POJ 3070 矩阵快速幂水题

Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 18002		Accepted: 12515

Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, and F_n = F_{n − 1} + F_{n − 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 F_n.

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 F_n. If the last four digits of F_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print F_n 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:

这道题用的快速幂，需要求矩阵乘法最高有1,000,000次。用快速幂时间复杂度log2N。矩阵乘法的使用需要对矩阵有一定的认识和对线性代数本质的理解。

#include<stdio.h>
#include<string.h>
long long n;
const int N=2;
int t[N][N];

void matrixm(int a[][N],int b[][N]){
	int i,j,k,temp[N][N];

	memset(temp,0,sizeof(temp));
	
	for(i=0;i<N;i++)
		for(j=0;j<N;j++)
			for(k=0;k<N;k++)
				temp[i][j]+=a[i][k]*b[k][j];
				
	for(i=0;i<N;i++)
		for(j=0;j<N;j++)
			b[i][j]=temp[i][j]%10000;
}

int main(){
	int a[N][N],i,j;
	
	while(scanf("%lld",&n)){
		if(n==-1)
			break;
		t[0][0]=1;      //t是行列式为1的矩阵
		t[0][1]=0;
		t[1][0]=0;
		t[1][1]=1;
		
		a[0][0]=1;      //a是标准模式矩阵
		a[0][1]=1;
		a[1][0]=1;
		a[1][1]=0;

		while(n){
			if(n&1==1)
				matrixm(a,t);
			matrixm(a,a);
			n/=2;
		}
		printf("%d\n",t[0][1]);
	}
	
	return 0;
}

这一道题中，可以注意一下const int N。 N在c语言中是不可以用来定义数组长度的比如a[N][N]，但是c++中是允许的。