矩阵快速幂求斐波那契数列第n项_弱的奋斗之旅

求斐波那契数列第 n 项

#include<cstdio>
#include<iostream>
#include<vector>

using namespace std;

typedef vector<int> vec;
typedef vector<vec> mat;

const int mod = 10000;

mat mul(mat A, mat B)
{
	mat C(A.size(), vec(B[0].size()));

	for(int i= 0; i< A.size(); i++)
		for(int k= 0; k< B.size(); k++)
			for(int j= 0; j< B[0].size(); j++)
				C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % mod;

	return C;
}

mat pow(mat A, int n)
{
	mat B(A.size(), vec(A.size()));
	for(int i= 0; i< A.size(); i++)
		B[i][i] = 1;

	while(n > 0)
	{
		if(n & 1) B = mul(B, A);
		A = mul(A, A);
		n >>= 1;
	}

	return B;
}

int main()
{
	int n;
	while(scanf("%d", &n) != EOF)
	{
		mat A(2, vec(2));
		A[0][0] = A[0][1] = A[1][0] = 1;
		A[1][1] = 0;

		A = pow(A, n);

		printf("%d\n", A[1][0]);
	}

	return 0;
}
#include<cstd