SGU 180 - Inversions（逆序数）

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

[submit]

[forum]

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逆序数对，树状数组+离散化

离散化方法：http://blog.youkuaiyun.com/gokou_ruri/article/details/7723378

先离散化使数据密集，然后对每个数进行处理，一个一个插入树状数组。比如插入一个数a[i]，就更新一遍a[i]置的值，也就是+1，表示出现了一个数字a[i]。Query(a[i])是当前小于等于a[i]的所有数的出现次数，也就是之前插入的数中有多少小于等于a[i]，那么 i-Query(a[i]) 就是当前大于a[i]的数的个数，是满足答案的解，加到sum中。

注意sum要定义成long long

#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;

const int maxn=70000;
int tree[maxn],n,a[maxn],b[maxn];
int Lowbit(int i)
{
    return i&(-i);
}

void Update(int i,int x)
{
    while(i<=n)
    {
        tree[i]+=x;
        i+=Lowbit(i);
    }
}

int Query(int i)
{
    int sum=0;
    while(i>0)
    {
        sum+=tree[i];
        i-=Lowbit(i);
    }
    return sum;
}

int main()
{
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
        b[i]=a[i];
    }
    sort(b,b+n);
    int size=unique(b,b+n)-b;
    long long sum=0;
    for(int i=0;i<n;i++)
    {
        int pos=upper_bound(b,b+size,a[i])-b;
        Update(pos,1);
        sum+=i+1-Query(pos);
    }
    printf("%I64d\n",sum);
    return 0;
}