POJ 3070 Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10530		Accepted: 7484

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:

.

 

矩阵乘法优化菲波那切数列，有意思

CODE:

#include <iostream>
#include <cstdio>
#include <cstring>
#define REP(i, s, n) for(int i = s; i <= n; i ++)
#define REP_(i, s, n) for(int i = n; i >= s; i --)
#define MAX_N 3
#define mod 10000
using namespace std;

int a, b, n;
struct node{
    int Mtx[MAX_N][MAX_N];
}tmp;

node operator+(node a,node b){
    node c;
    REP(i, 1, 2) REP(j, 1, 2){
        c.Mtx[i][j] = a.Mtx[i][j] + b.Mtx[i][j];
    }
    return c;
}

node operator*(node a,node b){
    node c;
    REP(i, 1, 2) REP(j, 1, 2){
        c.Mtx[i][j] = 0;
        REP(k, 1, 2) c.Mtx[i][j] = (c.Mtx[i][j] + a.Mtx[i][k] * b.Mtx[k][j]) % mod;
    }
    return c;
} 

node operator^(node a,int k){
    if(k == 0){
        memset(a.Mtx, 0, sizeof(a.Mtx));
        REP(i, 1, 2) a.Mtx[i][i] = 1;
        return a;
    }
    if(k == 1) return a;
    node c = a ^ (k >> 1);
    if(k & 1) return c * c * a;
    return c * c;
}
    
int main(){
    while(scanf("%d", &n) != EOF){
        if(n == -1) break;
        if(n == 0) { cout << 0 << endl; continue; }
        else {
            tmp.Mtx[1][1] = 1; tmp.Mtx[1][2] = 1; tmp.Mtx[2][1] = 1; tmp.Mtx[2][2] = 0;
            tmp = tmp ^ n;                
            cout << tmp.Mtx[1][2] << endl;
        }
    }
    return 0;
}