SGU180:Inversions(树状数组)

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

题意：

求逆序数

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <list>
#include <algorithm>
#include <climits>
using namespace std;

#define lson 2*i
#define rson 2*i+1
#define LS l,mid,lson
#define RS mid+1,r,rson
#define UP(i,x,y) for(i=x;i<=y;i++)
#define DOWN(i,x,y) for(i=x;i>=y;i--)
#define MEM(a,x) memset(a,x,sizeof(a))
#define W(a) while(a)
#define gcd(a,b) __gcd(a,b)
#define LL long long
#define N 67000
#define INF 0x3f3f3f3f
#define EXP 1e-8
#define lowbit(x) (x&-x)
const int mod = 1e9+7;

LL c
,n,tot,r
;

struct node
{
LL x,s,id;
} a
;

int cmp(node a,node b)
{
if(a.x!=b.x)
return a.x<b.x;
return a.id<b.id;
}

LL sum(LL x)
{
LL ret = 0;
while(x>0)
{
ret+=c[x];
x-=lowbit(x);
}
return ret;
}

void add(LL x,LL d)
{
while(x<=n)
{
c[x]+=d;
x+=lowbit(x);
}
}

int main()
{
LL i,j,k;
while(~scanf("%lld",&n))
{
MEM(c,0);
for(i = 1; i<=n; i++)
{
scanf("%lld",&a[i].x);
a[i].id = i;
}
sort(a+1,a+1+n,cmp);
for(i = 1; i<=n; i++)
{
r[a[i].id] = i;
}
LL ans = 0;
for(i = 1; i<=n; i++)
{
add(r[i],1);
ans+=(i-sum(r[i]));
}
printf("%lld\n",ans);
}
return 0;
}