Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 9342    Accepted Submission(s): 5739

Problem Description

The inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that satisfy i < j and ai > aj.

For a given sequence of numbers a1, a2, ..., an, if we move the first m >= 0 numbers to the end of the seqence, we will obtain another sequence. There are totally n such sequences as the following:

a1, a2, ..., an-1, an (where m = 0 - the initial seqence)

a2, a3, ..., an, a1 (where m = 1)

a3, a4, ..., an, a1, a2 (where m = 2)

...

an, a1, a2, ..., an-1 (where m = n-1)

You are asked to write a program to find the minimum inversion number out of the above sequences.

 

 

Input

The input consists of a number of test cases. Each case consists of two lines: the first line contains a positive integer n (n <= 5000); the next line contains a permutation of the n integers from 0 to n-1.

 

 

Output

For each case, output the minimum inversion number on a single line.

 

 

Sample Input

10

1 3 6 9 0 8 5 7 4 2

 

 

Sample Output

16

 

 

Author

CHEN, Gaoli

 

 

Source

 

代码：

1 //线段树实现单点更新，并求和 2 #include
       
         3 #define maxn 5001 4 struct node{ 5   int lef,rig,sum; 6   int mid(){ return lef+((rig-lef)>>1) ;} 7 }; 8  node seg[maxn<<2]; 9  int aa[maxn+3];10 void build(int left,int right,int p )11 {12     seg[p].lef=left;13     seg[p].rig=right;14     seg[p].sum=0;15     if(left==right) return ;16     int mid=seg[p].mid();17     build(left,mid,p<<1);18     build(mid+1,right,p<<1|1);19 }20 void updata(int pos,int p,int val)21 {22     if(seg[p].lef==seg[p].rig)23     {24       seg[p].sum+=val;25       return ;26     }27     int mid=seg[p].mid();28     if(pos<=mid) updata(pos,p<<1,val);29     else updata(pos,p<<1|1,val);30    seg[p].sum=seg[p<<1].sum+seg[p<<1|1].sum;31 }32 int query(int be ,int en,int p)33 {34     if(be<=seg[p].lef&&seg[p].rig<=en)35         return seg[p].sum;36     int mid=seg[p].mid();37     int res=0;38     if(be<=mid) res+=query(be ,en ,p<<1);39     if(mid
        
         ans) min=ans;60      }61      printf("%d\n",min);62     }63     return 0 ;64 }