傳送門

首先不難設\(f[i][j]\)表示跳到\((i,j)\)的方案數，那麼不難得到如下轉移
\[f[i][j]=\sum\limits_{k=1}^{\frac n2}f[i-2k+1][j-1]+f[i-2k+1][j]+f[i-2k+1][j+1]\]
然後維護兩個字首和\(s1,s2\)，分別表示與當前列相差為偶數的字首和以及與當前列相差為奇數的字首和，那麼可以這樣轉移
\[s1[i+1][j]=s2[i][j]+s1[i][j-1]+s1[i][j]+s1[i][j+1]\]
\[s2[i+1][j]=s1[i][j]\]
然而直接轉移會T，我們考慮用矩陣乘法來優化。構造一個\(1*2n\)

然後最後字首和減一減就好了

//minamoto
#include<bits/stdc++.h>
#define R register int
#define fp(i,a,b) for(R i=a,T=b+1;i<T;++i)
#define fd(i,a,b) for(R i=a,T=b-1;i>T;--i)
using namespace std;
const int P=30011;
int n,m;
struct node{
    int a[105][105];
    node(){memset(a,0,sizeof(a));}
    int *operator [](const R &x){return a[x];}
    node operator *(node &b){
        node res;
        fp(i,1,n)fp(j,1,n)fp(k,1,n)
        res[i][j]=(res[i][j]+a[i][k]*b[k][j])%P;
        return res;
    }
}I,A,B;
node ksm(node x,R y){
    node res;fp(i,1,n)res[i][i]=1;
    for(;y;y>>=1,x=x*x)if(y&1)res=res*x;
    return res;
}
int main(){
//  freopen("testdata.in","r",stdin);
    scanf("%d%d",&n,&m);
    fp(i,1,n)I[i][i]=I[i+n][i]=I[i][i+n]=1;
    fp(i,1,n-1)I[i+1][i]=I[i][i+1]=1;
    n<<=1,A=ksm(I,m-2),B=A*I;
    printf("%d\n",(B[1][n>>1]-A[1][n]+P)%P);
    return 0;
}