看了其他人博客，貌似i个盘子的方案数满足 f[i] = f[i-1] * x + y
？？？？？？？
神来之笔
貌似没有找到严格的证明……
牛逼……
如果这样的话暴力求出x和y然后递推完事
#include<cstdio>
#include<stack>
#define REP(i, a, b) for(int i = (a); i < (b); i++)
using namespace std;int s[10][2];
stack<int> st[3];bool judge(int a, int b, int pre)
{if(st[a].empty() || st[a].top() == pre) return false;if(st[b].empty() || st[a].top() < st[b].top()) return true;return false;
}int solve(int n)
{REP(i, 0, 3)while(!st[i].empty())st[i].pop();for(int i = n; i >= 1; i--) st[0].push(i);int ret = 0, pre = -1;while(1){if(st[1].size() == n || st[2].size() == n) return ret;REP(i, 0, 6){int a = s[i][0], b = s[i][1];if(judge(a, b, pre)){pre = st[a].top();st[b].push(pre);st[a].pop();ret++;break;}}}
}int main()
{int n, f2, f3;int x, y;while(~scanf("%d", &n)){REP(i, 0, 6){char str[5];scanf("%s", str);s[i][0] = str[0] - 'A';s[i][1] = str[1] - 'A';}f2 = solve(2); f3 = solve(3);x = (f3 - f2) / (f2 - 1);y = f2 - x;long long ans = 1;REP(i, 2, n + 1)ans = ans * x + y;printf("%lld\n", ans);}return 0;
}