SGU 180-Inversions(树状数组离散化求逆序对数)

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

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Author:Stanislav AngelyukResource:Saratov ST team Spring Contest #1Date:18.05.2003

 

/***************************** author:crazy_石头* date：2014/01/15* algorithm:BIT* Pro：SGU180-inversions***************************/#include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>#include <algorithm>#include <queue>#include <map>#include <string>using namespace std;#define INF 1<<29#define eps 1e-8#define A system("pause")#define rep(i,h,n) for(int i=(h);i<=(n);i++)#define ms(a,b) memset((a),(b),sizeof(a))#define LL __int64const int maxn=70000+10;struct Node{    LL id,num;}e[maxn];LL C[maxn],n,h[maxn];inline bool cmp(Node E,Node D){    return E.num<D.num;}inline int lowbit(int x){    return x&-x;}inline void update(int x,int d){    while(x<=n)    {        C[x]+=d;        x+=lowbit(x);    }}inline LL getsum(int x){    LL ret=0;    while(x>0)    {        ret+=C[x];        x-=lowbit(x);    }    return ret;}int main(){    while(~scanf("%d",&n))    {        ms(C,0);        rep(i,1,n)        {            scanf("%I64d",&e[i].num);            e[i].id=i;        }        sort(e+1,e+n+1,cmp);        h[e[1].id]=1;        int cnt=1;        rep(i,2,n)        {            if(e[i].num!=e[i-1].num) cnt++;            h[e[i].id]=cnt;        }        LL sum=0;        rep(i,1,n)        {            update(h[i],1);            sum+=i-getsum(h[i]);        }        printf("%I64d\n",sum);    }    return 0;}