrix c;
    c.init();
    for (int i = 0; i < width; ++i) {
        for (int j = 0; j < width; ++j) {
            for (int k = 0; k <width; ++k) {
                c.a[i][j] = (c.a[i][j] + a.a[i][k] * b.a[k][j]) % mod;
            }
        }
    }
    return c;
}

Matrix operator ^ (Matrix& a, int k) {
    Matrix c;
    for (int i = 0; i < width; ++i)
        for (int j = 0; j < width; ++j) {
            if (i == j)
                c.a[i][j] = 1;
            else
                c.a[i][j] = 0;
        }
    while (k) {
        if (k&1)
            c = a * c;
        k >>= 1;
        a = a * a;
    }
    return c;
}

Matrix operator + (const Matrix& a, const Matrix& b) {
    Matrix c;
    c.init();
    for (int i = 0; i < width; ++i)
        for (int j = 0; j < width; ++j)
            c.a[i][j] = (a.a[i][j] + b.a[i][j]) % mod;
    return c;
}

int main() {
    width = 2;
    int fib[45];
    fib[0] = 0;
    fib[1] = fib[2] = 1;
    for (int i = 3; i < 40; ++i)
        fib[i] = fib[i-1] + fib[i-2];
    int n;
    Matrix a;
    while (cin >> n) {
        if (n < 40) {
            cout << fib[n] << endl;
            continue;
        }
        double k = log10(1.0/sqrt(5.0)) + static_cast<double>(n * log10((1.0 + sqrt(5.0))/2.0));
        double num = k;
        num = k - static_cast<int>(num);
        a.init();
        a.a[0][0] = a.a[1][0] = a.a[0][1] = 1;
        
        Matrix temp = a ^ n;
        printf("%d...%0.4d\n",(int)(1000.0 * pow(10.0, num)), temp.a[0][1]%mod);

	//  注意补0； 
    }
}