考虑记\(f_{i,j,k}\)为\(k\)次操作后,\(i,j\)位置被调换的概率. 那么我们考虑枚举我们要算的答案即\((x,y)\). 那么有\(\frac{n * (n + 1)}{2}\)种调换顺序. 以此分类讨论: 一:不相交: 对答案不产生影响. 二:包含 因为是反转操作,考虑枚举枚举翻转移动的距离,从\(f_{i + q,j + q,k - 1}\)转移过来. 三:端点相交 同样考虑枚举反转距离 ,从\(f_{i + q,j,k - 1}\)还有\(f_{i,j + q,k -…

转自九野:http://blog.csdn.net/qq574857122/article/details/43643135 题目链接:点击打开链接 题意: 给定n ,k 下面n个数表示有一个n的排列, 每次操作等概率翻转一个区间,操作k次. 问: k次操作后逆序数对个数的期望. 思路: dp[i][j]表示 a[i] 在a[j] j前面的概率 初始就是 dp[i][j]  = 1( i < j ) 则对于翻转区间 [i, j], 出现的概率 P = 1 / ( n * (n+1) /2) 并且…

Problem Introduction An inversion of a sequence \(a_0,a_1,\cdots,a_{n-1}\) is a pair of indices \(0 \leq i < j < n\) such that \(a_i>a_j\). The number of inversions of a sequence in some sense measures how close the sequence is to being sorted. F…

A Game题意:A,B各自拥有两堆石子,数目分别为n1, n2,每次至少取1个,最多分别取k1,k2个, A先取,最后谁会赢. 分析:显然每次取一个是最优的,n1 > n2时,先手赢. 代码: #include <bits/stdc++.h> #define pb push_back #define mp make_pair #define esp 1e-14 #define lson l, m, rt<<1 #define rson m+1, r, rt<<1…

A. Game 题目大意:A有N1个球,B有N2个球,A每次可以扔1-K1个球,B每次可以扔1-K2个球,谁先不能操作谁熟 思路:．．．．．显然每次扔一个球最优．．．． #include<iostream> #include<cstdio> #include <math.h> #include<algorithm> #include<string.h> #include<queue> #define MOD 10000007 #def…

在Merge Sort的基础上改改就好了. public class Inversions { public static int inversions(int [] A,int p, int r) { if(p<r) { int q = (int) Math.floor( (p+r)/2 ); int left = inversions(A,p,q); int right = inversions(A,q+1,r); int c = combine(A,p,q,r); return left…

Dynamic Inversions II Time Limit: 6000/3000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others) SubmitStatus Problem Description 给出N个数a[1],a[2] ... a[N],a[1]...a[N]是1-N的一个排列,即1 <= a[i] <= N且每个数都不相同.有M个操作,每个操作给出x,y两个数,你将a[x],a[y]交换,然后求交换后数组的逆序…

Dynamic Inversions Time Limit: 30000/15000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others) SubmitStatus Problem Description 给出N个数a[1],a[2] ... a[N],有M个操作,每个操作给出x,y两个数,你将a[x],a[y]交换,然后求交换后数组的逆序对个数.逆序对的意思是1 <= i < j <= N 且a[i] > a[j].…

We have some permutation Aof [0, 1, ..., N - 1], where N is the length of A. The number of (global) inversions is the number of i < j with 0 <= i < j < N and A[i] > A[j]. The number of local inversions is the number of i with 0 <= i <…

We have some permutation A of [0, 1, ..., N - 1], where N is the length of A. The number of (global) inversions is the number of i < j with 0 <= i < j < N and A[i] > A[j]. The number of local inversions is the number of i with 0 <= i <…