poj3070Fibonacci【矩阵快速幂】

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Fibonacci

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11438 Accepted: 8133

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

图片来源：http://www.cnblogs.com/xudong-bupt/archive/2013/03/19/2966954.html

因此斐波那契数列可用矩阵表示

#include<cstdio>#include<cstdlib>#include<cstring>using namespace std;const int MOD=10000;int ans[2][2];int a[2][2]={1,1,1,0};void Mulit(int A[2][2],int B[2][2]){int D[2][2]={0};for(int i=0;i<2;++i){for(int k=0;k<2;++k){if(A[i][k]){for(int j=0;j<2;++j){D[i][j]=(D[i][j]+(A[i][k]*B[k][j]))%MOD;}}}}memcpy(A,D,sizeof(A)*4);}void Matrix(int A[2][2],int n){ans[0][0]=ans[1][1]=1;ans[0][1]=ans[1][0]=0;while(n){if(n&1)Mulit(ans,A);Mulit(A,A);n>>=1;}}int main(){int n,i,j,k;while(scanf("%d",&n)&&n!=-1){a[0][0]=a[0][1]=a[1][0]=1;a[1][1]=0;Matrix(a,n);printf("%d\n",ans[0][1]);}return 0;}