POJ 3070-Fibonacci

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 26421		Accepted: 17616

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 F_n. If the last four digits of F_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print F_n 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:

.

SourceStanford Local 2006

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斐波那契数列是这样一个数列：0,1,1,2,3,5…，其数学定义为：

$$\begin{equation}f(n)=\begin{cases} 0, & n=0;\ 1, & n=1;\ f(n-1)+f(n-2), & n>1.\end{cases}\end{equation}$$

最简单的递归解法为：

typedef long long LL;
LL Fib1(int n)
{
    if (n == 0)
        return 0;
    if (n == 1)
        return 1;
    return Fib1(n – 1) + Fib1(n – 2);
}

 这种算法的时间复杂度是 $O(2^n)$。 但是这种解法有很多重复计算，比如算f(8)=f(7)+f(6)时，f(7)重复计算了f(6)。所以我们可以从小开始算起，把f(n)的前两个值保存下来，每次滚动赋值，代码如下：

LL Fib2(int n)
{
    if (n == 0)
        return 0;
    if (n == 1)
        return 1;
    LL fn_2 = 0, fn_1 = 1, ans = 0;
    for (int i = 2; i <= n; ++i) {
        ans = fn_1 + fn_2;
        fn_2 = fn_1;
        fn_1 = ans;
    }
    return ans;
}

这种算法的时间复杂度是 $O(n)$。 最近遇到快速幂相关的题，发现求解斐波那契数列也可以用快速幂来求解，时间复杂度居然能降到 $O(lg(n))$。

题目的图中给出了斐波那契数列的矩阵形式，f(n)为矩阵[1,1;1,0]的n次幂的第一行第二个数。可以用归纳法证明。矩阵的n次幂和数的n次幂都可以用快速幂来求解，时间复杂度能降到 $O(lg(n))$。 代码如下：

const int MAXN = 2;
vector<vector<LL> > multi(const vector<vector<LL> >& mat1, const vector<vector<LL> >& mat2)
{
    vector<vector<LL> > mat(MAXN, vector<LL>(MAXN, 0));
    for (int i = 0; i < MAXN; ++i) {
        for (int j = 0; j < MAXN; ++j) {
            for (int k = 0; k < MAXN; ++k) {
                mat[i][j] += mat1[i][k] * mat2[k][j];
            }
        }
    }
    return mat;
}
vector<vector<LL> > fastPow(vector<vector<LL> >& mat1, int n)
{
    vector<vector<LL> > mat(MAXN, vector<LL>(MAXN, 0));
    for (int i = 0; i < MAXN; ++i)
        mat[i][i] = 1;
    while (n != 0) {
        if (n & 1)
            mat = multi(mat, mat1);
        mat1 = multi(mat1, mat1);
        n >>= 1;
    }
    return mat;
}
LL Fib3(int n)
{
    if (n == 0)
        return 0;
    if (n == 1)
        return 1;
    vector<vector<LL> > mat = { { 1, 1 }, { 1, 0 } };
    vector<vector<LL> > matN = fastPow(mat, n);
    return matN[0][1];
}

POJ这题的代码如下，古老的POJ居然不支持第37行的vector初始化，崩溃。

#include <iostream> #include<cstdio>
#include <vector>
using namespace std;
typedef long long LL;
const int MAXN = 2;
const int MOD = 10000;
vector<vector<LL> > multi(const vector<vector<LL> >& mat1, const vector<vector<LL> >& mat2)
{
    vector<vector<LL> > mat(MAXN, vector<LL>(MAXN, 0));
    for (int i = 0; i < MAXN; ++i) {
        for (int j = 0; j < MAXN; ++j) {
            for (int k = 0; k < MAXN; ++k) {
                mat[i][j] = (mat[i][j] + mat1[i][k] * mat2[k][j]) % MOD;
            }
        }
    }
    return mat;
}
vector<vector<LL> > fastPow(vector<vector<LL> >& mat1, int n)
{
    vector<vector<LL> > mat(MAXN, vector<LL>(MAXN, 0));
    for (int i = 0; i < MAXN; ++i)
        mat[i][i] = 1;
    while (n != 0) {
        if (n & 1)
            mat = multi(mat, mat1);
        mat1 = multi(mat1, mat1);
        n >>= 1;
    }
    return mat;
}
LL Fib3(int n)
{
    if (n == 0)
        return 0;
    if (n == 1)
        return 1;
    //vector<vector<LL>> mat = { { 1,1 },{ 1,0 } }; //———————–
    vector<vector<LL> > mat;
    vector<LL> r1;
    r1.push_back(1), r1.push_back(1);
    vector<LL> r2;
    r2.push_back(1), r2.push_back(0);
    mat.push_back(r1);
    mat.push_back(r2); //———————-
    vector<vector<LL> > matN = fastPow(mat, n);
    return matN[0][1];
}
int main()
{
    int n;
    while (scanf("%d", &n) && n != -1) {
        LL ans = Fib3(n);
        printf("%d\n", ans % MOD);
    }
    return 0;
}

本代码提交AC，用时32MS，真的好快呀。