POJ-3070 Fibonacci(矩阵快速幂)

Fibonacci

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 14992 Accepted: 10542

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:

.

Source

Stanford Local 2006

矩阵快速幂入门题，连构造都不需要构造。

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;const int mod =1e4;const int l=2;struct matrix{    int mat[l][l];    matrix friend operator * (matrix a,matrix b){        matrix tp;        memset(tp.mat,0,sizeof(tp.mat));        for(int i=0;i<l;i++){            for(int j=0;j<l;j++){                if(!a.mat[i][j]){                    continue;                }                for(int k=0;k<l;k++){                    tp.mat[i][k]+=(a.mat[i][j]*b.mat[j][k])%mod;                    tp.mat[i][k]%=mod;                }            }        }        return tp;    }};int work(int n){    matrix base={1,1,1,0},ans={0};    ans.mat[0][0]=1;ans.mat[0][1]=0;    ans.mat[1][0]=0;ans.mat[1][1]=-1;    while(n){        if(n&1){            ans=ans*base;        }        base=base*base;        n>>=1;    }    return ans.mat[0][1];}int main(){    int n;    while(~scanf("%d",&n)&&n!=-1)    {        printf("%d\n",work(n));    }    return 0;}