HDU 2855（由递推式构造矩阵+矩阵快速幂）

最新推荐文章于 2020-11-11 23:26:07 发布

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最新推荐文章于 2020-11-11 23:26:07 发布

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分类专栏：数论 HDU 文章标签：递推式构造矩阵矩阵快速幂打表找规律 HDU 2855

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本文介绍了一种基于Fibonacci数列的检查算法——FibonacciCheck-up，该算法用于生成特定编号，并通过矩阵快速幂的方式进行优化计算。

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Fibonacci Check-up

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1356 Accepted Submission(s): 778

Problem Description

Every ALPC has his own alpc-number just like alpc12, alpc55, alpc62 etc.
As more and more fresh man join us. How to number them? And how to avoid their alpc-number conflicted?
Of course, we can number them one by one, but that’s too bored! So ALPCs use another method called Fibonacci Check-up in spite of collision.

First you should multiply all digit of your studying number to get a number n (maybe huge).
Then use Fibonacci Check-up!
Fibonacci sequence is well-known to everyone. People define Fibonacci sequence as follows: F(0) = 0, F(1) = 1. F(n) = F(n-1) + F(n-2), n>=2. It’s easy for us to calculate F(n) mod m.
But in this method we make the problem has more challenge. We calculate the formula

, is the combination number. The answer mod m (the total number of alpc team members) is just your alpc-number.

Input

First line is the testcase T.
Following T lines, each line is two integers n, m ( 0<= n <= 10^9, 1 <= m <= 30000 )

Output

Output the alpc-number.

Sample Input

Sample Output

1
3

分析：

先暴力打表找出规律，得出递推式为：F(n) = 3*F(n-1) - F(n-2)

由于N很大，线性时间会超时，要优化到 log（N)，所以将递推公式变为矩阵，由矩阵快速幂求解即可。

形如： F(n) = a*F(n-1) + b*F(n-2)的式子，可以转化为：

a b

1 0

形式的矩阵。

代码：

#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <iostream>
#include <cstring>

using namespace std;

typedef long long int ll;

/**暴力打表找规律 
int arr[10];
int C(int n,int m){
	int ans = 1;
	for(int i=0; i<m; i++)
		ans *= (n-i);
	for(int i=1; i<=m; i++)
		ans /= i;
	return ans;
}

void make_list(){
	int Fac[10];
	Fac[0] = 0;
	Fac[1] = Fac[2] = 1;
	for(int i=3; i<10; i++)
		Fac[i] = Fac[i-1]+Fac[i-2];
	arr[0] = 0;
	for(int i=1; i<10; i++){
		for(int j=0; j<=i; j++)
			arr[i] += C(i,j)*Fac[j]; 
	}
	for(int i=0; i<10; i++)
		cout << arr[i] << endl;
}  */

int N,M; 
struct matrix{
	int a[2][2];
	matrix operator*(const matrix& tmp){
		matrix ans;
		memset(ans.a,0,sizeof(ans.a));
		for(int i=0; i<2; i++)
			for(int j=0; j<2; j++)
				for(int k=0; k<2; k++)
					ans.a[i][j] = (ans.a[i][j] + a[i][k]*tmp.a[k][j])%M;
		return ans;
	}
};

matrix quick_pow(matrix a,int n){
	matrix ans;
	ans.a[0][0] = ans.a[1][1] = 1;
	ans.a[0][1] = ans.a[1][0] = 0;
	while(n){
		if(n%2)
			ans = ans*a;
		a = a*a;
		n /= 2; 
	}
	return ans;
}

int main(){
	
	//make_list();
	int T;
	scanf("%d",&T);
	matrix ans;
	while(T--){ 
		ans.a[0][0] = 3;
		ans.a[0][1] = -1;
		ans.a[1][0] = 1;
		ans.a[1][1] = 0;
		scanf("%d %d",&N,&M);
		if(N==0){
			printf("0\n");
			continue;
		}
		ans = quick_pow(ans,N-1);
		printf("%d\n",(ans.a[0][0]+M)%M);
	}
	return 0;
}