Codeforces Round #285 (Div.1 B & Div.2 D) Misha and Permutations Summation --二分+树状数组

题意：给出两个排列，求出每个排列在全排列的排行，相加，模上n!（全排列个数)得出一个数k，求出排行为k的排列。

解法：首先要得出定位方法，即知道某个排列是第几个排列。比如 (0, 1, 2), (0, 2, 1), (1, 0, 2), (1, 2, 0), (2, 0, 1), (2, 1, 0).

拿排列(1,2,0)来说，首位是1,前面有cnt=1个小于1的没被用过的数(0)，所以它的排行要加上(cnt=1)*2!，第二位为2，因为1已经放了，所以小于2的只有0了，即cnt=1个，所以，排行又要加上(cnt=1)*1!,所以排行为3.

推出一般性结论:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
using namespace std;
#define N 200107

int p1
,p2
,p3
,c
;
int n;

int lowbit(int x) { return x&-x; }
void modify(int x,int val)
{
while(x <= n+10)
{
c[x] += val;
x += lowbit(x);
}
}
int getsum(int x)
{
int res = 0;
while(x > 0)
{
res += c[x];
x -= lowbit(x);
}
return res;
}

int main()
{
int i,j,x;
while(scanf("%d",&n)!=EOF)
{
memset(c,0,sizeof(c));
for(i=1;i<=n;i++) modify(i,1);
for(i=1;i<=n;i++)
{
scanf("%d",&x);
x++;
p1[i] = getsum(x-1);
modify(x,-1);
}
memset(c,0,sizeof(c));
for(i=1;i<=n;i++) modify(i,1);
for(i=1;i<=n;i++)
{
scanf("%d",&x);
x++;
p2[i] = getsum(x-1);
modify(x,-1);
}
memset(p3,0,sizeof(p3));
for(i=n;i>=1;i--)
{
p3[i] += p1[i]+p2[i];
if(p3[i] >= (n-i+1))
{
p3[i] = p3[i]-(n-i+1);
if(i != 1) p3[i-1]++;
}
}
memset(c,0,sizeof(c));
for(i=1;i<=n;i++) modify(i,1);
//        for(i=1;i<=n;i++)
//            cout<<p3[i]<<" ";
//        cout<<endl;
for(i=1;i<=n;i++)
{
int low = 1, high = n;
while(low <= high)
{
int mid = (low+high)/2;
if(getsum(mid-1) > p3[i])
high = mid-1;
else if(getsum(mid-1) == p3[i] && getsum(mid)-getsum(mid-1) == 1)
high = mid-1;
else if(getsum(mid-1) == p3[i] && getsum(mid)-getsum(mid-1) < 1)
low = mid+1;
else if(getsum(mid-1) < p3[i])
low = mid+1;
}
modify(low,-1);
printf("%d ",low-1);
}
puts("");
}
return 0;
}

View Code
比赛中写的代码，没有最简化，有很多冗余和多此一举的地方。