HDU 5895 Mathematician QSC

矩阵快速幂，欧拉定理。

$g(n)$递推式：$g(n)=5g(n-1)+5g(n-2)-g(n-3)$，可以构造矩阵快速求递$n$项，指数很大，可以利用欧拉定理降幂。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0),eps=1e-;
void File()
{
    freopen("D:\\in.txt","r",stdin);
    freopen("D:\\out.txt","w",stdout);
}
template <class T>
inline void read(T &x)
{
    char c=getchar(); x=;
    while(!isdigit(c)) c=getchar();
    while(isdigit(c)) {x=x*+c-''; c=getchar();}
}

LL MOD;

struct Matrix
{
    long long A[][];
    int R, C;
    Matrix operator*(Matrix b);
};

Matrix X, Y, Z;

Matrix Matrix::operator*(Matrix b)
{
    Matrix c;
    memset(c.A, , sizeof(c.A));
    int i, j, k;
    for (i = ; i <= R; i++)
        for (j = ; j <= b.C; j++)
            for (k = ; k <= C; k++)
                c.A[i][j] = (c.A[i][j] + (A[i][k] * b.A[k][j]) % MOD) % MOD;
    c.R = R; c.C = b.C;
    return c;
}

void init()
{
    memset(X.A, , sizeof X.A);
    memset(Y.A, , sizeof Y.A);
    memset(Z.A, , sizeof Z.A);

    Y.R = ; Y.C = ;
    for (int i = ; i <= ; i++) Y.A[i][i] = ;

    X.R = ; X.C = ;
    X.A[][]=-;
    X.A[][]=; X.A[][]=;
    X.A[][]=; X.A[][]=;

    Z.R = ; Z.C = ;
    Z.A[][]=; Z.A[][]=; Z.A[][]=;
}

LL work(LL tt)
{
    if(tt==) return ;
    if(tt==) return ;
    if(tt==) return ;

    tt=tt-; init();
    while (tt)
    {
        if (tt %  == ) Y = Y*X;
        tt = tt >> ;
        X = X*X;
    }
    Z = Z*Y;
    return Z.A[][];
}

int T;
LL n,x,y,s;

LL phi(LL x)
{
     LL res=x,a=x;
     for(int i=;i*i<=a;i++)
     {
         if(a%i==)
         {
             res=res/i*(i-);
             while(a%i==) a/=i;
         }
     }
     if(a>) res=res/a*(a-);
     return res;
}

LL pow(LL a,LL b,LL mod)
{
    LL c=;
    while(b!=)
    {
        if(b%==) c=(c*a)%mod,b--;
        else a=(a*a)%mod,b=b/;
    }
    return c;
}

int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lld%lld%lld%lld",&n,&y,&x,&s);
        MOD=phi(s+);
        LL gn=work(n*y)+MOD;
        LL ans=pow(x,gn,s+);
        printf("%lld\n",ans);
    }
    return ;
}