POJ 3070 Fibonacci

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6065		Accepted: 4246

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

Sample Output

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

//利用二进制的快速运算、好算法噢

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

void js(int (*a)[2],int (*c)[2])
{
    int b[2][2];
    b[0][0]=(a[0][0]*c[0][0]+a[0][1]*c[1][0])%10000;
    b[0][1]=(a[0][0]*c[0][1]+a[0][1]*c[1][1])%10000;
    b[1][0]=(a[1][0]*c[0][0]+a[1][1]*c[1][0])%10000;
    b[1][1]=(a[1][0]*c[0][1]+a[1][1]*c[1][1])%10000;
    a[0][0]=b[0][0];
    a[0][1]=b[0][1];
    a[1][0]=b[1][0];
    a[1][1]=b[1][1];
}
int main()
{
    int a[2][2],b[2][2];
    int n;
    while(scanf("%d",&n),n>=0)
    {
        if(n==0)
          {printf("0\n");
          continue;}
        a[0][0]=1;
        a[0][1]=1;
        a[1][0]=1;
        a[1][1]=0;
        b[0][0]=1;
        b[0][1]=0;
        b[1][0]=0;
        b[1][1]=1;

        for(;n>0;n=n>>1,js(a,a))//短小精悍呀、呵呵
            if(n&1) js(b,a);
        printf("%d\n",b[0][1]);
    }
    return 0;
}