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