Inversions

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

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#define MAXN 65540

using namespace std;

struct node
{
    long long v;
    int id;
    bool operator<(const node&p)const
    {return v<p.v;}
};
node a[MAXN+10];
long long c[MAXN+10];
long long b[MAXN+10];
int n;

long long lowbit(long long a)
{
    return a&(-a);
}

long long sum(long long a)
{
    long long s=0;
    while(a>0)
    {
        s+=c[a];
        a-=lowbit(a);
    }
    return s;
}

void update(long long a)
{
    while(a<=n+1)
    {
        c[a]++;
        a+=lowbit(a);
    }
}

int main(void)
{
    while(scanf("%d",&n)==1)
    {
        memset(a,0,sizeof(a));
        memset(c,0,sizeof(c));
        memset(b,0,sizeof(b));
        int i;
        for(i=1;i<=n;i++)
        {
            scanf("%lld",&(a[i].v));
            a[i].id=i;
        }
        sort(a+1,a+n+1);
        int pre=-1;
        int prevalue=0;
        for(i=1;i<=n;i++)
        {
            if(pre!=a[i].v)
            {
                pre=a[i].v;
                a[i].v=++prevalue;
            }
            else 
            {
                a[i].v=prevalue;
            }
        }
        for(i=1;i<=n;i++)
        {
            b[a[i].id]=a[i].v;
        }
        long long s=0;
        for(i=n;i>=1;i--)
        {
            update(b[i]);
            s+=sum(b[i]-1);
        }
        printf("%I64d
",s);            
    }
    return 0;
}

如果最多可以交换K  次相邻的数，则最后最小的逆序书是多少

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
//#define MAXN 65540
#define MAXN 100001
using namespace std;

struct node
{
    long long v;
    int id;
    bool operator<(const node&p)const
    {return v<p.v;}
};
node a[MAXN+10];
long long c[MAXN+10];
long long b[MAXN+10];
int n,k; 

long long lowbit(long long a)
{
    return a&(-a);
}

long long sum(long long a)
{
    long long s=0;
    while(a>0)
    {
        s+=c[a];
        a-=lowbit(a);
    }
    return s;
}

void update(long long a)
{
    while(a<=n+1)
    {
        c[a]++;
        a+=lowbit(a);
    }
}

int main(void)
{
    while(scanf("%d%d",&n,&k)==2)
    {
        memset(a,0,sizeof(a));
        memset(c,0,sizeof(c));
        memset(b,0,sizeof(b));
        int i;
        for(i=1;i<=n;i++)
        {
            scanf("%lld",&(a[i].v));
            a[i].id=i;
        }
        sort(a+1,a+n+1);
        int pre=-1;
        int prevalue=0;
        for(i=1;i<=n;i++)
        {
            if(pre!=a[i].v)
            {
                pre=a[i].v;
                a[i].v=++prevalue;
            }
            else 
            {
                a[i].v=prevalue;
            }
        }
        for(i=1;i<=n;i++)
        {
            b[a[i].id]=a[i].v;
        }
        long long s=0;
        for(i=n;i>=1;i--)
        {
            update(b[i]);
            s+=sum(b[i]-1);
        }
        if(s<=k)
            puts("0");
        else
            printf("%I64d
",s-k);         
    }
    return 0;
}