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;}