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[][];
     ma()
     {
         memset(m,,sizeof(m));
     }
 };
 ma ma_m(ma a,ma b,ll p)
 {
     ma c;
     for(int i=;i<;i++)
     {
         for(int j=;j<;j++)
         {            

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

         }
     }
     return ;
 }