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],n,tot,r[N];struct node{    LL x,s,id;} a[N];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;}