POJ 3070 Fibonacci——————矩阵快速幂解法

**Fibonacci **
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 19481 Accepted: 13467
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

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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 <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;

const int INF = 0x3f3f3f3f;
const int MAXN = 17;
const double PI = acos(-1.0);
const int mod = 10000;
int N;
ll b_n = 0;
ll C[MAXN];
ll h[MAXN];

struct  matrix{	ll m[MAXN][MAXN];};

matrix multi(matrix a,matrix b)//矩阵乘法
{
	matrix tmp;
	for(int i=1;i<=N;i++)
		for(int j=1;j<=N;j++)
		{
			tmp.m[i][j] = 0;
			for(int k=1;k<=N;k++)
				tmp.m[i][j] = (tmp.m[i][j] + a.m[i][k]*b.m[k][j] )%mod;
 		}
 		return tmp;

}

matrix pow_mod(matrix a,int n)//矩阵快速幂
{
	matrix res;
	for(int i=1;i<=N;i++)
		for(int j=1;j<=N;j++)
			res.m[i][j] = (i==j);
	while(n)
	{
		if(n&1)	res = multi(res,a);
		a = multi(a,a);
		n>>=1;
	}
	return res;
}

void init(matrix &res,matrix &H)//初始化
{
	for(int i=1;i<=N;i++)	res.m[1][i]=C[i];//第一列赋值
	res.m[1][N] = b_n;
	for(int i=2;i<=N;i++)//构造单位矩阵
		for(int j=1;j<=N;j++)
			res.m[i][j] = (i==j+1);
	res.m[N][N-1] = 0;
	res.m[N][N] = 1;

	for(int i=1;i<=N;i++)
	{
		H.m[i][1] = h[N-i];
		for(int j=2;j<=N;j++)
			H.m[i][j] = 0;
	}
	H.m[N][1]=1;
}

void slove(int k,int n)
{
	matrix res,H;
	init(res,H);
	res = pow_mod(res,n-k+1);
	res = multi(res,H);
	ll ans = res.m[2][1];
	printf("%lld\n",ans);
}


int main()
{
	int k,n;
	k=2;
	C[1]=1; C[2]=1;
	h[1]=1; h[2]=1;
	while(~scanf("%d",&n)&& -1!=n)
	{
		if(!n)
		{
			printf("0\n");
			continue;
		}
		N = k+1;
		slove(k, n);
	}
	return 0;
}