【poj3070】Fibonacci（矩阵）

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<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
struct Matrix
{
    long long a[5][5];
    int h,w;
}pr,ne;
long long n;
void init()
{
    pr.a[2][1]=pr.a[1][2]=0;//初始化单位矩阵（a[i][i]值为1，其他值为0）
    pr.a[1][1]=pr.a[2][2]=1;//任何矩阵乘以单位矩阵，其值不变 
    pr.h =2,pr.w =2;
    ne.a[1][1]=ne.a[1][2]=ne.a[2][1]=1;
    ne.a[2][2]=0;//初始化初始矩阵（用来进行n次幂的矩阵） 
    ne.h =2,ne.w =2;
}
Matrix Matrix_multiply(Matrix x,Matrix y)//矩阵乘法 
{
    Matrix t;
    memset(t.a,0,sizeof(t.a));
    t.h =x.h ;//新矩阵的长宽 
    t.w =y.w ;
    for(int i=1;i<=x.h ;i++)
    {
        for(int j=1;j<=y.h;j++)//y.h==x.w
        {
            if(x.a[i][j]==0)
            continue;
            for(int l=1;l<=y.w ;l++)
            {
                t.a[i][l]=(t.a[i][l]+x.a[i][j]*y.a[j][l]%10000)%10000;
            }
        }
    }
    return t;

 } 
void Matrix_mod(int n)
{
    while(n)
    {
        if(n&1) pr=Matrix_multiply(ne,pr);
        ne=Matrix_multiply(ne,ne);
        n>>=1;
    }
    /*
    主要是初始矩阵的n次幂，若n为奇数，则n&1值为1，
    就要pr自己算一次，若为偶数则pr就要进行n次幂了 
    */
}
int main()
{
    while(~scanf("%lld",&n)&&n!=-1)
    {
        init();//初始化 
        Matrix_mod(n); //快速幂 
        printf("%lld\n",pr.a[1][2]%10000);
    }
    return 0;
}