2875: [Noi2012]随机数生成器

题目链接

题目大意：Xn+1 = (aXn + c) mod m ，求Xn mod g

题解：模板题，随手构造一发矩阵

加了特技快速乘

注意矩乘的时候mod m，最后再mod g

我的收获：233

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define Rep(i,x) for(int i=0;i<x;i++)
#define N 2

ll a,m,n,g,c,x0;

ll mul(ll x,ll y)
{
    ll tmp=(x*y-(ll)((long double)x/m*y+1e-8)*m);
    return tmp<0?tmp+m:tmp;
}

struct Matrix{
    int x,y;ll v[N][N];
    ll *operator [] (int x){return v[x];}
    Matrix(int _x,int _y,int w=0){
        x=_x,y=_y;
        Rep(i,N) Rep(j,N) v[i][j]=(i==j)?w:0;
    }
};

Matrix operator * (Matrix a,Matrix b){
    Matrix c(a.x,b.y);
    Rep(i,a.x) Rep(j,b.y) Rep(k,a.y) c[i][j]=(c[i][j]+mul(a[i][k],b[k][j]))%m;
    return c;
}

Matrix operator ^ (Matrix a,ll b){
    Matrix c(a.x,a.y,1);
    for(;b;b>>=1,a=a*a)
    if(b&1) c=c*a;
    return c;
}

Matrix F(1,N),T(N,N);

void work()
{
    T=T^n;F=F*T;
    printf("%lld\n",F[0][0]%g);
}

void init()
{
    scanf("%lld%lld%lld%lld%lld%lld",&m,&a,&c,&x0,&n,&g);
    F[0][0]=x0;F[0][1]=1;
    T[0][0]=a;T[1][0]=c;T[0][1]=0;T[1][1]=1;
}

int main()
{
    init();
    work();
    return 0;
}