最近事情好多，数据库作业，没天要学2个小时java，所以更新的sgu就比较少了

SGU 131

题意：给你两种小块一种，1*1，一种2*2-1*1，问你填满一个m*n的矩形有多少钟方法，n和m小于等于9,

收获：状态压缩，每一行都最多由上一行转移过来，因为上一行，那么最多有7情况，详情看代码

#include
      
       #define de(x) cout<<#x<<"="<
       
        <
        
         =(b);--i)#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)#define ll long long#define mt(a,b) memset(a,b,sizeof(a))#define fi first#define se second#define inf 0x3f3f3f3f#define INF 0x3f3f3f3f3f3f3f3f#define pii pair
         
          #define pdd pair
          
           #define pdi pair
           
            #define mp(u,v) make_pair(u,v)#define sz(a) (int)a.size()#define ull unsigned long long#define ll long long#define pb push_back#define PI acos(-1.0)#define qc std::ios::sync_with_stdio(false)#define db double#define all(a) a.begin(),a.end()const int mod = 1e9+7;const int maxn = 11;const double eps = 1e-6;using namespace std;bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }bool ls(const db &a, const db &b) { return a + eps < b; }bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }ll gcd(ll a,ll b) { return a==0?b:gcd(b%a,a); };ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }ll kpow(ll a,ll b) {ll res=1;a%=mod; if(b<0) return 1; for(;b;b>>=1){ if(b&1)res=res*a%mod;a=a*a%mod;}return res;}ll read(){ ll x=0,f=1;char ch=getchar(); while (ch<'0'||ch>'9'){ if(ch=='-')f=-1;ch=getchar();} while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f;}//inv[1]=1;//for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod;int n,m;ll dp[maxn][1<