【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;}