簡易的歸並排序處理出每一層可能的逆序數,對於反轉q層,相當於q層及以下的數都反轉。。。然後求所有的逆序對和。。。。
C. Mashmokh and Reverse Operation time limit per test 4 seconds memory limit per test 512 megabytes input standard input output standard outputMashmokh'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.
InputThe 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++.
OutputOutput m lines. In the i-th line print the answer (the number of inversions) for the i-th query.
Sample test(s) input2 2 1 4 3 4 1 2 0 2output
0 6 6 0input
1 1 2 3 0 1 1output
0 1 0Note
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].
#include#include #include #include using namespace std; typedef long long int LL; int n,m,a[1<<21]; LL dp[21][2]; void merge_sort(int l,int r,int dep) { if(l>=r) return ; int m=(l+r)/2; merge_sort(l,m,dep-1); merge_sort(m+1,r,dep-1); for(int i=l;i<=m;i++) { dp[dep][0]+=lower_bound(a+m+1,a+r+1,a[i])-(a+m+1); dp[dep][1]+=r-m-(upper_bound(a+m+1,a+r+1,a[i])-(a+m+1)); } sort(a+l,a+r+1); } int main() { scanf("%d",&n); for(int i=1,sz=(1<