[ACM] POJ 3070 Fibonacci （矩阵幂运算）

Fibonacci

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9517 Accepted: 6767

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

解题思路：

题目中给了提示矩阵。

网赛时才接触矩阵幂运算，自己掌握的知识还是太少了。。。。

代码：

#include <iostream>#include <string.h>#include <stdio.h>#include <algorithm>using namespace std;const int maxn=2;const int mod=10000;int n=2;//矩阵大小int num;//幂大小struct mat{    int arr[maxn][maxn];    mat()    {        memset(arr,0,sizeof(arr));    }};mat mul(mat a,mat b){    mat ans;    for(int i=0;i<n;i++)    {        for(int k=0;k<n;k++)        {            if(a.arr[i][k])                for(int j=0;j<n;j++)            {                ans.arr[i][j]+=a.arr[i][k]*b.arr[k][j];                if(ans.arr[i][j]>=mod)                    ans.arr[i][j]%=mod;            }        }    }    return ans;}mat power(mat p,int k){    if(k==1) return p;    mat e;    for(int i=0;i<n;i++)        e.arr[i][i]=1;    if(k==0) return e;    while(k)    {        if(k&1)            e=mul(e,p);        p=mul(p,p);        k>>=1;    }    return e;}int main(){    mat ori;//题目中的原始矩阵    ori.arr[0][0]=ori.arr[0][1]=ori.arr[1][0]=1;    while(cin>>num&&num!=-1)    {        mat ans;        ans=power(ori,num);        cout<<ans.arr[0][1]<<endl;    }    return 0;}