hdu 4911Inversion

链接：http://acm.hdu.edu.cn/showproblem.php?pid=4911

题意:给定一个序列和k，求在k次交换之后序列的逆序数，只能相邻两个数交换且只有左边的数大于右边时才能交换

代码如下

#include<cstdio>  
#include<iostream>  
#include<algorithm>
#include<math.h> 
#include<string.h>  
#include<vector> 
#include<queue>
#include<iterator>
#include<vector>
#include<set>
#define dinf 0x3f3f3f3f
typedef long long ll;
//const int Max=(1<<16)+10;
using namespace std;
#define SIZE 100000005


ll a[10000005],s[10000005],ans,k;

ll merge(int l1,int r1,int l2,int r2)
{
    ans=0l;
    int i,j,k;
    for(i=l1;i<=r2;i++)
        s[i]=a[i];
    
    i=l1;j=l2;k=l1;
    while(i<=r1 && j<=r2)
    {
        if(s[i]<=s[j])
            a[k++]=s[i++];
        else
        {
            a[k++]=s[j++];
            ans+=(r1-i+1);
        }
    }
    while(i<=r1)
        a[k++]=s[i++];
    while(j<=r2)
        a[k++]=s[j++];
    return ans;
 } 
 
ll merge_s(int l,int r)
{
    ll ans=0l;
     if(l<r)
     {
         int mid=(l+r)/2;
         ans+=merge_s(l,mid);
         ans+=merge_s(mid+1,r);
         ans+=merge(l,mid,mid+1,r);
    }
    return ans;
 }


int main()
{
    int n;
    while(~scanf("%d %lld",&n,&k))
    {
        for(int i=1;i<=n;i++)
            scanf("%lld",&a[i]);
        ans=merge_s(1,n);
        if(k>=ans)
            printf("0
");
        else
            printf("%lld
",ans-k);
    }
    return 0;
}