sgu180：Inversions

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

裸的求逆序对，果断树状数组+离散化。
ps：1.排序的条件写<,不要写<=
    2.相同的数，优先级也相同

#include <cstdio>
#include <algorithm>
using namespace std;
const int MAXN = 65540;
int N = 0;
struct number
{
  int v;	
  int node;
}A[MAXN] = {0};
int c[MAXN] = {0}, t[MAXN] = {0};
int top = 0;

bool cmp(const number &P, const number &Q) { return P.v < Q.v;	}

int find(int cur)
{
  int re = 0;
  for(; cur; cur -= cur&(-cur)) re += c[cur];
  return re;	
}
void insert(int cur)
{
  for(; cur <= top; cur += cur&(-cur))
    c[cur] += 1;	
}

int main()
{
  scanf("%d", &N);
  for(int i = 1; i <= N; ++i)
  {
    scanf("%d", &A[i].v);	
    A[i].node = i;
  }
  sort(A+1, A+N+1, cmp);
  A[0].v = -20;
  for(int i = 1; i <= N; ++i)
    t[A[i].node] = A[i-1].v!=A[i].v?++top:top;	

  long long ans = 0;
  for(int i = 1; i <= N; ++i)
  {
    insert(t[i]);
	ans += (long long)(i-find(t[i]));	
  }
  printf("%I64d", ans);
  return 0;	
}