POJ 3070 Fibonacci (矩阵快速幂)

题目链接：POJ 3070

Problem Description

In the Fibonacci integer sequence, (F_0 = 0), (F_1 = 1), and (F_n = F_{n − 1} + F_{n − 2}) for (n ge 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 le n le 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

Solution

题意

给定正整数 (n)，求斐波那契数列第 (n) 项。

题解

矩阵快速幂

矩阵快速幂模板题。

[ left[ egin{matrix} F_{n + 1} & F_n \ F_n & F_{n - 1} end{matrix} ight] = left[ egin{matrix} 1 & 1 \ 1 & 0 end{matrix} ight] ^ n ]

Code

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
typedef long long ll;
const int maxn = 1e2 + 5;
const ll mod = 1e4;

struct Matrix {
    int n, m;
    ll a[maxn][maxn];
    Matrix(int n = 0, int m = 0) : n(n), m(m) {}
    void input() {
        for(int i = 1; i <= n; ++i) {
            for(int j = 1; j <= m; ++j) {
                scanf("%lld", &a[i][j]);
            }
        }
    }
    void output() {
        for(int i = 1; i <= n; ++i) {
            for(int j = 1; j <= m; ++j) {
                printf("%lld", a[i][j]);
                printf("%s", j == m? "
": " ");
            }
        }
    }
    void init() {
        memset(a, 0, sizeof(a));
    }
    void unit() {
        init();
        for(int i = 1; i <= n; ++i) {
            a[i][i] = 1;
        }
    }
    Matrix operator *(const Matrix b) {
        Matrix c(n, b.m);
        c.init();
        for(int i = 1; i <= c.n; ++i) {
            for(int k = 1; k <= m; ++k) {
                for(int j = 1; j <= c.m; ++j) {
                    c.a[i][j] = (c.a[i][j] + a[i][k] * b.a[k][j]) % mod;
                }
            }
        }
        return c;
    }
    Matrix qmod(ll b) {
        Matrix ans = Matrix(n, n);
        ans.unit();
        if(!b) return ans;
        while(b) {
            if(b & 1) ans = ans * (*this);
            *this = (*this) * (*this);
            b >>= 1;
        }
        return ans;
    }
};

int main() {
    int n;
    while(~scanf("%d", &n)) {
        if(n == -1) break;
        ll ans = 0;
        if(n) {
            Matrix m(2, 2);
            m.a[1][1] = 1;
            m.a[1][2] = 1;
            m.a[2][1] = 1;
            m.a[2][2] = 0;
            m = m.qmod(n - 1);
            ans = m.a[1][1] % mod;
        } 
        printf("%lld
", ans);
    }
    return 0;
}