HDU 5667 矩阵快速幂关于指数的递推

Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1498    Accepted Submission(s): 505


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

 
根据题意写成递推式 ，然后可以看出

指数有递推式，可以通过矩阵快速幂来求解。再用下面这公式快速幂取模即可。


（C是素数）

C是素数所以   x只需要对(C-1) 取模就行了

对于指数 ， 求可以得到递推式

f1=0;

f2=b;

f3=c*b;

f(n)=c*f(n-1)+f(n-2)+b;

/*
| c 1 b|    | Fn-1|         | Fn  |
| 1 0 0| *  | Fn-2| -》  | Fn-1|
| 0 0 1|    | 1      |         | 1   |

*/

#include<bits/stdc++.h>
#define LL long long
#define INF 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof(a))
#define bug puts("***********")
using namespace std;
const int N =510;
LL mod,eul;
struct Mat{
LL mat[3][3];
}sta,fi;
Mat mul(Mat a,Mat b){
Mat c;
mem(c.mat,0);
for(int i=0;i<3;i++){
for(int j=0;j<3;j++){
for(int k=0;k<3;k++){
c.mat[i][j]=(c.mat[i][j]+a.mat[i][k]*b.mat[k][j]%(mod-1))%(mod-1);
}
}
}
return c;
}
Mat quick(Mat a,LL k){
Mat c;
mem(c.mat,0);
for(int i=0;i<3;i++)
c.mat[i][i]=1;
while(k){
if(k&1) c=mul(c,a);
a=mul(a,a);
k>>=1;
}
return c;
}
LL quick(LL a,LL k){
LL ans=1;
while(k){
if(k&1)ans=(ans*a)%mod;
a=(a*a)%mod;
k>>=1;
}
return ans%mod;
}
int main(){
int t;
// cout<<(LL)pow(201,3)*27*27%233<<endl;
// cout<<(LL)pow(220,3)*27*201%233<<endl;
scanf("%d",&t);
LL n,a,b,c;

while(t--){
scanf("%lld%lld%lld%lld%lld",&n,&a,&b,&c,&mod);
mem(sta.mat,0);
mem(fi.mat,0);
LL temp=quick(a,b);
fi.mat[2][0]=1;
fi.mat[1][0]=temp;
fi.mat[0][0]=temp*quick(temp,c)%mod;

sta.mat[0][0]=c;
sta.mat[0][1]=1;
sta.mat[0][2]=b;
sta.mat[1][0]=sta.mat[2][2]=1;
if(n<=3){
printf("%lld\n",fi.mat[3-n][0]);
continue;
}
// puts("***");

sta=quick(sta,n-2);
LL ans=sta.mat[0][0]*b+sta.mat[0][2];
//cout<<fi.mat[0][0]<<endl;
ans=quick(a,ans);
printf("%lld\n",ans);
}
return 0;
}