POJ 3070 Fibonacci

Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, andF_n = F_{n − 1} + F_{n − 2} forn ≥ 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 ofF_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., printF_n mod 10000).

Sample Input

099999999991000000000-1

Sample Output

0346266875

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]);    }}