codeforces 501D Misha and Permutations Summation（康拓展开+数据结构）
题目链接
Let's define the sum of two permutations p and q of numbers 0, 1, ..., (n - 1) as permutation 
, where Perm(x) is the x-th lexicographically permutation of numbers 0, 1, ..., (n - 1) (counting from zero), and Ord(p) is the number of permutation p 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, ..., an - 1) is called to be lexicographically smaller than permutation b = (b0, b1, ..., bn - 1), if for some kfollowing conditions hold: a0 = b0, a1 = b1, ..., ak - 1 = bk - 1, ak < bk.
The first line contains an integer n (1 ≤ n ≤ 200 000).
The second line contains n distinct integers from 0 to n - 1, separated by a space, forming permutation p.
The third line contains n distinct integers from 0 to n - 1, separated by spaces, forming permutation q.
Print n distinct integers from 0 to n - 1, forming the sum of the given permutations. Separate the numbers by spaces.
2 0 1 0 1
0 1
2 0 1 1 0
1 0
3 1 2 0 2 1 0
1 0 2
Permutations of numbers from 0 to 1 in the lexicographical order: (0, 1), (1, 0).
In the first sample Ord(p) = 0 and Ord(q) = 0, so the answer is 
.
In the second sample Ord(p) = 0 and Ord(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 and Ord(q) = 5, so the answer is 
.
题意：已知两个0到n-1的n个数的排列a，b。ord（a）在0到n-1的数的所有排列中的按字典序排序后的位置。ord（b）同理。输出排列x，ord（x）=（ord（a）+ord（b））%n！。
题解：首先要了解康拓展开：康拓展开。
x=a[n]*(n-1)!+a[n-1]*(n-2)!+.......a[1]*0!。，0<=a[i]<i。
由于数据范围太大，直接求出ord显然是不行的。我们可以根据a，b的康拓展开式，直接求出x的康拓展开式。
因为a[i]<=i-1。
所以，排列a康拓展开有可能的最大值为：(n-1)*(n-1)!+(n-2)*(n-2)!+.....(i-1)*(i-1)!+...0*0!
同理排列b康拓展开的最大值为：(n-1)*(n-1)!+(n-2)*(n-2)!+.....(i-1)*(i-1)!+...0*0!
将两个排列的康拓展开相加可得：n!+(n-1)*(n-1)!+(n-2)*(n-2)!+....(i-1)*(i-1)!+....0*0!
上式%n!，则可得：(n-1)*(n-1)!+(n-2)*(n-2)!+.........(i-1)*(i-1)!+....0*0!
由此可见，两个排列的康拓展开之和，最大到达n!的数量级，对n!取模过后,剩下的就是一个排列的康拓展开。我们把这个康拓展开逆展开，就解决问题了。
我们可以用类似高精度加法的方法，求出两个排列的康拓展开的和%n!的康拓展开。然后用树状数组+二分的方法，将康拓展开逆展开。复杂度为O(n*lgn*lgn)。当然也可以用平衡树或者线段树，不用二分，复杂度为O(n*lgn)。
我的代码是树状数组+二分的方法，代码如下：
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<string>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<stdlib.h>
#include<vector>
#define inff 0x3fffffff
#define nn 210000
#define mod 1000000007
typedef long long LL;
const LL inf64=inff*(LL)inff;
using namespace std;
int n;
int a[nn],b[nn];
int ans[nn];
int c[nn];
int dp[nn];
int lowbit(int x)
{
return x&(-x);
}
void modify(int x,int y)
{
for(int i=x;i<=n;i+=lowbit(i))
{
c[i]+=y;
}
}
int getsum(int x)
{
int sum=0;
for(int i=x;i>=1;i-=lowbit(i))
sum+=c[i];
return sum;
}
int main()
{
int i;
while(scanf("%d",&n)!=EOF)
{
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
for(i=1;i<=n;i++)
{
scanf("%d",&b[i]);
}
int ix;
memset(c,0,sizeof(c));
for(i=1;i<=n;i++)
{
ix=getsum(a[i]);
modify(a[i]+1,1);
a[i]=a[i]-ix;
}
memset(c,0,sizeof(c));
for(i=1;i<=n;i++)
{
ix=getsum(b[i]);
modify(b[i]+1,1);
b[i]=b[i]-ix;
}
ix=0;
for(i=n;i>=1;i--)
{
ans[i]=(a[i]+b[i]+ix)%(n-i+1);
ix=(a[i]+b[i]+ix)/(n-i+1);
}
for(i=1;i<=n+1;i++)
dp[i]=i;
memset(c,1,sizeof(1));
int l,r,mid;
for(i=1;i<=n;i++)
{
l=1,r=n;
while(l<r)
{
mid=(l+r)/2;
if(mid-getsum(mid)<ans[i]+1)
l=mid+1;
else
r=mid;
}
ans[i]=l;
modify(l,1);
}
for(i=1;i<=n;i++)
{
printf("%d%c",ans[i]-1,i==n?'\n':' ');
}
}
return 0;
}
线段树做法代码如下：
#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<math.h>
#define nn 210000
#define eps 1e-8
#define inff 0x7fffffff
#define lson rt<<1,l,m
#define rson rt<<1|1,m+1,r
#define mod 20071027
using namespace std;
typedef __int64 LL;
typedef unsigned long long LLU;
int n;
int a[nn],b[nn],c[nn];
int L[nn*4],R[nn*4],sum[nn*4];
void push_up(int id)
{
sum[id]=sum[id<<1]+sum[id<<1|1];
}
void build(int id,int l,int r)
{
L[id]=l,R[id]=r;
if(l==r)
{
sum[id]=1;
return ;
}
int m=(l+r)>>1;
build(2*id,l,m);
build(2*id+1,m+1,r);
push_up(id);
}
void update(int id,int x)
{
if(L[id]==R[id])
{
sum[id]=0;
return ;
}
int m=(L[id]+R[id])>>1;
if(x<=m)
update(2*id,x);
else
update(2*id+1,x);
push_up(id);
}
int query(int id,int l,int r)
{
if(L[id]>=l&&R[id]<=r)
{
return sum[id];
}
int m=(L[id]+R[id])>>1;
int re=0;
if(l<=m)
re+=query(2*id,l,r);
if(r>m)
re+=query(2*id+1,l,r);
return re;
}
int ask(int id,int x)
{
if(L[id]==R[id])
{
sum[id]=0;
return L[id];
}
int re;
if(sum[2*id]>=x)
re=ask(2*id,x);
else
re=ask(2*id+1,x-sum[2*id]);
push_up(id);
return re;
}
int main()
{
int i,x;
while(scanf("%d",&n)!=EOF)
{
build(1,0,n-1);
for(i=1;i<=n;i++)
{
scanf("%d",&x);
update(1,x);
a[i]=query(1,0,x);
}
build(1,0,n-1);
for(i=1;i<=n;i++)
{
scanf("%d",&x);
update(1,x);
b[i]=query(1,0,x);
}
int ix=0;
build(1,0,n-1);
for(i=n;i>=1;i--)
{
c[i]=(a[i]+b[i]+ix)%(n-i+1);
ix=(a[i]+b[i]+ix)/(n-i+1);
}
for(i=1;i<=n;i++)
{
printf("%d%c",ask(1,c[i]+1),i==n?'\n':' ');
}
}
return 0;
}