POJ3070——矩阵快速幂——Fibonacci

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

/*
快速幂模板应用
*/
/************************************************
Author        :powatr
Created Time  :2015-8-5 21:06:30
File Name     :b.cpp
************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
const int N = 2;
typedef long long ll;
const int MAX = 50;

ll n;
const int INF = 0x3f3f3f3f;
const int mod = 1e4;

struct Matrix
{
    ll a[2][2];
   void inti()
    {
        a[0][0] = a[0][1] = a[1][0] = 1;
        a[1][1] = 0;
    }
}matrix;

Matrix mul(Matrix a, Matrix b)//矩阵乘法
{
    Matrix ans;
    for(int i = 0 ; i < N; i++)
        for(int j = 0 ; j <N ;j++){
            ans.a[i][j] = 0;
            for(int k = 0; k < N; k++){
                if(a.a[i][k] == 0 || b.a[k][j] == 0) continue;
                ans.a[i][j] += a.a[i][k] * b.a[k][j];
            ans.a[i][j] %= mod;
            }
        }
    return ans;
}

Matrix add(Matrix a, Matrix b) //矩阵加法
{
    int i, j, k;
    Matrix ans;
    for(int i = 0 ; i < N; i++)
        for(int j = 0 ; j < N; j++){
            ans.a[i][j] = a.a[i][j] + b.a[i][j];
            ans.a[i][j] %= mod;
        }
    return ans;
}

Matrix pow(Matrix a, int n) // 矩阵快速幂
{
    Matrix ret;
    ret.a[0][0] = ret.a[1][1] = 1;
    ret.a[0][1] = ret.a[1][0] = 0;
    while(n){
        if(n&1) ret = mul(ret, a);
        n>>= 1;
        a = mul(a, a);
    }
    return ret;
}

Matrix sum(Matrix a, int n)//矩阵幂和
{
    int m;
    Matrix ans, pre;
    if(n == 1) return a;
    m = n >> 1;
    pre = sum(a, m);
    ans = add(pre, mul(pre, pow(a, m)));
    if(n&1) ans = add(ans, pow(a, n));
    return ans;
}
void output(Matrix a) //输出
{
    for(int i = 0 ; i < n; i++)
        for(int j = 0 ; j < n ; j++)
            printf("%d%c",a.a[i][j], j == n -1 ? ‘\n‘ : ‘ ‘);
}

int main()
{
    Matrix ans;
    while(~scanf("%I64d", &n) && ~n){
        ans.inti();
        Matrix d = pow(ans, n);
        printf("%I64d\n", d.a[0][1]);
        }
   return 0;
}