Minimum Inversion Number

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

 

Sample Output

 

Author

CHEN, Gaoli

 1 #include<cstdio> 2  3 int t[5010]; 4 int n,ans,sum; 5 int main() 6 { 7     while (scanf("%d",&n)!=EOF) 8     { 9         sum = 0;10         for (int i=1; i<=n; ++i)11             scanf("%d",&t[i]);12         for (int i=1; i<n; ++i)13             for (int j=i+1; j<=n; ++j)14                 if (t[i]>t[j]) sum++;15         ans = sum;16         for (int i=n; i>=1; --i)17         {18             sum -= n-1-t[i];19             sum += t[i];20             if (sum < ans) ans = sum;21         }22         printf("%d\n",ans);23     }24     return 0;25 }