POJ 3070 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:

.

题意：要求f[n]%10000

分析：此题可以用斐波拉契数列及取余的性质 找到循环节m则f[n]%10000=f[n%m]%10000;

#include<stdio.h>
#include<string.h>
#define mod 10000
#include<vector>
using namespace std;
int main()
{
    vector<int>fib;
    fib.push_back(0);
    fib.push_back(1);
    int i;
    for(i=2;;i++)
    {
        fib.push_back((fib[i-1]+fib[i-2])%10000);
        if(fib[i]==1&&fib[i-1]==0)
            break;
    }
    int cnt=i-1;
    __int64 a;
    while(scanf("%I64d",&a)&&a!=-1)
    {
        printf("%d\n",fib[a%cnt]);
    }
}