UVa 10229 Modular Fibonacci (矩阵快速幂求斐波那契)

10229 - Modular Fibonacci

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=115&page=show_problem&problem=1170

The Fibonacci numbers (0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...) are defined by the recurrence:

F₀ = 0
F₁ = 1
F_i = F_i-1 + F_i-2 for i>1

Write a program which calculates M_n = F_n mod 2^m for given pair of n and m. 0< =n< =2147483647 and 0<=m< 20. Note that a mod b gives the remainder when a is divided by b.

Input and Output

Input consists of several lines specifying a pair of n and m. Output should be corresponding M_n, one per line.

Sample Input

11 7
11 6

Sample Output

89
25

思路：

如上图，使用矩阵快速幂搞定。

注意：(1<<19)^2会超int，所以要用long long

完整代码：

/*0.015s*/

#include<cstdio>
#include<cstring>
typedef long long ll;

ll mod;

struct mat
{
	ll v[2][2];
	mat()
	{
		memset(v, 0, sizeof(v));
	}
} m1;

mat operator * (mat a, mat b)
{
	mat c;
	for (int i = 0; i < 2; i++)
		for (int j = 0; j < 2; j++)
			for (int k = 0; k < 2; k++)
				c.v[i][j] = (c.v[i][j] + a.v[i][k] * b.v[k][j]) % mod;
	return c;
}

inline mat quickpow(mat a, int k)
{
	mat b;
	b.v[0][0] = b.v[1][1] = 1;
	while (k)
	{
		if (k & 1)
			b = b * a;
		k >>= 1;
		a = a * a;
	}
	return b;
}

int main()
{
	m1.v[0][0] = m1.v[0][1] = m1.v[1][0] = 1, m1.v[1][1] = 0;
	int n, m;
	while (~scanf("%d%d", &n, &m))
	{
		if (n == 0 || m == 0) puts("0");
		else
		{
			mod = 1 << m;
			printf("%d\n", quickpow(m1, n - 1).v[0][0]);
		}
	}
	return 0;
}