sgu 180 - Inversions （离散化+树状数组）

Posted by 111qqz on Thursday, August 6, 2015

TOC

• Inversions

**Time Limit:**250MS **Memory Limit:**4096KB 64bit IO Format:%I64d & %I64u

Submit Status

Description

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<=iA[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

/*************************************************************************
> File Name: code/sgu/180.cpp
> Author: 111qqz
> Email: rkz2013@126.com
> Created Time: 2015年08月06日 星期四 16时40分53秒
************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#define y0 abc111qqz
#define y1 hust111qqz
#define yn hez111qqz
#define j1 cute111qqz
#define tm crazy111qqz
#define lr dying111qqz
using namespace std;
#define REP(i, n) for (int i=0;i<int(n);++i)
typedef long long LL;
typedef unsigned long long ULL;
const int inf = 0x7fffffff;
const int N=7E4+7;
struct Q
{
int val;
int id;
}q[N];
int c[N];
int ref[N];
int n;
bool cmp(Q a,Q b)
{
if (a.val<b.val)
return true;
return false;
}

int lowbit( int x)
{
return x&(-x);
}
void update( int x,int delta)
{
for ( int i = x; i < N ; i=i+lowbit(i) )
{
c[i] = c[i] + delta;
}
}
int Sum( int x)
{
int res  =0;
for ( int i = x; i >= 1 ; i = i-lowbit(i))
{
res = res + c[i];
}
return res;
}
int main()
{
while (scanf("%d",&n)!=EOF)
{

memset(c,0,sizeof(c));
for ( int i = 1; i <= n ; i++ )
{
scanf("%d",&q[i].val);
q[i].id  = i ;
}
sort(q+1,q+n+1,cmp);
for ( int i = 1; i <= n ; i++ )
{
if (q[i].val!=q[i-1].val)
{
ref[q[i].id]=i;
}
else
{
ref[q[i].id]=ref[q[i-1].id];
}
}

for ( int i = 1 ;i <= n ; i ++) cout<<ref[i]<<endl;
LL  ans = 0;
for ( int i = 1 ; i <= n ; i++ )
{
update(ref[i],1);
ans = ans + i-Sum(ref[i]);
}
cout<<ans<<endl;
}

return 0;
}

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