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+7;
long long n,A,B,x;

struct Matrix
{
    long long A[5][5];
    int R, C;
    Matrix operator*(Matrix b);
};

Matrix X, Y, Z;

Matrix Matrix::operator*(Matrix b)
{
    Matrix c;
    memset(c.A,0,sizeof c.A);
    int i, j, k;
    for (i = 1; i <= R; i++)
        for (j = 1; j <= C; j++)
            for (k = 1; 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,0,sizeof X.A);
    memset(Y.A,0,sizeof Y.A);
    memset(Z.A,0,sizeof Z.A);

    for(int i=1;i<=2;i++) Y.A[i][i]=1;
    Y.R = 2; Y.C = 2;

    X.A[1][1]=1; X.A[1][2]=B%mod;
    X.A[2][1]=0; X.A[2][2]=A%mod;
    X.R=2; X.C=2;

    Z.A[1][1]=1; Z.A[1][2]=x%mod;
    Z.R=1; Z.C=2;
}

void work()
{
    while (n)
    {
        if (n % 2 == 1) Y = Y*X;
        n = n >> 1;
        X = X*X;
    }
    Z = Z*Y;

    printf("%lld\n", Z.A[1][2]);
}

int main()
{
    scanf("%lld%lld%lld%lld",&A,&B,&n,&x);
    init();
    work();
    return 0;
}