AcWing96 奇怪的汉诺塔 (递推)

题目链接：https://www.acwing.com/problem/content/98/

设\(d[n]\)表示\(n\)盘\(3\)塔问题的最小步数，\(f[n]\)表示\(n\)盘\(4\)塔问题的最小步数
则\(f[n] = min_{1 \leq i< n}\{2*f[i]+d[n-i]\}\)
其中\(f[1] = 1\)
含义是先将\(i\)个盘子在四塔模式下移到\(B\)柱,再将\(n-i\)个盘子在三塔模式下移到\(D\)柱

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<stack>
#include<queue>
using namespace std;
typedef long long ll;

const int maxn = 20;
const int inf = 100000007;

int n;
int d[maxn],f[maxn];
int sta[maxn],tail = 0; 

ll read(){ ll s=0,f=1; char ch=getchar(); while(ch<'0' || ch>'9'){ if(ch=='-') f=-1; ch=getchar(); } while(ch>='0' && ch<='9'){ s=s*10+ch-'0'; ch=getchar(); } return s*f; }

int main(){
	n = 12;
	d[0] = 0;
	d[1] = 1;
	for(int i=1;i<=n;++i) d[i] = d[i-1] * 2 + 1;
	
	f[1] = 1;
	for(int i=2;i<=n;++i){
		f[i] = inf;
		for(int j=1;j<i;++j){
			f[i] = min(f[i],2 * f[j] + d[i-j]);
		}
	}
	
	for(int i=1;i<=n;++i) printf("%d\n",f[i]);
	
	return 0;
}