CodeForces 678D Iterated Linear Function

简单矩阵快速幂。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<stack>
using namespace std;

long long mod=1e9+;
long long n,A,B,x;

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 <= 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);

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

    X.A[][]=; X.A[][]=B%mod;
    X.A[][]=; X.A[][]=A%mod;
    X.R=; X.C=;

    Z.A[][]=; Z.A[][]=x%mod;
    Z.R=; Z.C=;
}

void work()
{
    while (n)
    {
        if (n %  == ) Y = Y*X;
        n = n >> ;
        X = X*X;
    }
    Z = Z*Y;

    printf("%lld\n", Z.A[][]);
}

int main()
{
    scanf("%lld%lld%lld%lld",&A,&B,&n,&x);
    init();
    work();
    return ;
}