SGU_180_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

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

乍一看是poj2299

但是有一点这个问题输入并没有保证数字不相同

所以做坐标变换的时候应该在比较值的同时比较一下位置信息

#include <iostream>
#include <stdio.h>
#include <algorithm>
using namespace std;

const int M=65540;
int tree[M];

struct NODE
{
int v,p;
}node[M];

bool cmp1(NODE a,NODE b)
{
if(a.v==b.v)
return a.p<b.p;    //值相同的时候比较下位置信息，这样把相同值离散成不同的小的值得时候
return a.v<b.v;        //不会产生同样数值变成的大的数字，位置到前面去影响结果
}

bool cmp2(NODE a,NODE b)
{
return a.p<b.p;
}

int lowbit(int x)
{
return x&(-x);
}

int getsum(int x)
{
int s=0;
for(int i=x;i;i-=lowbit(i))
s+=tree[i];
return s;
}
void add(int x,int v)
{
for(int i=x;i<M;i+=lowbit(i))
tree[i]+=v;
}
int main()
{
int n;
long long co=0;
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%d",&node[i].v);
node[i].p=i;
}
sort(node,node+n,cmp1);
for(int i=0;i<n;i++)
{
node[i].v=i+1;
}
sort(node,node+n,cmp2);
for(int i=0;i<n;i++)
{
co+=i-getsum(node[i].v-1);
add(node[i].v,1);
}
printf("%I64d\n",co);
return 0;
}