Simulation game Chino number of columns

Matrix accelerated recurrence

N range is relatively small, a large range of k, we can consider starting from n.

1. First, we know that any matrix * matrix will not change.

So for the exchange operation, we can create such a matrix:

Addition of s, m rows, every other row is f [i] [i] = 1;

S-line: f [s] [m] = 1; m-th row: f [m] [s] = 1;

Then we completed the exchange operation.

2. For the left shift operation, we can create such a matrix:

In addition to n-th row, every other row is f [i] [i + 1] = 1;

N-th row: f [n] [1] = 1;

3. Our initial matrix f [i] [1] = a [i];

Because the binding matrix in line with the law, so we can speed up a similar rapid power method.

Then it can be, the time complexity O ( \ (n-^. 3 \ Times log (K) \) ).

I offer long and stinking code

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
long long n,s,m,k;
struct jz
{
    long long c[85][85];
}f,base,lin1,lin2;
jz operator * (const jz &a,const jz &b)//矩阵重载乘号 
{
    jz lin;
    for(int i=1;i<=80;++i)
    {
        for(int j=1;j<=80;++j)
        {
            lin.c[i][j]=0;
            for(int k=1;k<=80;++k)
            {
                lin.c[i][j]+=(a.c[k][j]*b.c[i][k]);
            }
        }
    }
    return lin;
}
jz ksm(jz a,long long b)
{
    jz anss;
    memset(anss.c,0,sizeof(anss.c));
    for(int i=1;i<=n;++i)anss.c[i][i]=1;
    for(;b;b>>=1,a=a*a)
    {
        if(b&1)
        anss=a*anss;
    }
    return anss;
}
void dy(jz x)//调试用的，可以忽略
{
    for(int i=1;i<=n;++i)
    {
        for(int j=1;j<=n;++j)
        cout<<x.c[i][j]<<" ";
        cout<<endl;
    }
}
int main()
{
    cin>>n>>m>>s>>k;
    for(int i=1;i<=n;++i)scanf("%lld",&f.c[i][1]);
    for(int i=1;i<=n;++i)if(i!=s&&i!=m)lin1.c[i][i]=1;
    lin1.c[s][m]=lin1.c[m][s]=1;
    for(int i=1;i<=n-1;++i)lin2.c[i][i+1]=1;
    lin2.c[n][1]=1;
    base=lin1*lin2;
    base=ksm(base,k);
    f=f*base;
    for(int i=1;i<=n;++i)printf("%lld ",f.c[i][1]);
    return 0;
}