Atcoder description 给你\(n\)和\(\{a_i\}\),你需要求所有满足\(p_i\le a_i\)的\(1-n\)排列的逆序对个数之和模\(10^9+7\). \(n \le 2\times10^5\) sol 首先考虑一下所有满足要求的排列总数.记\(cnt_i\)表示有多少个\(a_k\ge i\),从大到小填数,方案数就是\[S=\prod_{i=1}^ncnt_i-(n-i)\] 考虑枚举两个位置\(i,j\),计算满足\(p_i>p_j\)的排列数.分两种情况…

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…

                                                                E. Infinite Inversions                                                                                          time limit per test 2 seconds                                            …

Inversions After Shuffle time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You are given a permutation of integers from 1 to n. Exactly once you apply the following operation to this permutat…

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 <…

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 <…

［抄题］: 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 <=…

题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=6318 Problem Description Long long ago, there was an integer sequence a.Tonyfang think this sequence is messy, so he will count the number of inversions in this sequence. Because he is angry, you will ha…

Swaps and Inversions Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2315    Accepted Submission(s): 882 Problem Description Long long ago, there was an integer sequence a.Tonyfang think this se…

[agc023E]Inversions(线段树,动态规划) 题面 AT 给定\(a_i\),求所有满足\(p_i\le a_i\)的排列\(p\)的逆序对数之和. 题解 首先如何计算排列\(p\)的个数. 设\(cnt[i]\)表示\(a_k\ge i\)的个数,那么满足条件的\(p\)的总数就是\(\prod cnt[i]-(n-i)\) 大概就是从\(n\)开始填数,对于每个数字\(i\)而言,它一共有\(cnt[i]\)个位置可以填,但是后面的数字一共占用了\(n-i\)个位置,所以还剩下…

Given a list of N integers A1, A2, A3,...AN, there's a famous problem to count the number of inversions in it. An inversion is defined as a pair of indices i < j such that Ai > Aj. Now we have a new challenging problem. You are supposed to count the…

A-Combination Lock  B-School Marks   C-Ice Cave   D-Bad Luck Island   E-Infinite Inversions E:Infinite Inversions 题意就是有一个序列1,2,3,4..... 现在有n次交换,每次都把ab交换求最终形成的序列的逆序数: 逆序数分为两部分.一部分是交换过位置的,另一部分是没有交换过的. 离散化后,利用树状数组求出交换过的位置的逆序数的个数. 第二部分: 看一个样例: 2 1 6 9 5…

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 <…

Swaps and Inversions Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3588    Accepted Submission(s): 976 Problem Description Long long ago, there was an integer sequence a.Tonyfang think this se…

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…

题目:   time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output There is an infinite sequence consisting of all positive integers in the increasing order: p = {1, 2, 3, ...}. We performed n swapoper…

题目原文: An inversion in an array a[] is a pair of entries a[i] and a[j] such that i<j but a[i]>a[j]. Given an array, design a linearithmic algorithm to count the number of inversions. 分析: 如果没有性能限制,用插入排序算法可以实现.题目性能被限制在nlogn,又是归并排序的练习题,很显然要实现个归并排序,并在里面计…

先上题目: A - 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…

Swaps and Inversions Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2787    Accepted Submission(s): 1071 Problem Description Long long ago, there was an integer sequence a.Tonyfang think this s…

题目如下: 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 <=…

For a given sequence A={a0,a1,...an−1}A={a0,a1,...an−1}, the number of pairs (i,j)(i,j) where ai>ajai>aj and i<ji<j, is called the number of inversions. The number of inversions is equal to the number of swaps of Bubble Sort defined in the fol…

For a given sequence A={a0,a1,...an−1}A={a0,a1,...an−1}, the number of pairs (i,j)(i,j) where ai>ajai>aj and i<ji<j, is called the number of inversions. The number of inversions is equal to the number of swaps of Bubble Sort defined in the fol…

题目一:E. Infinite Inversions 这个题目没什么思维量,还比较简单,就是离散化要加上每一个值的后面一个值,然后每一个值放进去的不是1 ,而是这个值与下一个点的差值. 因为这个数代表了一堆数,然后每一次的找到了的逆序对都要乘以这个num. #include <cstdio> #include <cstdlib> #include <cstring> #include <queue> #include <vector> #incl…

思路及代码参考:https://blog.csdn.net/u014800748/article/details/45420085 There is an infinite sequence consisting of all positive integers in the increasing order: p = {1, 2, 3, ...}. We performed n swap operations with this sequence. A swap(a, b) is an ope…

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 -…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址: https://leetcode.com/problems/global-and-local-inversions/description/ 题目描述: We have some permutation A of [0, 1, ..., N - 1], where N is the length of A. The number of (global) inversions…