mes LG 減少 輸出 where cas ID seq turn

描述

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.

輸入

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.

輸出

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

樣例輸入

10
1 3 6 9 0 8 5 7 4 2

樣例輸出

16

解題思路：用歸並排序先求出原來序列的逆對數，循環從0到n-1,把前面的n個元素移動到後面，

因為元素的大小是從0到n-1，所以移動到最後逆序對減少了a[i]個，增加了（n-1-a[i]）個，記錄最小的就行了、

#include <iostream>
#include <algorithm>
#include <cstdio>
using
 
 namespace std;
#define Max 5000
int a[Max+5],temp[Max+5],c[Max+5];
long long sum;
void memrge(int left,int mid,int right)
{
    int i=left,j=mid+1,k=left;
    while(i<=mid&&j<=right)
    {
        if(a[i]>a[j])
        {
            temp[k++]=a[j++];
            sum+=mid-i+1;
        }
        else temp[k++]=a[i++];
    }
    while(i<=mid) temp[k++]=a[i++];
    while(j<=right) temp[k++]=a[j++];  
    for(int i=left;i<=right;i++) a[i]=temp[i];
}
int fen(int left,int right)
{
    if(left<right)
    {
        int mid=(left+right)/2;
        fen(left,mid);
        fen(mid+1,right);
        memrge(left,mid,right);
    }
}
int main()
{
   int n,m,k;
   while(cin>>n)
   {
        sum=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            c[i]=a[i];
        }
        fen(0,n-1);
        int Min=sum;
        for(int i=0;i<n;i++)
        {
            sum=sum-c[i]+(n-1-c[i]);
            if(sum<Min) Min=sum;
        } //將元素移到最後
        cout<<Min<<endl;
   }
    return 0;
}

TOJ1254: Minimum Inversion Number