Fibonacci

Time Limit: 1000MS	?	Memory Limit: 65536K
Total Submissions: 18084	?	Accepted: 12572

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 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 Fn.

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 Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn mod 10000).

Sample Input

0 9 999999999 1000000000 -1

Sample Output

0 34 626 6875

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

?

?

矩陣快速冪的知識

點擊打開鏈接

?

#include<iostream> using namespace std; typedef long long ll; const int MOD=1e4; #define mod(x) ((x)%MOD) struct mat{ int m[2][2]; }unit; mat operator *(mat a,mat b) {mat ret;ll x;for(int i=0;i<2;i++)for(int j=0;j<2;j++){x=0;for(int k=0;k<2;k++)x+=mod( (ll) (a.m[i][k]*b.m[k][j]) );ret.m[i][j]=mod(x);}return ret; } void init_unit() {for(int i=0;i<2;i++)unit.m[i][i]=1;//單位矩陣，主對角線上元素為1，其余元素為0return; } mat pow_mat(mat a,ll n) {mat ret=unit;while(n){if(n&1) ret=ret*a;a=a*a;n>>=1;}return ret; } int main() {ios::sync_with_stdio(false);ll n;init_unit();while(cin>>n&&n!=-1){mat a;a.m[0][0]=a.m[0][1]=a.m[1][0]=1;a.m[1][1]=0;a=pow_mat(a,n);cout<<a.m[0][1]<<endl;}return 0; }

?

總結

以上是生活随笔為你收集整理的poj3070 Fibonacci 矩阵快速幂的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。