Minimum Inversion Number Array

Minimum Inversion numberTime limit:1000MS Memory Limit:32768KB 64bit IO Format:%i64d &%i64 U SubmitStatusPracticeHDU 1394

Description

The inversion number of a given number sequence A1, A2, ..., 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'll Obtain another sequence. There is 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 is 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 the 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 in a single line.

Sample Input

101 3 6 9 0 8 5 7 4 2

Sample Output

16

1#include <stdio.h>2#include <string.h>3#include <algorithm>4 using namespacestd;5 6 intn,c[5050];7 8 intLowbit (intx)9 {Ten     returnx& (-x); One } A  - voidAddintIintval) - { the      while(i<=N) -     { -c[i]=c[i]+Val; -i=i+lowbit (i); +     } - } +  A intSuminti) at { -     ints=0; -      while(i) -     { -s=s+C[i]; -i=i-lowbit (i); in     } -     returns; to } +  - intMain () the { *     inti,j,k; $     inta[5050];Panax Notoginseng      while(SCANF ("%d", &n)! =EOF) -     { the         intCnt=0; +Memset (c,0,sizeof(c)); A          for(i=1; i<=n;i++) the         { +scanf"%d",&a[i]); -a[i]++; $Cnt=cnt+sum (N)-sum (a[i]); $Add (A[i],1); -         } -         intMi=CNT; the          for(i=1; i<n;i++) -         {Wuyicnt=cnt-(a[i]-1) + (na[i]); the             if(cnt<mi) -Mi=CNT; Wu         } -printf"%d\n", MI); About     } $     return 0; -}

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Minimum Inversion Number array