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[],s[],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+);
         }
     }
     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)/;
          ans+=merge_s(l,mid);
          ans+=merge_s(mid+,r);
          ans+=merge(l,mid,mid+,r);
     }
     return ans;
  }

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