1. 程式人生 > >hdu 5667 Sequence(矩陣快速冪+費馬小定理+快速冪)

hdu 5667 Sequence(矩陣快速冪+費馬小定理+快速冪)

#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);
	}
}