POJ 3070 Fibonacci 矩阵快速幂模板

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18607		Accepted: 12920

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:

.

Source

Stanford Local 2006

 

题意：求斐波拉契数列，只不过n值很大要用到矩阵快速幂

分析：这是矩阵快速幂的入门题，借此题写下模板。

首先我们很容易的得到递推式：f(n) = f(n-1)+f(n-2)

也很容易的得到他们的矩阵式：

| f(n-1)  f(n-2)  |   x  | 1  1 |   =  | f(n)  f(n-1) |

|    0         0     |       | 1  0 |       |   0       0    |

          a                      b                  c

写下简单的推导过程：首先把右边式子写在矩阵a第一行，把右边式子可能得到的结果写在矩阵c的第一行，a和c剩下的每行都为0，接下来根据矩阵a和矩阵c写出矩阵b。

得到矩阵式后，就是简单的套用矩阵快速幂的模板了，下面是我的代码

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 1e4 + 10;
const int mod = 10000;
typedef long long ll;
struct matrix {
    ll a[10][10];
};
matrix base, ans;
matrix multip( matrix x, matrix y ) { //求c矩阵的过程
    matrix tmp;
    for( ll i = 0; i < 2; i ++ ) {
        for( ll j = 0; j < 2; j ++ ) {
            tmp.a[i][j] = 0;
            for( ll k = 0; k < 2; k ++ ) {
                tmp.a[i][j] = ( tmp.a[i][j] + x.a[i][k] * y.a[k][j] ) % mod;
            }
        }
    }
    return tmp;
}
ll qow( ll a, ll b ) { //求数的快速幂，与此题无关
    ll sum = 1;
    while( b ) {
        if( b&1 ) {
            sum = sum*a%mod;
        }
        a = a*a%mod;
        b /= 2;
    }
    return sum;
}
ll f( ll x ) {  //矩阵快速幂的运算
    while( x ) {
        if( x&1 ) {
            ans = multip( ans, base );
        }
        base = multip( base, base );
        x /= 2;
    }
    return ans.a[0][0];
}
int main() {
    ll n;
    while( cin >> n ) {
        if( n == -1 ) {
            break;
        }
        memset( ans.a, 0, sizeof(ans.a) ); //初始化a矩阵和b矩阵，根据你所得到的矩阵式初始化
        memset( base.a, 0, sizeof(base.a) );
        ans.a[0][0] = 1, ans.a[0][1] = 0;
        base.a[0][0] = base.a[0][1] = base.a[1][0] = 1;
        if( n == 0 ) {
            cout << 0 << endl;
        } else if( n == 1 ) {
            cout << 1 << endl;
        } else {
            cout << f(n-1) << endl;
        }
    }
    return 0;
}