POJ 3070 Fibonacci（简单矩阵快速幂）

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题目链接：可以看原题目http://poj.org/problem?id=3070

Fibonacci
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 12736 Accepted: 9063
Description

In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 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 Fn.

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 Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn 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:

Source

很水很水的一道题目，直接敲出来就好了。
下面是AC代码：

#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;struct Mat{    int a[2][2];    void init()    {        memset(a,0,sizeof(a));        a[0][0]=1;        a[1][1]=1;    }};Mat mul(Mat a,Mat b){    Mat ans;    ans.init();    for(int i=0;i<2;i++)    {        for(int j=0;j<2;j++)        {            ans.a[i][j]=0;            for(int k=0;k<2;k++)            {                ans.a[i][j]+=a.a[i][k]*b.a[k][j];            }            ans.a[i][j]%=10000;        }    }    return ans;}Mat power(Mat a,int num){    Mat ans;    ans.init();    while(num)    {        if(num&1)        {            ans=mul(ans,a);        }        num/=2;        a=mul(a,a);    }    return ans;}int main(){    int n;    while(~scanf("%d",&n))    {        if(n==-1)        {            break;        }        if(n==0)        {            printf("0\n");            continue;        }        Mat a,ans;        a.a[0][0]=1,a.a[0][1]=1,a.a[1][0]=1,a.a[1][1]=0;        ans=power(a,n);        printf("%d\n",ans.a[1][0]%10000);    }}

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