hdu 5667 Sequence(矩阵快速幂+费马小定理+快速幂)

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.

题意：

问f(n)对p取模的结果;

思路：

(ab)p
= ab * ((ab)p[n-1])c *
((ab)p[n-2]);递推式子可以这样写;
合并后变为(ab)p
=(ab)(c*p[n-1]+p[n-2]+1);
可以得到p
=c*p[n-1]+p[n-2]+1;
这样的递推式可以用矩阵乘法得到第n项;这是矩阵乘法的一个应用,给matrix67大神的博客地址可以学习,
构造矩阵乘法:

p
　　　　c 1 1　　　　p[n-1]
p[n-1] = 1 0 0 * 　p[n-2]
1　　　　　 0 0 1　　　　　　1

然后这中间还有一个问题,就是取模的问题;ab*p
%mod=ab*p
%(mod-1)%mod;
这是根据费马小定理得到的;a(p-1)Ξ1%p;

ab*p
%mod=ab*p
/(mod-1)*(mod-1)+b*p
%(mod-1)%mod;

令x=b*p
/(mod-1)则ab*p
%mod=ax*(mod-1)+b*p
%(mod-1)%mod=ab*p
%(mod-1)%mod;
这就是取模的结果啦;
然后就是快速幂算答案啦;
还有要注意的就是a%p==0的情况;

#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <list>
#include <queue>
#include <map>
#include <stack>
using namespace std;
#define L(i) i<<1
#define R(i) i<<1|1
#define INF  0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-9
#define maxn 100010
#define MOD 1000000007
int N;
long long p,a,b,c;
struct Matrix
{
long long a[22][22];
Matrix()
{
memset(a,0,sizeof(a));
for(int i = 0; i < N; i++)
a[i][i] = 1;
}
};
Matrix operator*(Matrix A,Matrix B)
{
Matrix ans;
for(int i = 0; i < N; i++)
for(int j = 0; j < N; j++)
{
ans.a[i][j] = 0;
for(int k = 0; k < N; k++)
ans.a[i][j] += (A.a[i][k] * B.a[k][j])%(p-1);
ans.a[i][j] %= (p-1);
}
return ans;
}
Matrix Pow(Matrix A,long long n)
{
Matrix ans;
while(n)
{
if(n % 2)
ans = ans * A;
n /= 2;
A = A * A;
}
return ans;
}
long long mod_pow(long long aa,long long bb,long long mod)
{
long long ans = 1;
while(bb)
{
if(bb&1)
ans *= aa;
ans %= mod;
bb /= 2;
aa *= aa;
aa %= mod;
}
return ans;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
N = 3;
long long n;
scanf("%lld%lld%lld%lld%lld",&n,&a,&b,&c,&p);
if(n==1)
{
printf("1\n");
continue;
}
if(n==2)
{
printf("%lld\n",mod_pow(a,b,p));
continue;
}
Matrix A;
memset(A.a,0,sizeof(A.a));
A.a[0][0] = c;
A.a[0][1] = 1;
A.a[0][2] = b;
A.a[1][0] = 1;
A.a[2][2] = 1;
A = Pow(A,n-2);
long long k = A.a[0][0] * b + A.a[0][2];
k = mod_pow(a,k,p);
printf("%lld\n",k);
}
}