矩阵快速幂 [POJ 3070 NYOJ 148] Fibonacci

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:

.

纯模板题、- -

#include <iostream>
#include <cstdio>
#include <time.h>
using namespace std;
#define MOD 10000
#define N 2

void mul(int a[N][N],int b[N][N])
{
    int i,j,k;
    int c[N][N]={0};
    for(i=0;i<N;i++)
    {
        for(j=0;j<N;j++)
        {
            for(k=0;k<N;k++)
            {
                c[i][j]=(c[i][j]+a[i][k]*b[k][j])%MOD;
            }
        }
    }
    for(i=0;i<N;i++)
    {
        for(j=0;j<N;j++)
        {
            a[i][j]=c[i][j];
        }
    }
}
int main()
{
    int n;
    while(scanf("%d",&n),n+1)
    {
        int a[N][N]={{0},{1}},b[N][N]={{1,1},{1,0}};     //首项和递推矩阵
        while(n)                                         //二分快速幂
        {
            if(n&1) mul(a,b);
            mul(b,b);
            n>>=1;
        }
        printf("%d\n",a[1][1]);
    }
    return 0;
}