POJ 3070 斐波那契数列（矩阵快速幂（版题））

链接：http://poj.org/problem?id=3070

• Fibonacci
  Time Limit: 1000MS Memory Limit: 65536K
  Total Submissions: 16351 Accepted: 11470

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:

.

.

#include <iostream>#include <string.h>using namespace std;const int maxn =15;const int mod =10000;struct matrix{    long long int a[2][2];};matrix multply(matrix x,matrix y){    int i,j,k;    matrix result;    memset(result.a,0,sizeof(result.a));//**    for (k=0; k<2; k++)        for (i=0; i<2; i++)        {            if(x.a[i][k])                for (j=0; j<2; j++)                {                    if (y.a[k][j])                    {                        result.a[i][j]+=x.a[i][k]*y.a[k][j];                        if (result.a[i][j]>=mod)                            result.a[i][j]%=mod;                    }                }        }    return result;}void  matrix_pow(matrix a,int nn){    matrix result;    memset(result.a,0,sizeof(result.a));    int i,j,k;    for (i=0; i<2; i++)        result.a[i][i]=1;    while (nn)    {        if (nn&1)            result=multply(result,a);        nn>>=1;        a=multply(a,a);//        int i,j;//        for (i=0; i<2; i++)//            for (j=0; j<2; j++)//                if(!j)//                    cout<<a.a[i][j]<<" ";//                else//                    cout <<a.a[i][j]<<endl;    }    cout<<result.a[0][1]<<endl;}int main (){    long long int nn;    while (cin>>nn)    {        if (nn==-1)            break;        matrix a;        a.a[0][0]=a.a[0][1]=a.a[1][0]=1;        a.a[1][1]=0;//        int i,j;//        for (i=0; i<2; i++)//            for (j=0; j<2; j++)//                if(!j)//                    cout<<a.a[i][j]<<" ";//                else//                    cout <<a.a[i][j]<<endl;        matrix_pow(a,nn);    }    return 0;}