hdu 2256 Problem of Precision

(5 + sqrt(6))^n = an + bn * sqrt(6)

關鍵就是推出公式

ans = 2 * an - 1

#include<stdio.h>
#include<string.h>
#define MOD 1024
struct mat{
	int m[2][2];
};
mat x,y;
int n;

void init()
{
	memset(y.m,0,sizeof(y.m));
	for(int i = 0;i < 2;i++)
		y.m[i][i] = 1;
}

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

mat Pow(mat a,mat b,int k)
{
	while(k)
	{
		if(k&1)
			b = a*b;
		a = a * a;
		k >>= 1;
	}
	return b;
}

int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
	    memset(x.m,0,sizeof(x.m));
	    init();
		scanf("%d",&n);
		x.m[0][0] = 5;
		x.m[0][1] = 12;
		x.m[1][0] = 2;
		x.m[1][1] = 5;
		mat ret = Pow(x,y,n-1);
		//printf("%d %d\n%d %d\n",ret.m[0][0],ret.m[0][1],ret.m[1][0],ret.m[1][1]);
		int a = ret.m[0][0]*5 + ret.m[0][1]*2;
		int ans = 2 * a - 1;
		printf("%d\n",ans%MOD);
	}
	return 0;
}