矩阵乘法入门

BZOJ3231

http://www.lydsy.com/JudgeOnline/problem.php?id=3231

luogu2461

https://www.luogu.org/problemnew/show/2461

这题代码在本地过编译，但在BZOJ过不了编译，只好在luogu上交

大致就是构造这样一个矩阵，然后快速幂

复杂度O(k³log n)

#include<cstdio>
#define ROF(i,s,t) for(register int i=s;i>=t;--i)
#define FOR(i,s,t) for(register int i=s;i<=t;++i)
typedef long long ll;
int k;
ll n,m;
int p,ansn,ansm;
struct matrix{
    static int a[17][17];
    inline matrix(){FOR(i,0,k)FOR(j,0,k)a[i][j]=0;}
    inline int *operator[](int x){return a[x];}
    inline matrix operator*(matrix A)const{
        matrix ret;
        FOR(i,0,k)
            FOR(j,0,k)
                FOR(l,0,k){
                    ret[i][j]+=(int)(1ll*a[i][l]*A[l][j]%p);
                    ret[i][j]%=p;
                }
        return ret;
    }
}A,B,C,D;
inline void fp(ll b){
    while(b){
        if(b&1)C=C*D;
        D=D*D;
        b>>=1;
    }
}
inline int getans(ll x){
    C=A;D=B;
    ll ret=0;
    if(x<=k){
        FOR(i,0,x-1)
            ret+=A[0][i],ret%=p;
        return ret;
    }
    fp(x-k);
    return C[0][k];
}
int main(){
    scanf("%d",&k);
    FOR(i,0,k-1)scanf("%d",&A.a[0][i]);
    ROF(i,k-1,0)scanf("%d",&B.a[i][k-1]);
    scanf("%lld%lld%d",&m,&n,&p);
    FOR(i,0,k-1)A[0][i]%=p,A[0][k]+=A[0][i],A[0][k]%=p;
    FOR(i,0,k-1)B[i][k-1]%=p,B[i][k]=B[i][k-1];
    B[k][k]=1;
    FOR(i,0,k-2)B[i+1][i]=1;
    ansm=getans(m-1ll);
    ansn=getans(n);
    printf("%d
",(ansn-ansm+p)%p);
    return 0;
}

　　

BZOJ2875

http://www.lydsy.com/JudgeOnline/problem.php?id=2875

都是套路题用矩乘优化线性递推式

#include<cstdio>
#include<iostream>
#define FOR(i,s,t) for(register int i=s;i<=t;++i)
using namespace std;
typedef long long ll;
ll n,m,x,g,a,c;
inline ll mul(ll a,ll b){
    ll res=0;
    while(b){
        if(b&1)res=(res+a)%m;
        a=(a<<1)%m;
        b>>=1;
    }
    return res;
}
struct matrix{
    ll a[3][3];
    inline void clean(){
        FOR(i,1,2)
            FOR(j,1,2)
                a[i][j]=0;
    }
    inline matrix operator*(matrix A)const{
        matrix B;
        B.clean();
        FOR(i,1,2)
            FOR(j,1,2)
                FOR(k,1,2){
                    B.a[i][j]+=1ll*mul(a[i][k],A.a[k][j])%m;
                    B.a[i][j]%=m;
                }
        return B;
    }
}A,B;
inline matrix fp(matrix B,ll b){
    matrix ret;
    ret.clean();
    ret.a[1][1]=1;
    ret.a[2][2]=1;
    while(b){
        if(b&1)ret=ret*B;
        B=B*B;
        b>>=1;
    }
    return ret;
}
int main(){
    cin>>m>>a>>c>>x>>n>>g;
    B.a[1][1]=a;
    B.a[2][1]=c;
    B.a[2][2]=1;
    B=fp(B,n);
    A.a[1][1]=x;
    A.a[1][2]=1;
    A=A*B;
    cout<<A.a[1][1]%g<<'
';
    return 0;
}