SGU 180-Inversions(树状数组离散化求逆序对数)

180. Inversions

time limit per test: 0.25 sec.
memory limit per test: 4096 KB

input: standard
output: standard

 

There are N integers (1<=N<=65537) A1, A2,.. AN (0<=Ai<=10^9). You need to find amount of such pairs (i, j) that 1<=i<j<=N and A[i]>A[j].

Input

The first line of the input contains the number N. The second line contains N numbers A1...AN.

Output

Write amount of such pairs.

Sample test(s)

Input

5 2 3 1 5 4

Output

3

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Author:	Stanislav Angelyuk
Resource:	Saratov ST team Spring Contest #1
Date:	18.05.2003

 

/****************************
* author:crazy_石头
* date：2014/01/15
* algorithm:BIT
* Pro：SGU180-inversions
***************************/
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <map>
#include <string>

using namespace std;

#define INF 1<<29
#define eps 1e-8
#define A system("pause")
#define rep(i,h,n) for(int i=(h);i<=(n);i++)
#define ms(a,b) memset((a),(b),sizeof(a))
#define LL __int64
const int maxn=70000+10;
struct Node
{
    LL id,num;
}e[maxn];
LL C[maxn],n,h[maxn];

inline bool cmp(Node E,Node D)
{
    return E.num<D.num;
}

inline int lowbit(int x)
{
    return x&-x;
}

inline void update(int x,int d)
{
    while(x<=n)
    {
        C[x]+=d;
        x+=lowbit(x);
    }
}

inline LL getsum(int x)
{
    LL ret=0;
    while(x>0)
    {
        ret+=C[x];
        x-=lowbit(x);
    }
    return ret;
}

int main()
{
    while(~scanf("%d",&n))
    {
        ms(C,0);
        rep(i,1,n)
        {
            scanf("%I64d",&e[i].num);
            e[i].id=i;
        }
        sort(e+1,e+n+1,cmp);
        h[e[1].id]=1;
        int cnt=1;
        rep(i,2,n)
        {
            if(e[i].num!=e[i-1].num) cnt++;
            h[e[i].id]=cnt;
        }
        LL sum=0;
        rep(i,1,n)
        {
            update(h[i],1);
            sum+=i-getsum(h[i]);
        }
        printf("%I64d\n",sum);
    }
    return 0;
}