洛谷 P2044 [NOI2012]随机数生成器

题意

读入X[0], m, a, c, n和g
$ X[n+1]=(a*X[n]+c)\mod m $
求X数列的第n项对g取余的值。

题解

矩阵加速
设\[ F=\begin{bmatrix} a&0\\1&1\end{bmatrix}, G=\begin{bmatrix} X[0]\\c\end{bmatrix}\]
则\[ \begin{bmatrix} X[n]\\c\end{bmatrix} = G * F ^ n (n > 0)\]
乘法用快速乘

代码

#include <cstdio>
typedef long long ll;
int T, N, g;
ll n, m, aa, cc, x0, mod;
struct Matrix
{
    ll a[4][4];
    Matrix& operator =(const Matrix& x)
    {
        for (register int i = 1; i <= N; ++i)
            for (register int j = 1; j <= N; ++j)
                a[i][j] = x.a[i][j];
        return *this;
    }
};
Matrix a, b, c;
ll _mul(ll x, ll y, ll s = 0)
{
    for (; y; y >>= 1, x = (x + x) % mod)
        if (y & 1)
            s = (s + x) % mod;
    return s;
}
Matrix Mul(const Matrix& x, const Matrix& y)
{
    Matrix s;
    for (register int i = 1; i <= N; ++i)
        for (register int j = 1; j <= N; ++j)
            s.a[i][j] = 0;
    for (register int i = 1; i <= N; ++i)
        for (register int j = 1; j <= N; ++j)
            for (register int k = 1; k <= N; ++k)
                s.a[i][j] = (s.a[i][j] + _mul(x.a[i][k], y.a[k][j])) % mod;
    return s;
}
Matrix _pow(Matrix x, ll y)
{
    Matrix s;
    for (register int i = 1; i <= N; ++i)
        for (register int j = 1; j <= N; ++j)
            s.a[i][j] = (i == j);
    for (; y; y >>= 1, x = Mul(x, x)) if (y & 1) s = Mul(s, x);
    return s;
}
int main()
{
    N = 2;
    scanf("%lld%lld%lld%lld%lld%d", &m, &aa, &cc, &x0, &n, &g);
    mod = m;
    c.a[1][1] = x0; c.a[1][2] = cc;
    a.a[1][1] = aa; a.a[2][1] = a.a[2][2] = 1;
    if (n <= 0) {printf("%lld\n", x0); return 0; }
    b = Mul(c, _pow(a, n));
    printf("%lld\n", (b.a[1][1] + mod) % mod % g);
}