hdu 1384 Minimum Inversion Number

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11797    Accepted Submission(s): 7240

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

101 3 6 9 0 8 5 7 4 2

 

Sample Output

16

#include<iostream>using namespace std;typedef long long ll;int main(){    int n;    while(cin>>n){        ll i,j,a[5005];        for(i=0;i<n;i++)  cin>>a[i];        ll sum=0,t;        for(i=0;i<n;i++){            t=0;            for(j=i+1;j<n;j++)                if(a[i]>a[j])  t++;            sum+=t;        }        ll min=sum;        for(i=0;i<n;i++){            sum=sum-a[i]+(n-1)-a[i];            if(sum<min)  min=sum;        }        cout<<min<<endl;    }    return 0;}