hdu 5667

Posted 2021-01-20 tingtin

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

 

 

  1 //https://blog.csdn.net/V5ZSQ/article/details/51302603
  2 
  3 #include <iostream>
  4 #include <cstdio>
  5 #include <algorithm>
  6 #include <cstring>
  7 #include <string>
  8 using namespace std;
  9 typedef long long ll;
 10 struct ma{
 11     ll m[3][3];
 12     ma()
 13     {
 14         memset(m,0,sizeof(m));
 15     }
 16 };
 17 ma ma_m(ma a,ma b,ll p)
 18 {
 19     ma c;
 20     for(int i=0;i<3;i++)
 21     {
 22         for(int j=0;j<3;j++)
 23         {            
 24             
 25                 for(int k=0;k<3;k++)
 26                 {
 27                     c.m[i][j] = (c.m[i][j]+(a.m[i][k]*b.m[k][j])%p+p)%p;
 28                 }
 29 
 30         }
 31     }
 32     return c;
 33 }
 34 ma ma_q(ma a,ll k,ll p)
 35 {
 36     ma ret;
 37     for(int i=0;i<3;i++)
 38     {
 39         ret.m[i][i]=1;
 40     }
 41     while(k)
 42     {
 43         if(k&1)
 44         {
 45             ret=ma_m(ret,a,p);
 46         }
 47         a=ma_m(a,a,p);
 48         k>>=1;
 49     }
 50     return ret;
 51 }
 52 ll qu(ll a,ll b,ll p)
 53 {
 54     a%=p;
 55     ll ret=1;
 56     while(b)
 57     {
 58         if(b&1)
 59         {
 60             ret=ret*a%p;
 61         }
 62         a=a*a%p;
 63         b>>=1;
 64     }
 65     return ret%p;
 66 }
 67 int  main()
 68 {
 69      ll t;
 70     ll n,a,b,c,p;
 71     scanf("%lld",&t);
 72     while(t--)
 73     {
 74         scanf("%lld %lld %lld %lld %lld",&n,&a,&b,&c,&p);
 75         if(a%p==0)
 76         {
 77             printf("0
");
 78         }
 79         else  if(n==1)
 80         {
 81             printf("1
");
 82         }
 83         else if(n==2)
 84         {
 85             printf("%lld
",qu(a,b,p));
 86         }
 87         else {
 88             p--;
 89             ma aa;
 90             aa.m[0][0]=c;aa.m[0][1]=1;aa.m[0][2]=b;
 91             aa.m[1][0]=1;aa.m[1][1]=0;aa.m[1][2]=0;
 92             aa.m[2][0]=0;aa.m[2][1]=0;aa.m[2][2]=1;
 93             aa=ma_q(aa,n-2,p);
 94             ll t=aa.m[0][0]*b+aa.m[0][2];
 95             p++;
 96             printf("%lld
",qu(a,t,p));
 97 
 98         }
 99     }
100     return 0;
101 }