洛谷 P1306 斐波那契公约数

题意简述

求斐波那契数列第n项和第m项的最大公约数

题解思路

设斐波那契数列第x项为F[x]
则有结论(Gcd(F[n], F[m]) = F[Gcd(n, m)])

证明：
不妨设n < m
则(F[m])
(= F[m-1] +F[m-2])
(= 2*F[m-2] + F[m-3])
(= 3*F[m-3] + 2*F[m-4])
(= ......)
(= F[x+1]*F[m-x] + F[x]*F[m-x-1])
(= F[m-n+1]*F[n] + F[m-n]*F[n-1])
所以
(Gcd(F[n], F[m]))
(= Gcd(F[n], F[m-n+1]*F[n] + F[m-n]*F[n-1]))
(= Gcd(F[n], F[m-n]*F[n-1]))
又因为(Gcd(F[n], F[n-1]) = Gcd(F[n-2], F[n-1]) = ...... = Gcd(F[1], F[2]) = 1)
所以
(Gcd(F[n], F[m]))
(= Gcd(F[n], F[m-n]))
同理可得
(Gcd(F[n], F[m]))
(= Gcd(F[n], F[m-n]))
(= Gcd(F[n], F[m-2*n]))
(= Gcd(F[n], F[m mod n]))
然后可以发现，这个过程实际上就是用更相减损术求(Gcd(m, n))
所以(Gcd(F[n], F[m]) = Gcd(F[Gcd(n, m)], F[Gcd(n, m)]) = F[Gcd(n, m)])

最后用矩阵求出斐波那契数列第(Gcd(m, n))项即可

代码

#include <cstdio>
typedef long long ll;
const int mod = 100000000;
int n, m, N, t;
struct Matrix
{
	int a[4][4];
	Matrix& operator =(const Matrix& x)
	{
		for (register int i = 1; i <= N; ++i)
			for (register int j = 1; j <= N; ++j)
				a[i][j] = x.a[i][j];
		return *this;
	}
};
Matrix a, b, c;
int _gcd(int x, int y, int z = 0)
{
	while ((z = x % y)) x = y, y = z;
	return y;
}
Matrix Mul(const Matrix& x, const Matrix& y)
{
	Matrix s;
	for (register int i = 1; i <= N; ++i)
		for (register int j = 1; j <= N; ++j)
			s.a[i][j] = 0;
	for (register int i = 1; i <= N; ++i)
		for (register int j = 1; j <= N; ++j)
			for (register int k = 1; k <= N; ++k)
				s.a[i][j] = (s.a[i][j] + (ll)x.a[i][k] * y.a[k][j] % mod) % mod;
	return s;
}
Matrix _pow(Matrix x, ll y)
{
	Matrix s;
	for (register int i = 1; i <= N; ++i)
		for (register int j = 1; j <= N; ++j)
			s.a[i][j] = (i == j);
	for (; y; y >>= 1, x = Mul(x, x)) if (y & 1) s = Mul(s, x);
	return s;
}
int main()
{
	N = 3;
	scanf("%d%d", &n, &m);
	c.a[1][1] = c.a[1][2] = 1;
	a.a[1][1] = a.a[1][2] = a.a[2][1] = 1;
	n = _gcd(n, m);
	if (n <= 2) {printf("1
"); return 0; }
	b = Mul(c, _pow(a, n - 2));
	printf("%d
", b.a[1][1]);
}