洛谷-2044

洛谷-2044

思路

首先对与递推式$x_n=(a\ast x_{n-1}+c)$ % $mod$，发现$c$的存在使得不能直接使用整数快速幂求出$x_n$，因为$x_n$是关于$a,c,x0$的$n$次多项式。并且$n$是$1e18$大小的。

于是考虑使用矩阵快速幂加速，但是显然不能使用一维矩阵（没意义），考虑通过二维矩阵
$\left(\begin{array}{cc}{x}_n& c\end{array}\right)$
来得到关系式。($x_{n-1}$和$c$均与$x_n$有关)

首先有

$$\begin{array}{}\text{(1)}& \left(\begin{array}{cc}{x}_n& c\end{array}\right)=\left(\begin{array}{cc}{x}_{n-1}& c\end{array}\right)\ast \left(\begin{array}{cc}A& B\\ C& D\end{array}\right)\end{array}$$

将$(1)$展开，有

$$\begin{array}{}\text{(2)}& \{\begin{array}{l}{x}_n=x_{n-1}\ast A+c\ast C\\ c=x_{n-1}\ast B+c\ast D\end{array}\end{array}$$

由于$x_n=a\ast x_{n-1}+c$，替换$(2)$中第一个方程的$x_n$，得到

$$\begin{array}{}\text{(3)}& \{\begin{array}{l}a\ast x_{n-1}+c=x_{n-1}\ast A+c\ast C\\ c=x_{n-1}\ast B+c\ast D\end{array}\end{array}$$

由$(3)$得到

$$\left(\begin{array}{cc}A=a& B=0\\ C=1& D=1\end{array}\right)$$

再由于

$$\left(\begin{array}{cc}{x}_n& c\end{array}\right)=\left(\begin{array}{cc}{x}_{n-1}& c\end{array}\right)\ast \left(\begin{array}{cc}a& 0\\ 1& 1\end{array}\right)=\left(\begin{array}{cc}{x}_0& c\end{array}\right)\ast {\left(\begin{array}{cc}a& 0\\ 1& 1\end{array}\right)}^n$$

可知，只需要求出
${\left(\begin{array}{cc}a& 0\\ 1& 1\end{array}\right)}^n$
即可得到答案。

Code

#include <bits/stdc++.h>
using namespace std;
#define _u_u_ ios::sync_with_stdio(false), cin.tie(nullptr)
#define cf int _o_o_;cin>>_o_o_;for (int Case = 1; Case <= _o_o_;Case++)
#define SZ(x) (int)(x.size())
inline void _A_A_();
signed main() {_A_A_();return 0;}

using ll = long long;
#define int long long
int mod = 1e9 + 7;
const int maxn = 2e5 + 10;
const int N = 2, M = 5010;
const int inf = 0x3f3f3f3f;

struct mat {
    int m[N][N];
};

ll n, c, x0,  g;

int mul(int a,int b) {
    int res = 0;
    a %= mod;
    b %= mod;
    while(b) {
        if (b & 1) {
            res = (res + a) % mod;
        }
        a = (a + a) % mod;
        b >>= 1;
    }
    return res;
}

mat operator * (const mat&a, const mat&b) {
    mat c;
    memset(c.m, 0, sizeof c.m);
    for (int i = 0;i < N;i++) {
        for (int j = 0;j < N;j++) {
            for (int k = 0;k < N;k++) {
                c.m[i][j] = (c.m[i][j] + mul(a.m[i][k] , b.m[k][j])) % mod;
            }
        }
    }
    return c;
}

mat qpow(mat a, int b ) {
    mat c;
    memset(c.m, 0,sizeof c.m);
    for (int i = 0;i < N;i++) c.m[i][i] = 1;
    while (b) {
        if (b & 1) {
            c = c * a;
        }
        a = a * a;
        b >>= 1;
    }
    return c;
}

inline void _A_A_() {
    #ifdef LOCAL
    freopen("in.in", "r", stdin);
    #endif
    _u_u_;
    mat s;
    memset(s.m, 0, sizeof s.m);
    s.m[1][0] = s.m[1][1] = 1;
    cin >> mod >> s.m[0][0] >> c >> x0 >> n >> g;
    s = qpow(s, n);
    int xn = (mul(s.m[0][0] , x0) + mul(c , s.m[1][0])) % mod;// 注意xn可能会大于mod，要取模。
    cout << (xn) % g << "\n";
}