HDU 5674 Function

Problem Description

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

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

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;
}