題目鏈接：
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:
- split the array into?2n?-?qi?parts, where each part is a subarray consisting of?2qi?numbers; the?j-th subarray?(1?≤?j?≤?2n?-?qi)?should contain the elements?a[(j?-?1)·2qi?+?1],?a[(j?-?1)·2qi?+?2],?...,?a[(j?-?1)·2qi?+?2qi];
- reverse each of the subarrays;
- join them into a single array in the same order (this array becomes new array?a);
- output the number of inversions in the new?a.
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.
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codeforces排名，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++.
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codeforces怎么提交？Output?m?lines. In the?i-th line print the answer (the number of inversions) for the?i-th query.
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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].
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codeforces積分。?
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codeforces卡，?
題意：
問2^n個數進行如題的操作每次操作后逆序對是多少;
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思路：
像歸并排序那樣先劃分,然后再求出每個劃分里面的逆序對,合并后再求兩個區間之間的逆序對,倒序把逆序對和正序對的數目交換了;
codeforces進不去，最近抄代碼抄的厲害,哎;
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AC代碼：
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#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; }
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