HDU 5674 Function

Problem Description

There is a function f(n)=(a+b√)n+(a−b√)n. a and b are integers (1≤a,b≤1,000,000).
Maybe the function looks complex but it is actually an integer.The question is to calculate f(xy).The answer can be
very large,so just output the answer mod 1,000,000,007.

 

Input

There are multiple test cases. The first line of input contains an integer T(1≤T≤200) indicating the number of test cases. For each test case:

One line contains four integers a,b  (1≤a,b≤1,000,000),  x(1≤x≤50),y(1≤y≤1018).

 

Output

For each test case, output one integer.

 

Sample Input

3
3 5 1 1
3 5 2 1
1 1 49 99999

 

Sample Output

6
28
160106184
循环节的证明真的是厉害啊，这种东西不知道的话实在是没什么办法。。
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
const int low(int x) { return x&-x; }
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 10;
int T;
LL a, b, x, y, z;

LL mul(LL a, LL b, LL mod) 
{
	LL ret;
	for (ret = 0; b; b >>= 1, (a <<= 1) %= mod)
	{
		if (b & 1) ret = (ret + a) % mod;
	}
	return ret;
}

LL get(LL x, LL y, LL z)
{
	LL ans = 1;
	for (; y; y >>= 1)
	{
		if (y & 1) ans = mul(ans, x, z);
		x = mul(x, x, z);
	}
	return ans;
}

struct martix
{
	LL a[2][2];
}A, B, e;

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

int main()
{
	scanf("%d", &T);
	while (T--)
	{
		cin >> a >> b >> x >> y;
		z = get(x, y, (LL)mod*mod - 1);

		A.a[0][0] = 2;	A.a[0][1] = 2 * a % mod;
		B.a[0][0] = 0;	B.a[0][1] = ((b - a*a) % mod + mod) % mod;
		B.a[1][0] = 1;	B.a[1][1] = 2 * a % mod;
		for (int i = 0; i < 3; i++)
		{
			for (int j = 0; j < 3; j++)
			{
				e.a[i][j] = i == j;
			}
		}
		for (--z; z; z >>= 1)
		{
			if (z & 1) e = e * B;
			B = B * B;
		}
		A = A*e;
		cout << A.a[0][1] << endl;
	}
	return 0;
}