poj3070 Fibonacci（矩阵快速幂）

思路：矩阵快速幂的入门题

#include<iostream>#include<cstdio>#include<cstring>using namespace std;struct Mat{int a[2][2];   //矩阵大小};int n;const int mod = 10000;Mat mul(Mat a,Mat b){Mat t;memset(t.a,0,sizeof(t.a));for(int i = 0;i<n;i++){for(int k = 0;k<n;k++){if(a.a[i][k])              for(int j = 0;j<n;j++)  {  t.a[i][j]+=a.a[i][k]*b.a[k][j];  if(t.a[i][j]>=mod)  t.a[i][j]%=mod;  }}}return t;}Mat expo(Mat p,int k){if(k==1)return p;Mat e;memset(e.a,0,sizeof(e.a));for(int i = 0;i<n;i++)   //初始化单位矩阵  e.a[i][i]=1;    if(k==0)return e;while(k){if(k&1)e = mul(p,e);p = mul(p,p);k>>=1;}return e;}int main(){    int k;n=2;while(scanf("%d",&k)!=EOF && k!=-1){        if(k==0)printf("0\n");else{Mat p;p.a[0][0]=p.a[0][1]=p.a[1][0]=1;p.a[1][1]=0;Mat ans = expo(p,k);/*for(int i = 0;i<n;i++,printf("\n"))for(int j = 0;j<n;j++)printf("%d ",ans.a[i][j]);*/printf("%d\n",ans.a[0][1]);}}}

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., printF_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:

.