hdu 5667

Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2887    Accepted Submission(s): 969

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

 

Source

BestCoder Round #80

 

//https://blog.csdn.net/V5ZSQ/article/details/51302603

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
using namespace std;
typedef long long ll;
struct ma{
    ll m[3][3];
    ma()
    {
        memset(m,0,sizeof(m));
    }
};
ma ma_m(ma a,ma b,ll p)
{
    ma c;
    for(int i=0;i<3;i++)
    {
        for(int j=0;j<3;j++)
        {            
            
                for(int k=0;k<3;k++)
                {
                    c.m[i][j] = (c.m[i][j]+(a.m[i][k]*b.m[k][j])%p+p)%p;
                }

        }
    }
    return c;
}
ma ma_q(ma a,ll k,ll p)
{
    ma ret;
    for(int i=0;i<3;i++)
    {
        ret.m[i][i]=1;
    }
    while(k)
    {
        if(k&1)
        {
            ret=ma_m(ret,a,p);
        }
        a=ma_m(a,a,p);
        k>>=1;
    }
    return ret;
}
ll qu(ll a,ll b,ll p)
{
    a%=p;
    ll ret=1;
    while(b)
    {
        if(b&1)
        {
            ret=ret*a%p;
        }
        a=a*a%p;
        b>>=1;
    }
    return ret%p;
}
int  main()
{
     ll t;
    ll n,a,b,c,p;
    scanf("%lld",&t);
    while(t--)
    {
        scanf("%lld %lld %lld %lld %lld",&n,&a,&b,&c,&p);
        if(a%p==0)
        {
            printf("0
");
        }
        else  if(n==1)
        {
            printf("1
");
        }
        else if(n==2)
        {
            printf("%lld
",qu(a,b,p));
        }
        else {
            p--;
            ma aa;
            aa.m[0][0]=c;aa.m[0][1]=1;aa.m[0][2]=b;
            aa.m[1][0]=1;aa.m[1][1]=0;aa.m[1][2]=0;
            aa.m[2][0]=0;aa.m[2][1]=0;aa.m[2][2]=1;
            aa=ma_q(aa,n-2,p);
            ll t=aa.m[0][0]*b+aa.m[0][2];
            p++;
            printf("%lld
",qu(a,t,p));

        }
    }
    return 0;
}