传送门
文章目录
- 题意：
- 思路：
题意：
思路：
首先肯定不能模n!n!n!，所以考虑先将a,ba,ba,b做一个逆康托展开，得到a′,b′a',b'a′,b′数组，以及a′+b′=suma'+b'=suma′+b′=sum数组，让后我们可以通过这个数组每个位置乘上权(n−i)!(n-i)!(n−i)!得到他的字典序排名，也就是需要模n!n!n!的数，考虑从低位向前递推，也就是sum[i−1]+=sum[i]mod(n−i+1),sum[i]=sum[i]mod(n−i+1)sum[i-1]+=sum[i]\bmod (n-i+1),sum[i]=sum[i]\bmod (n-i+1)sum[i−1]+=sum[i]mod(n−i+1),sum[i]=sum[i]mod(n−i+1)即可，最后再将其逆回来即可。
// Problem: D. Misha and Permutations Summation
// Contest: Codeforces - Codeforces Round #285 (Div. 2)
// URL: https://codeforces.com/problemset/problem/501/D
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
//#pragma GCC optimize(2)
#include<cstdio>
#include<iostream>
#include<string>
#include<cstring>
#include<map>
#include<cmath>
#include<cctype>
#include<vector>
#include<set>
#include<queue>
#include<algorithm>
#include<sstream>
#include<ctime>
#include<cstdlib>
#include<random>
#include<cassert>
#define X first
#define Y second
#define L (u<<1)
#define R (u<<1|1)
#define pb push_back
#define mk make_pair
#define Mid ((tr[u].l+tr[u].r)>>1)
#define Len(u) (tr[u].r-tr[u].l+1)
#define random(a,b) ((a)+rand()%((b)-(a)+1))
#define db puts("---")
using namespace std;//void rd_cre() { freopen("d://dp//data.txt","w",stdout); srand(time(NULL)); }
//void rd_ac() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//AC.txt","w",stdout); }
//void rd_wa() { freopen("d://dp//data.txt","r",stdin); freopen("d://dp//WA.txt","w",stdout); }typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> PII;const int N=1000010,mod=1e9+7,INF=0x3f3f3f3f;
const double eps=1e-6;int n;
int a[N],b[N];
LL fun[N];
struct Node {int l,r;int cnt;
}tr[N<<2];void pushup(int u) {tr[u].cnt=tr[L].cnt+tr[R].cnt;
}void build(int u,int l,int r) {tr[u]={l,r};if(l==r) {tr[u].cnt=1;return;}build(L,l,Mid); build(R,Mid+1,r);pushup(u);
}void change(int u,int pos) {if(tr[u].l==tr[u].r) {tr[u].cnt=0;return;}if(pos<=Mid) change(L,pos);else change(R,pos);pushup(u);
}int query1(int u,int l,int r) {if(tr[u].l>=l&&tr[u].r<=r) return tr[u].cnt;int ans=0;if(l<=Mid) ans+=query1(L,l,r);if(r>Mid) ans+=query1(R,l,r);return ans;
}int query2(int u,int cnt) {if(tr[u].l==tr[u].r) return tr[u].l;if(tr[L].cnt>cnt) return query2(L,cnt);else return query2(R,cnt-tr[L].cnt);
}int main()
{
// ios::sync_with_stdio(false);
// cin.tie(0); scanf("%d",&n);for(int i=1;i<=n;i++) scanf("%d",&a[i]),a[i]++;for(int i=1;i<=n;i++) scanf("%d",&b[i]),b[i]++;build(1,1,n);for(int i=1;i<=n;i++) {int cnt=query1(1,1,a[i]-1);change(1,a[i]);a[i]=cnt;}build(1,1,n);for(int i=1;i<=n;i++) {int cnt=query1(1,1,b[i]-1);change(1,b[i]);b[i]=cnt;}for(int i=n;i>=1;i--) {a[i]+=b[i];int t=a[i]/(n-i+1);a[i-1]+=t;a[i]%=(n-i+1);}build(1,1,n);for(int i=1;i<=n;i++) {int pos=query2(1,a[i]);change(1,pos);printf("%d%c",pos-1,i==n? '\n':' ');} return 0;
}
/**/
