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题目链接：
C. Mashmokh and Reverse Operation
Mashmokh‘s boss, Bimokh, didn‘t like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh‘s team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn‘t able to solve them. That‘s why he asked you to help him with these tasks. One of these tasks is the following.
You have an array a of length 2n and m queries on it. The i-th query is described by an integer qi. In order to perform the i-th query you must:
Given initial array a and all the queries. Answer all the queries. Please, note that the changes from some query is saved for further queries.
The first line of input contains a single integer n (0 ≤ n ≤ 20).
The second line of input contains 2n space-separated integers a[1], a[2], ..., a[2n] (1 ≤ a[i] ≤ 109), the initial array.
The third line of input contains a single integer m (1 ≤ m ≤ 106).
The fourth line of input contains m space-separated integers q1, q2, ..., qm (0 ≤ qi ≤ n), the queries.
Note: since the size of the input and output could be very large, don‘t use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.
Output m lines. In the i-th line print the answer (the number of inversions) for the i-th query.
2
2 1 4 3
4
1 2 0 2
0
6
6
0
1
1 2
3
0 1 1
0
1
0
If we reverse an array x[1], x[2], ..., x[n] it becomes new array y[1], y[2], ..., y[n], where y[i] = x[n - i + 1] for each i.
The number of inversions of an array x[1], x[2], ..., x[n] is the number of pairs of indices i, j such that: i < j and x[i] > x[j].
题意：
问2^n个数进行如题的操作每次操作后逆序对是多少;
思路：
像归并排序那样先划分,然后再求出每个划分里面的逆序对,合并后再求两个区间之间的逆序对,倒序把逆序对和正序对的数目交换了;
最近抄代码抄的厉害,哎;
AC代码：
#include <bits/stdc++.h> using namespace std; #define Riep(n) for(int i=1;i<=n;i++) #define Riop(n) for(int i=0;i<n;i++) #define Rjep(n) for(int j=1;j<=n;j++) #define Rjop(n) for(int j=0;j<n;j++) #define mst(ss,b) memset(ss,b,sizeof(ss)); typedef long long LL; const LL mod=1e9+7; const double PI=acos(-1.0); const int inf=0x3f3f3f3f; const int N=1e6+5e5; LL sum[22][2]; int n,a[1<<21]; void dfs(int l,int r,int deep) { if(l>=r)return ; int mid=(l+r)>>1; dfs(l,mid,deep-1); dfs(mid+1,r,deep-1); for(int i=l;i<=mid;i++) { int temp=lower_bound(a+mid+1,a+r+1,a[i])-(a+mid+1); sum[deep][0]+=(LL)temp; temp=r-mid-(upper_bound(a+mid+1,a+r+1,a[i])-(a+mid+1)); sum[deep][1]+=(LL)temp; } sort(a+l,a+r+1); } int main() { scanf("%d",&n); int y=(1<<n); Riep(y)scanf("%d",a+i); dfs(1,y,n); int q,x; scanf("%d",&q); while(q--) { scanf("%d",&x); while(x) { swap(sum[x][0],sum[x][1]); x--; } LL ans=0; Riep(n)ans+=sum[i][0]; printf("%I64d\n",ans); } return 0; }
codeforces 414C C. Mashmokh and Reverse Operation(归并排序求逆序对)
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原文地址：http://www.cnblogs.com/zhangchengc919/p/5467305.html
