POJ_3070_Fibonacci

Fibonacci

Time Limit: 1000MS
Memory Limit: 65536KTotal Submissions: 15397
Accepted: 10803

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

• 普通的常系数线性齐次递推
• 建立矩阵快速幂加速

#include <iostream>#include <string>#include <cstdio>#include <cstring>#include <algorithm>#include <climits>#include <cmath>#include <vector>#include <queue>#include <stack>#include <set>#include <map>using namespace std;typedef long long           LL ;typedef unsigned long long ULL ;const int    maxn = 1000 + 10  ;const int    inf  = 0x3f3f3f3f ;const int    npos = -1         ;const int    mod  = 1e4        ;const int    mxx  = 100 + 5    ;const double eps  = 1e-6       ;struct Matrix{int n;int a[mxx][mxx];void reset(int x){n=x;memset(a,0,sizeof(a));}Matrix operator * (const Matrix X){Matrix Y;Y.reset(n);for(int k=0;k<n;k++)for(int i=0;i<n;i++)if(a[i][k])for(int j=0;j<n;j++)if(X.a[k][j])Y.a[i][j]=((Y.a[i][j] + (a[i][k]*X.a[k][j])%mod)%mod + 2*mod)%mod;return Y;}};Matrix pow_mod(Matrix X, int k){int n=X.n;Matrix Y;Y.reset(X.n);for(int i=0;i<n;i++)Y.a[i][i]=1;while(k){if(1&k)Y=Y*X;X=X*X;k>>=1;}return Y;}int cal(int k){if(k<2){return k;}else{Matrix Y;Y.reset(2);Y.a[0][0]=1; Y.a[0][1]=1;Y.a[1][0]=1; Y.a[1][1]=0;Y=pow_mod(Y,k-1);return Y.a[0][0];}}int n, ans, f[2]={0,1};int main(){// freopen("in.txt","r",stdin);// freopen("out.txt","w",stdout);while(~scanf("%d",&n) && -1!=n){printf("%d\n",cal(n));}return 0;}