poj 3070

http://poj.org/problem?id=3070

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

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:

.

题目大意：求Fibonacci数列的项

解题思路：矩阵连乘

#include <stdio.h>#include <iostream>#include <string.h>#include <math.h>using namespace std;typedef long long LL;const int N=2;const LL MOD=10000;struct Matrix{    LL m[N][N];};Matrix I={    1,0,    0,1,};Matrix multi(Matrix a,Matrix b){    Matrix c;    for(int i=0; i<N; i++)        for(int j=0; j<N; j++)        {            c.m[i][j]=0;            for(int k=0; k<N; k++)            {                c.m[i][j]+=a.m[i][k]*b.m[k][j]%MOD;            }            c.m[i][j]=c.m[i][j]%MOD;        }    return c;}Matrix quick_mod(Matrix a,LL k){    Matrix ans=I;    while(k!=0)    {        if(k&1)        {            ans=multi(ans,a);        }        k>>=1;        a=multi(a,a);    }    return ans;}int main(){    LL n;    while(~scanf("%lld",&n))    {        if(n==-1)            break;        Matrix A= {1,1,                   1,0                  };        Matrix x1=quick_mod(A,n);        printf("%lld\n",x1.m[0][1]);    }    return 0;}