康托 Codeforces Round #285 (Div. 2) D. Misha and Permutations Summation

然后Rank和为(p1[i]+p2[i])*(n-1)! + ... + (p1[n]+p2[n])*0! = p3[1]*(n-1)! + ... + p3[n]*0! ，但是得出的表达式可能不是规整的形式，这是我们需要检测一边，从后往前扫，如果p3[i] >= (n-i+1), 说明第 i 项已经超过 (n-i+1)*(n-i), 那么就应进位到(n-i+1)!, 即p3[i-1]+=1，依此类推。

D. Misha and Permutations Summation

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Let's define the sum of two permutations p andq of numbers0, 1, ..., (n - 1) as permutation, wherePerm(x) is the x-th lexicographically permutation of numbers0, 1, ..., (n - 1) (counting from zero), andOrd(p) is the number of permutationp in the lexicographical order.

For example, Perm(0) = (0, 1, ..., n - 2, n - 1),Perm(n! - 1) = (n - 1, n - 2, ..., 1, 0)

Misha has two permutations, p and q. Your task is to find their sum.

Permutation a = (a₀, a₁, ..., a_n - 1) is called to be lexicographically smaller than permutationb = (b₀, b₁, ..., b_n - 1), if for somek following conditions hold: a₀ = b₀, a₁ = b₁, ..., a_k - 1 = b_k - 1, a_k < b_k.

Input

The first line contains an integer n (1 ≤ n ≤ 200 000).

The second line contains n distinct integers from0 ton - 1, separated by a space, forming permutationp.

The third line contains n distinct integers from0 ton - 1, separated by spaces, forming permutationq.

Output

Print n distinct integers from 0 to n - 1, forming the sum of the given permutations. Separate the numbers by spaces.

Sample test(s)

Input

20 10 1

Output

0 1

Input

20 11 0

Output

1 0

Input

31 2 02 1 0

Output

1 0 2

Note

Permutations of numbers from 0 to 1 in the lexicographical order: (0, 1), (1, 0).

In the first sample Ord(p) = 0 andOrd(q) = 0, so the answer is.

In the second sample Ord(p) = 0 andOrd(q) = 1, so the answer is.

Permutations of numbers from 0 to 2 in the lexicographical order: (0, 1, 2), (0, 2, 1), (1, 0, 2), (1, 2, 0), (2, 0, 1), (2, 1, 0).

In the third sample Ord(p) = 3 andOrd(q) = 5, so the answer is.

#include<iostream>#include<map>#include<string>#include<cstring>#include<cstdio>#include<cstdlib>#include<cmath>#include<queue>#include<vector>#include<algorithm>using namespace std;int bit[200010],n;void update(int x,int val){for(int i=x;i<=n;i+=i&-i)bit[i]+=val;}int psum(int x){int sum=0;for(int i=x;i>0;i-=i&-i)sum+=bit[i];return sum;}int p[200010],a[200010];int main(){cin>>n;for(int i=1;i<=n;i++)update(i,1);for(int i=0;i<n;i++){cin>>p[i];a[i]=psum(p[i]);update(p[i]+1,-1);}for(int i=1;i<=n;i++)update(i,1);for(int i=0;i<n;i++){cin>>p[i];a[i]+=psum(p[i]);update(p[i]+1,-1);}for(int i=n-1;i>=0;i--){if(i)a[i-1]+=a[i]/(n-i);a[i]%=n-i;}memset(bit,0,sizeof(bit));for(int i=1;i<=n;i++)update(i,1);for(int i=0;i<n;i++){int l=1,r=n;while(l<=r){int m=(l+r)>>1;if(psum(m)-1<a[i])l=m+1;elser=m-1;}if(i)cout<<" ";cout<<l-1;update(l,-1);}}