Fibonacci(矩阵快速幂)

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 24331		Accepted: 16339

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

解题思路：因为n很大所以不能用递推来算;

a[2]  a[1]  a[0]   ------> a[3]    a[2]     a[1]

  1       1     0 　　　  　2         1         1               

构造一个矩阵 1行3列的矩阵*（莫一个矩阵）=1行3列的矩阵  ----->   矩阵大小为3行3列

递推式=  a[n]=a[n-1]]+a[n-2];                      a[n+1]=a[n]+a[n-1];

后一个由第一个怎么变换到达  -----> 构造的矩阵为

1 1 0

1 0 1

0 0 0

#include <iostream>
#include <cstring>
#include <algorithm>
#define ll long long
#define mod(x) ((x)%MOD)
using namespace std;

ll n;
const int maxn=3;
const int MOD=10000;

struct mat{
    int m[maxn][maxn];
}unit;

mat operator *(mat a,mat b){
    mat ret;
    ll x;
    for(int i=0;i<maxn;i++)
    for(int j=0;j<maxn;j++){
        x=0;
        for(int k=0;k<maxn;k++){
            x+=mod(1LL*a.m[i][k]*b.m[k][j]);
        }
        ret.m[i][j]=mod(x);
    }
    return ret;
}

void init(){
    for(int i=0;i<maxn;i++)
        unit.m[i][i]=1;
    return;
}

mat pow_mat(mat a,ll m){
    mat ret=unit;
    while(m){
        if(m&1) ret=ret*a;
        a=a*a;
        m=m/2;
    }
    return ret;
}

int main(){
    ios::sync_with_stdio(false);
    init();
    while(cin>>n&&n!=-1){
        if(n==0) cout << 0 << endl;
        else if(n==1) cout << 1 << endl;
        else if(n==2) cout << 1 << endl;
        else{
            mat a,b;
            a.m[0][0]=1,a.m[0][1]=1,a.m[0][2]=0;

            b.m[0][0]=1,b.m[0][1]=1,b.m[0][2]=0;
            b.m[1][0]=1,b.m[1][1]=0,b.m[1][2]=1;
            b.m[2][0]=0,b.m[2][1]=0,b.m[2][2]=0;

            b=pow_mat(b,n-2);
            a=a*b;
            cout << mod(a.m[0][0]) << endl;
        }
    }
    return 0;
}