POJ 3070 Fibonacci 矩阵快速幂

Fibonacci

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

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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:

.

#include<stdio.h>
#include<string>
#include<cstring>
#include<queue>
#include<algorithm>
#include<functional>
#include<vector>
#include<iomanip>
#include<math.h>
#include<iostream>
#include<sstream>
#include<stack>
#include<set>
#include<bitset>
using namespace std;
const int INF=0x3f3f3f3f;
const int MOD=10000;
const int SIZE=4;
typedef long long ll;
struct Matrix
{
int n;
ll Mat[SIZE][SIZE];
Matrix(int a):n(a)
{
for (int i=0;i<n;i++)
for (int j=0;j<n;j++)
Mat[i][j]=0;
}
Matrix operator * (Matrix& x)
{
Matrix result(n);
for (int k=0;k<n;k++)
for (int i=0;i<n;i++)
for (int j=0;j<n;j++)
result.Mat[i][j]=(result.Mat[i][j]+Mat[i][k]%MOD*x.Mat[k][j]%MOD)%MOD;
return result;
}
Matrix operator ^ (ll x)
{
Matrix temp(n),a(*this);
for (int i=0;i<n;i++)
temp.Mat[i][i]=1;
while (x)
{
if (x&1)
temp=a*temp;
a=a*a;
x>>=1;
}
return temp;
}
};
int main()
{
cin.sync_with_stdio(false);
int n;
Matrix F(2);
F.Mat[0][0]=1,F.Mat[0][1]=1,F.Mat[1][0]=1,F.Mat[1][1]=0;
while (cin>>n&&n!=-1)
{
Matrix Ans=F^n;
cout<<Ans.Mat[0][1]<<endl;
}
return 0;
}