https://codeforces.com/problemset/problem/1239/A

---

思路:

如果真在考场一时间自己又推不出来性质，打表找找规律

nm复杂度不能接受，那么如果结论的话应该到Log或者o1.

这个范围就猜一手横纵坐标独立。

先打表观察。

#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<cstdio>
#include<algorithm>
#define debug(a) cout<<#a<<"="<<a<<endl;
using namespace std;
const int maxn=50;
typedef int LL;
inline LL read(){LL x=0,f=1;char ch=getchar();	while (!isdigit(ch)){if (ch=='-') f=-1;ch=getchar();}while (isdigit(ch)){x=x*10+ch-48;ch=getchar();}
return x*f;}
LL a[maxn][maxn];
LL dx[4]={1,0,-1,0};
LL dy[4]={0,1,0,-1};
LL ans=0;
LL check(LL n,LL m){
    for(LL i=1;i<=n;i++){
        for(LL j=1;j<=m;j++){
            LL t0=0;LL t1=0;
            for(LL k=0;k<4;k++){
                LL nx=i+dx[k];LL ny=j+dy[k];
                if(nx<1||nx>n||ny<1||ny>m) continue;
                if(a[nx][ny]==0) t0++;
                if(a[nx][ny]==1) t1++;
            }
            if(a[i][j]==0&&t0>1) return 0;
            if(a[i][j]==1&&t1>1) return 0;
        }
    }
    return 1;
}
void dfs(LL x,LL y,LL n,LL m){
    if(x==n+1){
        ans+=check(n,m);
        return;
    }
    for(LL i=0;i<=1;i++){
        a[x][y]=i;
        LL nx=x;LL ny=y;
        if(ny==m){
            dfs(nx+1,1,n,m);
        }
        else dfs(nx,ny+1,n,m);
    }
}
int main(void)
{
  ///cin.tie(0);std::ios::sync_with_stdio(false);
  for(LL i=1;i<=5;i++){
    for(LL j=1;j<=5;j++){
        ans=0;
        memset(a,0,sizeof(a));
        dfs(1,1,i,j);
        cout<<ans<<"\n";
    }
  }
return 0;
}

/// 2 4 6 10 16
/// 4 6 8 12 18
/// 6 8 10 14 20
///10 12 14 18 24
///16 18 20 24 30

第一列满足一个广义斐波那契

然后每一行也遵守一个增量为2,2,4,6,10的广义斐波那契

#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<cstdio>
#include<algorithm>
#define debug(a) cout<<#a<<"="<<a<<endl;
using namespace std;
const int maxn=1e5+1000;
typedef long long LL;
const LL mod=1e9+7;
inline LL read(){LL x=0,f=1;char ch=getchar();	while (!isdigit(ch)){if (ch=='-') f=-1;ch=getchar();}while (isdigit(ch)){x=x*10+ch-48;ch=getchar();}
return x*f;}
LL dp[maxn];
LL row[maxn];
/// 2 4 6 10 16
/// 4 6 8 12 18
/// 6 8 10 14 20
///10 12 14 18 24
///16 18 20 24 30
int main(void)
{
  cin.tie(0);std::ios::sync_with_stdio(false);
  LL n,m;cin>>n>>m;
  row[1]=2;row[2]=4;row[3]=6;
  for(LL i=4;i<maxn;i++) row[i]=(row[i-1]%mod+row[i-2]%mod)%mod;
  dp[1]=2;dp[2]=2;dp[3]=4;
  for(LL i=4;i<maxn;i++) dp[i]=(dp[i-1]%mod+dp[i-2]%mod)%mod;
  LL ans=row[n];
  for(LL i=2;i<=m;i++){
    ans=(ans%mod+dp[i-1]%mod)%mod;
  }
  cout<<ans%mod<<"\n";
return 0;
}