NOIP2018 Day2 T2 填数游戏 - 打表 - 搜索 - 找规律

真尼玛是个好题
弄个爆搜跑出(7,8)。
然后发现规律;
规律是:
f ( n , m ) = f ( m , n ) f ( n , m ) = 3 f ( n , m 1 ) , m > n + 1 f ( 1 , n ) = 2 n f ( 2 , 2 ) = 12 , f ( 2 , 3 ) = 36 , f ( 3 , 3 ) = 112 , f ( 3 , 4 ) = 336 f ( n + 1 , n + 1 ) = 8 f ( n , n ) 10 2 n , n 4 f ( n , n + 1 ) = 3 f ( n , n ) 3 2 n , n 4 f(n,m)=f(m,n)\\ f(n,m)=3f(n,m-1),m>n+1\\ f(1,n)=2^n\\ f(2,2)=12,f(2,3)=36,f(3,3)=112,f(3,4)=336\\ f(n+1,n+1)=8f(n,n)-10\cdot2^n,n\geq4\\ f(n,n+1)=3f(n,n)-3\cdot2^n,n\geq4
然后综合这些规律就可以获得100分的好成绩。

#include<bits/stdc++.h>
#define lint long long
#define mod 1000000007
using namespace std;
inline int fast_pow(int x,int k,int ans=1)
{
	for(;k;k>>=1,x=(lint)x*x%mod)
		if(k&1) ans=(lint)ans*x%mod;
	return ans;
}
int f[100];
int main()
{
	int n,m;cin>>n>>m;if(n>m) swap(n,m);
	if(n==1) return !printf("%d\n",fast_pow(2,m));f[2]=12,f[3]=112,f[4]=912;
	for(int i=4;i<n;i++) f[i+1]=(8ll*f[i]%mod-10ll*fast_pow(2,i)%mod+mod)%mod;
	if(n==m) return !printf("%d\n",f[n]);
	int g=(3ll*(f[n]-fast_pow(2,n))%mod+mod)%mod;
	if(n==2) g=36;if(n==3) g=336;
	return !printf("%lld\n",(lint)g*fast_pow(3,m-n-1)%mod);
}

打表:

#include<bits/stdc++.h>
using namespace std;
const int N=100;int val[N][N],Rv[N],a[N][N],ans,no_ans,las;
int ok(int x,int y,int n,int m,int cur)
{
	if(no_ans) return 0;cur<<=1,cur|=a[x][y];
	if(x==n&&y==m) return no_ans|=(cur<las),las=cur;
	if(y==m) ok(x+1,y,n,m,cur);
	else if(x==n) ok(x,y+1,n,m,cur);
	else ok(x,y+1,n,m,cur),ok(x+1,y,n,m,cur);
	return 0;
}
inline int can(int x,int y,int v) { return a[x][y]=v,no_ans=0,las=0,ok(1,1,x,y,0),!no_ans; }
int dfs(int x,int y,int n,int m)
{
	if(x==n+1) return ++ans;
	for(int i=0;i<=1;i++) if(can(x,y,i))
	{
		a[x][y]=i;
		if(y>1) val[x][y]=(val[x][y-1]<<1)|a[x][y];
		else val[x][y]=(val[x-1][y]<<1)|a[x][y];
		if(y==m) dfs(x+1,1,n,m);
		else dfs(x,y+1,n,m);
	}
	return 0;
}
inline int calc(int n,int m) { return ans=0,dfs(1,1,n,m),ans; }
int main()
{
	for(int n=2;n<=7;n++)
	{
		printf("ans(%d, %d) = %d\n",n,n,calc(n,n));
		printf("ans(%d, %d) = %d\n",n,n+1,calc(n,n+1));
//		printf("ans(%d, %d) = %d\n",n,n+2,calc(n,n+2));
//		printf("ans(%d, %d) = %d\n",n,n+3,calc(n,n+3));
	}
	return 0;
}

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转载自blog.csdn.net/Mys_C_K/article/details/84314044