题解：[NOIP 2018 提高组] 填数游戏

题目传送门

打表

容易发现 \(n,m\) 没有什么区别，即 \(n\) 行 \(m\) 列的棋盘等价于 \(m\) 行 \(n\) 列的棋盘。

观察数据范围，可以发现 \(n\leq8\)，而 \(m\leq10^6\)。

因此可以考虑打表，然后找一下规律。

打出来表得到：

int f[9][9]={
	{0,0,0,0,0,0,0,0,0},
	{0,2,4,8,16,32,64,128,256},
	{0,4,12,36,108,324,972,2916,8748},
	{0,8,36,112,336,1008,3024,9072,27216},
	{0,16,108,336,912,2688,8064,24192,72576},
	{0,32,324,1008,2688,7136,21312,63936,191808},
	{0,64,972,3024,8064,21312,56768,170112,510336},
	{0,128,2916,9072,24192,63936,170112,453504,1360128},
	{0,256,8748,27216,72576,191808,510336,1360128,3626752}
};

记 \(\textit{ans}_{n,m}\) 为 \(n\) 行 \(m\) 列的棋盘的答案，钦定 \(n\leq m\)，则有：

• \(n=1\) 时，\(\textit{ans}_{n,m}=2^m\)。
• \(n>1\) 时，\(\textit{ans}_{n,m}=3\cdot\textit{ans}_{n,m-1}\)。

于是就可以愉快地 AC 了。

AC 代码

//#include<bits/stdc++.h>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<iomanip>
#include<cstdio>
#include<string>
#include<vector>
#include<cmath>
#include<ctime>
#include<deque>
#include<queue>
#include<stack>
#include<list>
using namespace std;
typedef long long ll;
constexpr const int N=8,M=1e6,P=1e9+7;
int n,m;
int f[N+1][N+1]={
	{0,0,0,0,0,0,0,0,0},
	{0,2,4,8,16,32,64,128,256},
	{0,4,12,36,108,324,972,2916,8748},
	{0,8,36,112,336,1008,3024,9072,27216},
	{0,16,108,336,912,2688,8064,24192,72576},
	{0,32,324,1008,2688,7136,21312,63936,191808},
	{0,64,972,3024,8064,21312,56768,170112,510336},
	{0,128,2916,9072,24192,63936,170112,453504,1360128},
	{0,256,8748,27216,72576,191808,510336,1360128,3626752}
};
int qpow(int base,int n){
	int ans=1;
	while(n){
		if(n&1){
			ans=1ll*ans*base%P;
		}
		base=1ll*base*base%P;
		n>>=1;
	}
	return ans;
}
int main(){
	/*freopen("test.in","r",stdin);
	freopen("test.out","w",stdout);*/
	
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	
	cin>>n>>m;
	if(n>m){
		swap(n,m);
	}
	if(n==1){
		cout<<qpow(2,m)<<'\n';
	}else if(m-1<=N){
		cout<<f[n][m]<<'\n';
	}else{
		int ans;
		if(n!=8){
			ans=f[n][n+1];
		}else{
			ans=10879488;
		}
		for(int i=1;i<=m-n-1;i++){
			ans=3ll*ans%P;
		}
		cout<<ans<<'\n';
	}
	
	cout.flush();
	 
	/*fclose(stdin);
	fclose(stdout);*/
	return 0;
}