POJ 3070 Fibonacci矩陣快速冪模數模板題

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13728		Accepted: 9724

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

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

原題連結：http://poj.org/problem?id=3070

遞推的式子一般都可以化成矩陣相乘的形式。關鍵的就是構造矩陣，

但這題題目已經構造好了，所以就成了模（shui）板（shu）題了。

AC代碼：

/**    blog:http://blog.csdn.net/hurmishine*/#include <iostream>#include <cstdio>using namespace std;typedef long long LL;const int mod= 10000;LL n;struct Mat{    int m[2][2];};Mat P={    1,1,    1,0};Mat I={    1,0,    0,1};Mat mul (Mat x,Mat y){    Mat z;    for(int i=0; i<2; i++)    {        for(int j=0; j<2; j++)        {            z.m[i][j]=0;            for(int k=0; k<2; k++)            {                z.m[i][j]+=(x.m[i][k]*y.m[k][j])%mod;                z.m[i][j]%=mod;            }        }    }    return z;}Mat quick_Mod(Mat a,LL b){    Mat ans=I;    Mat tmp=a;    while(b>0)    {        if(b&1)            ans=mul(ans,tmp);        tmp=mul(tmp,tmp);        b>>=1;    }    return ans;}int main(){    while(cin>>n)    {        if(n==-1) break;        Mat tmp=quick_Mod(P,n);        cout<<tmp.m[0][1]<<endl;    }    return 0;}

尊重原創，轉載請註明出處：http://blog.csdn.net/hurmishine