HDU 5895 Mathematician QSC

题目地址

欧拉函数+矩阵快速幂

#include<cstdio>
#include<algorithm>
#include<string.h>
#include<queue>
#define LL long long
using namespace std;
const int Nmax=10;
LL n,y,x,s,tmp;
int mod;
int oula_mod;

struct Matrix
{
    int n,m;
    long long map[Nmax][Nmax];
    Matrix(int x,int y)
    {
        n=x;m=y;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
                map[i][j]=0;
    }
    Matrix operator * (const Matrix b)
    {
        Matrix c(n,b.m);
        if(m==b.n)
        {
            for(int i=1;i<=c.n;i++)
                for(int j=1;j<=c.m;j++)
                    for(int k=1;k<=m;k++)
                        c.map[i][j]=(c.map[i][j]+(map[i][k]*b.map[k][j])%oula_mod)%oula_mod;
            return c;
        }
        printf("error!!!!!!!!!!!!!!
");    
    }
};


int oula(int n)
{
    int ret=1,i;
    for(i=2;i*i<=n;i++)
    {
        if(n%i==0)
        {
            n/=i,ret*=i-1;
            while(n%i==0) n/=i,ret*=i;
        }
    }
    if(n>1) ret*=n-1;
    return ret;
}




Matrix get(long long n)
{
    Matrix base(4,4);
    base.map[1][1]=1;base.map[1][2]=1;base.map[1][3]=0;base.map[1][4]=0;
    base.map[2][1]=0;base.map[2][2]=4;base.map[2][3]=1;base.map[2][4]=4;
    base.map[3][1]=0;base.map[3][2]=1;base.map[3][3]=0;base.map[3][4]=0;
    base.map[4][1]=0;base.map[4][2]=2;base.map[4][3]=0;base.map[4][4]=1;
    Matrix ans(4,4);
    for(int i=1;i<=ans.n;i++)
        ans.map[i][i]=1;
    
    while(n>0)
    {
        if(n & 1)
            ans=ans*base;
        base=base*base;
        n>>=1;
    }
    
    return ans;
}

long long get_ans(long long times)
{
    long long ans=1;
    long long base=x;
    while(times>0)
    {
        if(times & 1)
            ans=(ans*base)%mod;
        base=(base*base)%mod;
        times>>=1;
    }
    return ans;
}


int main()
{
    
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld%lld%lld%lld",&n,&y,&x,&s);
        mod=s+1;
        oula_mod=oula(mod);
        //printf("oula_mod:%d
",oula_mod);
        Matrix base(4,1);
        base.map[1][1]=0;
        base.map[2][1]=1;
        base.map[3][1]=0;
        base.map[4][1]=0;
        Matrix ans=get(n*y)*base;
        long long mi=ans.map[1][1]+oula_mod;
        //printf("mi:%lld
",mi );
        //continue;
        printf("%lld
",get_ans(mi));
    }
    return 0;
}