对于n个数，可以做k次移动，每次移动可以互换相邻位置的两个数，问最少 number of pair (i,j) where 1≤i<j≤n and ai>aj. 如果不移动的话，ans=’n个数的逆序对数‘，移动k次会减少k个 归并排序求逆序对数： #include "stdio.h"#include "string.h"#include "math.h"int b[1

网页链接 Inversion Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 1962    Accepted Submission(s): 765 Problem Description bobo ha

Problem Description bobo has a sequence a 1,a 2,…,a n. He is allowed to swap two adjacent numbers for no more than k times. Find the minimum number of inversions after his swaps. Note: The nu

题目链接：hdu 4911 Inversion 题目大意：给定一个序列，有k次机会交换相邻两个位置的数，问说最后序列的逆序对数最少为多少。 解题思路：每交换一次一定可以减少一个逆序对，所以问题转换成如何求逆序对数。 #include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef

上题目 Inversion Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 2924 Accepted Submission(s): 1081 Problem Description bobo has a sequence a1,a