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「NOI2012」随机数生成器-矩阵快速幂

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给定正整数 $n,m,a,c,X[0],g$ ,求按照 $X[n+1] = (a X[n] + c) \bmod m$ 生成出的第 $n$ 项 $X[n] \bmod g$ 的值。

数据范围: $n,m,a,c,X[0] \leq 10^{18}$

链接

Luogu P2044

题解

构造转移矩阵:

$$
\left[\begin{matrix}
a & c \
0 & 1
\end{matrix}\right]
\times
\left[\begin{matrix}
x_{n}\
1
\end{matrix} \right]

\left[\begin{matrix}
a x_n + c \
0 + 1
\end{matrix} \right]

\left[\begin{matrix}
x_{n+1} \
1
\end{matrix} \right]
$$

快速幂即可。

这里的乘法需要快速乘。

代码

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#include <cstdio>
#include <cstring>
using namespace std;
#define ld long double
#define ll long long

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

ll mul(ll a,ll b){
    a%=m,b%=m;
    return ((a*b - (ll)((ll)(((ld)a/m) * b + 1e-3) * m))%m+m)%m;
}

struct Matrix{
    ll a[3][3];
    Matrix(){
        memset(a,0,sizeof(a));
    }
};

Matrix mul(Matrix &_a,Matrix &_b){
    Matrix ans;
    for(int i = 1;i<=2;i++){
        for(int j = 1;j<=2;j++){
            for(int k = 1;k<=2;k++){
                (ans.a[i][j] += mul(_a.a[i][k],_b.a[k][j]))%=m;
            }
        }
    }
    return ans;
}

Matrix pow(Matrix x,ll p){
    Matrix ans;
    ans.a[1][1] = ans.a[2][2] = 1;
    for(ll i = p;i;x = mul(x,x),i>>=1) if(i & 1) ans = mul(ans,x);
    return ans;
}

int main(){
    scanf("%lld %lld %lld %lld %lld %lld",&m,&a,&c,&x0,&n,&g);x0 %= m;
    Matrix tmp;
    tmp.a[1][1] = a % m, tmp.a[1][2] = c % m,tmp.a[2][2] = 1;
    tmp = pow(tmp,n);
    ll ans = ((mul(tmp.a[1][1],x0) + tmp.a[1][2])%m)%g;
    printf("%lld\n",ans);
    return 0;
}


cqqqwq
WRITTEN BY
cqqqwq
A student in Computer Science.

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