非打表题解

#include<bits/stdc++.h>
using namespace std; 
int main() 
{
	int n;
	scanf("%d",&n);
	long long a[2]={0,9},b[2]={0,1};
	for(int i=2;i<=n;i++)
	{
		a[i%2]=b[(i-1)%2]+9*a[(i-1)%2];
		b[i%2]=a[(i-1)%2]+9*b[(i-1)%2];
	}
	printf("%lld\n",a[n%2]-a[(n-1)%2]);
	return 0; 
}