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 mod
p.

 

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 mod
p.

 

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#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
LL n,a,b,c,p;
int T;
struct Matrix{
LL a[3][3];
}A,B;//A,初始矩阵，B，构造矩阵

Matrix operator*(const Matrix&a,const Matrix&b)//矩阵乘法加费马小定理
{
Matrix 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;
}
Matrix pow(Matrix a,LL x)
{
Matrix c;
for(int i=0; i<3; i++)
{
for(int j=0; j<3; j++)
{
if(i==j)c.a[i][j]=1;
else c.a[i][j]=0;
}
}
while(x)
{
if(x&1) c=c*a;
a=a*a;
x>>=1;
}
return c;
}
//LL get(LL x,LL y)
//{
// LL res=x;
// while(y)
// {
// if(y&1)res=(res*x)%p;
// x=((x%p)*(x%p))%p;
// y>>=1;
// }
// return res;
//}
LL powmod(LL a,LL b,LL c)
{
if(b==0)b=c-1;//没有这句话，会WA.
LL ans=1;
while (b)
{
if (b%2==1) ans=ans*a%c;
b/=2;
a=a*a%c;
}
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){printf("1\n");continue;}
B = pow(B,n-2);
A = A*B;
LL ans = powmod(a,A.a[0][1],p);
cout<<ans<<endl;
}
return 0;
}