poj3070 Fibonacci（矩阵乘法初学）

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Fibonacci

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9972 Accepted: 7114

Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, and F_n = F_{n − 1} + F_{n − 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 F_n.

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

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代码：

#include<iostream>#include<cstdio>#include<cstring>using namespace std;#define mod 10000struct mat{   int A[3][3];}s;mat matmul(mat x,mat y)       //x*y  {mat t;    t.A[0][0]=(x.A[0][0]*y.A[0][0]+x.A[0][1]*y.A[1][0])%mod;    t.A[0][1]=(x.A[0][0]*y.A[0][1]+x.A[0][1]*y.A[1][1])%mod;    t.A[1][0]=(x.A[1][0]*y.A[0][0]+x.A[1][1]*y.A[1][0])%mod;    t.A[1][1]=(x.A[1][0]*y.A[0][1]+x.A[1][1]*y.A[1][1])%mod;    return t;}mat matpow(mat ss,int k)      //ss^k{if(k==1) return s;ss=matpow(ss,k/2);if(k%2==0){return matmul(ss,ss);}elsereturn matmul(matmul(ss,ss),s);}int main(){int i,j,n;while(~scanf("%d",&n)&&n!=-1){        if(n==0){printf("0\n");continue;}s.A[0][0]=s.A[0][1]=s.A[1][0]=1;s.A[1][1]=0;        s=matpow(s,n);printf("%d\n",s.A[0][1]);}return 0;}