hdu5322 Hope(dp+FFT+分治) hdu 题目大意:n个数的排列,每个数向后面第一个大于它的点连边,排列的权值为每个联通块大小的平方,求所有排列的权值和. 思路: 考虑直接设dp[i]表示n=i时的答案. 我们考虑放完前n-1个数之后再插入n,会发现n前面所有数都和它联通. 于是dp方程就出来了: $dp[n]=\Sigma(dp[n-k]*k^{2}*(k-1)!*C_{n-1}^{k-1})$ 组合数倒腾过来变成: $\frac{dp[n]}{(n-1)!}=\Sigma(\f…

题目 Source http://acm.hdu.edu.cn/showproblem.php?pid=5322 Description Hope is a good thing, which can help you conquer obstacles in your life, just keep fighting, and solve the problem below. In mathematics, the notion of permutation relates to the ac…

设dp[n]为n个数字排列时候的答案,那么可以得到dp方程 dp[n]=Σdp[n-i]*c(n-1,i-1)*(i-1)!*i^2(1<=i<=n) 然后上式可以化成卷积形式,分治FFT即可.复杂度O(nlogn^2) 代码 #include <iostream> #include <cstring> #include <cstdio> using namespace std; typedef long long LL; *(<<)+; <…

转载请注明出处: http://www.cnblogs.com/fraud/          ——by fraud Hope Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 116    Accepted Submission(s): 44 Problem Description Hope is a good thing, which…