POJ 3070 Fibonacci(矩阵快速幂)

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13147		Accepted: 9348

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:


.

题意：求解斐波那契数列第n项对1e4取模的结果。

思路：运用矩阵快速幂即可
<pre name="code" class="cpp">#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<cctype>
using namespace std;
typedef long long ll;
const int M = 1e4;

struct Mar
{
int a[2][2];
Mar(int ha = 0) {
memset(a, 0, sizeof a);
}
void init() {
a[0][0] = a[0][1] = a[1][0] = 1;
a[1][1] = 0;
}
void dan() { //矩阵单位化
memset(a, 0, sizeof a);
a[0][0] = a[1][1] = 1;
}
Mar& operator = (const Mar &Rhs);
friend Mar operator * (const Mar &A, const Mar &B);
};

Mar& Mar::operator = (const Mar &Rhs)
{
for (int i = 0; i < 2; i++)
for (int j = 0; j < 2; j++)
a[i][j] = Rhs.a[i][j];
return *this;
}

Mar operator * (const Mar &A, const Mar &B)
{
Mar C;
for (int i = 0; i < 2; i++)
for (int j = 0; j < 2; j++)
for (int k = 0; k < 2; k++)
C.a[i][j] = (C.a[i][j] + A.a[i][k] * B.a[k][j] % M) % M;
return C;
}

Mar my_pow(Mar& A, ll n)
{
Mar Ans;
if (n == 0) {
Ans.dan();
return Ans;
}
Ans = my_pow(A, n/2);
Ans = Ans * Ans;
if (n % 2) Ans = Ans * A;
return Ans;
}

void solve(ll n)
{
Mar A;
A.init();
Mar Ans = my_pow(A
a00d
,n);
cout << Ans.a[1][0] << endl;
}

int main()
{
ll n;
while(cin >> n && n+1) {
solve(n);
}
}