| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 17171 | Accepted: 11999 |
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
#include <iostream> #include <cstddef> #include <cstring> #include <vector> using namespace std; typedef long long ll; const int mod=10000; typedef vector<ll> vec; typedef vector <vec> mat; mat mul(mat &a,mat &b) { mat c(a.size(),vec(b[0].size())); for(int i=0;i<2;i++) for(int j=0;j<2;j++) for(int k=0;k<2;k++){ c[i][j]+=a[i][k]*b[k][j]; c[i][j]%=mod; } return c; } mat Pow(mat a,ll n) { mat res(a.size(),vec(a.size())); for(int i=0;i<a.size();i++) res[i][i]=1; while(n) { if(n&1) res=mul(res,a); a=mul(a,a); n/=2; } return res; } ll solve(ll n) { mat a(2,vec(2)); a[0][0]=1; a[0][1]=1; a[1][0]=1; a[1][1]=0; a=Pow(a,n); return a[0][1]; } int main() { ll n; while(cin>>n&&n!=-1) { cout<<solve(n)<<endl; } }
