51nod - 1691 比大小

有两个数列A和B
已知A_0,a,b,N
A_n=A_(n-1)*a+b (n>=1)
B数列满足
B_n=2*B_(n/2) + 1 (n为偶数)
B_n=2*B_((n-1)/2) + (n+1)/2 (n为奇数)

现在问B数列的第A_N项和第(A_N)+1项的关系

T组数据
A_0,a,b,N<=1e15

T<=100

Input

一个数T，数据组数
每行四个数A_0,a,b,N

Output

一个字符，表示大小关系。

Input示例

1
0 5 5 1

Output示例

=

思路：

先将B数列的值打出来发现每四个以

'<','=','<','>'循环出现。第0个是特例等于第1个。

用矩阵快速幂的方式计算A_N的值。对于第0个需要特判。

#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
 
typedef long long ll;
const int MOD = 4;
ll a0, a, b, n;
int t;
 
struct Matrix
{
	short int val[2][2];
};
 
Matrix multi(const Matrix &left, const Matrix &right)
{
	Matrix result;
	result.val[0][0] = (left.val[0][0] * right.val[0][0] + left.val[0][1] * right.val[1][0]) % MOD;
	result.val[0][1] = (left.val[0][0] * right.val[0][1] + left.val[0][1] * right.val[1][1]) % MOD;
	result.val[1][0] = (left.val[1][0] * right.val[0][0] + left.val[1][1] * right.val[1][0]) % MOD;
	result.val[1][1] = (left.val[1][0] * right.val[0][1] + left.val[1][1] * right.val[1][1]) % MOD;
 
	return result;
}
 
Matrix fast_pow(ll a0, ll n)
{
	Matrix result;
	result.val[0][0] = a0 % MOD;
	result.val[0][1] = 1;
	result.val[1][0] = 0;
	result.val[1][1] = 0;
 
	Matrix base;
	base.val[0][0] = a % MOD;
	base.val[0][1] = 0;
	base.val[1][0] = b % MOD;
	base.val[1][1] = 1;
 
	while (n > 0)
	{
		if (n & 1)
		{
			result = multi(result, base);
		}
 
		n /= 2;
		base = multi(base, base);
	}
 
	return result;
}
 
 
char s[4] = {'<','=','<','>'};
 
int main() 
{
	cin >> t;
	for (int i = 0; i < t; i++)
	{
		cin >> a0 >> a >> b >> n;
		if (b == 0)
		{
			if (a0 == 0 || a == 0)
			{
				cout << "=" << endl;
				continue;
			}
		}
		Matrix result = fast_pow(a0, n);
		int index = result.val[0][0];
		cout << s[index] << endl;
    }
 
	return 0;
}