POJ 3070 Fibonacci

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9769		Accepted: 6959

Description

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 is:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...

An alternative formula for the Fibonacci sequence is

.

Given an integer n, your goal was to compute the last 4 digits of Fn.

Input

The input test file would contain multiple test cases. Each of the test case consists of a 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 is all zeros, print ' 0 '; Otherwise, omit any leading zeros (i.e., print Fn mod 10000).

Sample Input

099999999991000000000-1

Sample Output

0346266875

Hint

As a reminder, matrix multiplication is associative, and the product of the A/2x2 matrices is given by

.

Also, note that raising any 2×2 matrix to the 0th power gives the identity matrix:

.

The idea of solving problems: simple fast matrix power

Quick Matrix Power Templates:

Matrix Add (Matrix A,matrix b) {matrix ans;for (int i=0;i<2;i++) {for (int j=0;j<2;j++) {ans.mat[i][j]=a.mat[i][j]+ B.mat[i][j];if (ans.mat[i][j]>=m) {ans.mat[i][j]%=m;}}} return ans;} Matrix Mul (Matrix A,matrix b) {matrix ans;for (int i=0;i<2;i++) {for (int j=0;j<2;j++) {ans.mat[i][j]=0;for (int k=0; k<2;k++) {ans.mat[i][j]+=a.mat[i][k]*b.mat[k][j];if (ans.mat[i][j]>=m) {ans.mat[i][j]%=m;}}}} return ans;} Matrix Init () {matrix ans;for (int i=0;i<2;i++) {for (int j=0;j<2;j++) {if (i==j) ans.mat[i][j]=1;elseans.mat[i][j] = 0;}} return ans;} Matrix exp (Matrix A,int k) {matrix ans=init (); while (k) {if (k&1) Ans=mul (ans,a); A=mul (a,a); k>>=1;} return ans;}

Reference code:

#include <iostream>using namespace Std;int m=10000;struct matrix{long long mat[2][2];}; Matrix Add (Matrix A,matrix b) {matrix ans;for (int i=0;i<2;i++) {for (int j=0;j<2;j++) {ans.mat[i][j]=a.mat[i][j]+ B.mat[i][j];if (ans.mat[i][j]>=m) {ans.mat[i][j]%=m;}}} return ans;} Matrix Mul (Matrix A,matrix b) {matrix ans;for (int i=0;i<2;i++) {for (int j=0;j<2;j++) {ans.mat[i][j]=0;for (int k=0; k<2;k++) {ans.mat[i][j]+=a.mat[i][k]*b.mat[k][j];if (ans.mat[i][j]>=m) {ans.mat[i][j]%=m;}}}} return ans;} Matrix Init () {matrix ans;for (int i=0;i<2;i++) {for (int j=0;j<2;j++) {if (i==j) ans.mat[i][j]=1;elseans.mat[i][j] = 0;}} return ans;} Matrix exp (Matrix A,int k) {matrix ans=init (); while (k) {if (k&1) Ans=mul (ans,a); A=mul (a,a); k>>=1;} return ans;} int main () {Matrix a;a.mat[0][0]=a.mat[0][1]=a.mat[1][0]=1;a.mat[1][1]=0;int n;while (cin>>n&&n!=-1) { Matrix Ans=exp (a,n); Cout<<ans.mat[0][1]%m<<endl;} return 0;}

POJ 3070 Fibonacci