Fibonacci（矩阵）

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Fibonacci

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

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:

题意：求Fibonacci数列的第n项。

注意n很大，用矩阵，就可以了。

题目链接：http://poj.org/problem?id=3070 转载请注明出处：寻找&星空の孩子

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;#define LL __int64#define mod 10000struct matrix{    LL mat[2][2];};matrix multiply(matrix a,matrix b){    matrix c;    memset(c.mat,0,sizeof(c.mat));    for(int i=0;i<2;i++)    {        for(int j=0;j<2;j++)        {            if(a.mat[i][j]==0)continue;            for(int k=0;k<2;k++)            {                if(b.mat[j][k]==0)continue;                c.mat[i][k]+=a.mat[i][j]*b.mat[j][k]%mod;                c.mat[i][k]%=mod;            }        }    }    return c;}matrix quicklymod(matrix a,LL n){    matrix res;    memset(res.mat,0,sizeof(res.mat));    for(int i=0;i<2;i++) res.mat[i][i]=1;    while(n)    {        if(n&1)            res=multiply(a,res);        a=multiply(a,a);        n>>=1;    }    return res;}int main(){    LL n;    while(scanf("%I64d",&n)!=EOF)    {        if(n==-1)break;        matrix ans;        ans.mat[0][0]=1;        ans.mat[0][1]=1;        ans.mat[1][0]=1;        ans.mat[1][1]=0;        if(n==0)        {            printf("0\n");            continue;        } /*       else if(n==1)        {            printf("1\n");            continue;        }*/        else            ans=quicklymod(ans,n); /*       for(int i=0; i<2; i++)        {            for(int j=0; j<2; j++)                printf("%I64d\t",ans.mat[i][j]);            printf("\n");        }        printf("\n");*/        printf("%I64d\n",ans.mat[1][0]);    }    return 0;}

矩阵入门题！

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