HDU 5667 Sequence

Problem Description

    Holion August will eat every thing he has found.

    Now there are many foods,but he does not want to eat all of them at once,so he find a sequence.

fn=⎧⎩⎨⎪⎪1,ab,abfcn−1fn−2,n=1n=2otherwise

    He gives you 5 numbers n,a,b,c,p,and he will eat  fn foods.But there are only p foods,so you should tell him  fn

Input

    The first line has a number,T,means testcase.

    Each testcase has 5 numbers,including n,a,b,c,p in a line.

    1≤T≤10,1≤n≤1018,1≤a,b,c≤109, p is a prime number,and  p≤109+7.

Output

Output one number for each case,which is 
 
 fn

Sample Input

1
5 3 3 3 233

Sample Output

190

把次数拿下来就可以用矩阵递推了，要注意会有a%p=0的情况

#pragma comment(linker, "/STACK:102400000,102400000")
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<bitset>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
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 = 2e5 + 10;
LL n, a, b, c, p;
int T;

struct martix
{
  LL a[3][3];
}A, B;

martix operator*(const martix&a, const martix&b)
{
  martix c;
  for (int i = 0; i < 3; i++)
  {
    for (int j = 0; j < 3; j++)
    {
      c.a[i][j] = 0;
      for (int k = 0; k < 3; k++)
      {
        (c.a[i][j] += a.a[i][k] * b.a[k][j] % (p - 1)) %= (p - 1);
      }
    }
  }
  return c;
}

martix get(martix a, LL x)
{
  martix c;
  for (int i = 0; i < 3; i++)
  {
    for (int j = 0; j < 3; j++)
    {
      c.a[i][j] = i == j;
    }
  }
  for (x; x; x >>= 1)
  {
    if (x & 1) c = c*a;
    a = a*a;
  }
  return c;
}

LL get(LL x, LL y)
{
  if (n > 1 && y == 0) y = p - 1;
  LL ans = 1;
  for (y; y; y >>= 1)
  {
    if (y & 1) ans = ans*x%p;
    x = x*x%p;
  }
  return ans;
}

int main()
{
  scanf("%d", &T);
  while (T--)
  {
    cin >> n >> a >> b >> c >> p;
    A.a[0][0] = 0;    A.a[0][1] = b;    A.a[0][2] = b;
    B.a[0][0] = 0;    B.a[0][1] = 1;    B.a[0][2] = 0;
    B.a[1][0] = 1;    B.a[1][1] = c;    B.a[1][2] = 0;
    B.a[2][0] = 0;    B.a[2][1] = 1;    B.a[2][2] = 1;
    if (n == 1) { cout << 1 << endl; continue; }
    B = get(B, n - 1);
    A = A*B;
    cout << get(a, A.a[0][0]) << endl;
  }
  return 0;
}