SGU180（树状数组，逆序对，离散）

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

 

这道题需要离散，树状数组求逆序对是离散后，统计加入该元素时当前数组中

已经存在多少个比它大的数，这就是该数作为逆序对后者的贡献度，然后就可以

求解了，一般需要离散化。

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<iostream>
#define N 70007
using namespace std;

int n;
long long ans;
int b[N],c[N];
struct Node
{
    int zhi,id;
}a[N];

bool cmp(Node x,Node y)
{
    return x.zhi<y.zhi;
}
int lowbit(int x)
{
    return x&(-x);
}
void change(int x,int y)
{
    for (int i=x;i<=n;i+=lowbit(i))
        c[i]+=y;
}
int query(int x)
{
    int res=0;
    for (int i=x;i>=1;i-=lowbit(i))
        res+=c[i];
    return res;    
}
int main()
{
    scanf("%d",&n);
    for (int i=1;i<=n;i++)
    {
        scanf("%d",&a[i].zhi);
        a[i].id=i;    
    }
    sort(a+1,a+n+1,cmp);
    int cnt=0;
    for (int i=1;i<=n;i++)
    {
        if (a[i].zhi!=a[i-1].zhi) b[a[i].id]=++cnt;
        else b[a[i].id]=cnt;
    }
    for (int i=1;i<=n;i++)
    {
        change(b[i],1);
        ans=ans+query(n)-query(b[i]);    
    }
    printf("%lld",ans);
}