ACM暑期集训26
Minimum Inversion Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24289 Accepted Submission(s): 14383
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
分析：
可利用归并排序求出第一次的逆序数和，在用下面的公式求解最小的逆序数和；
如果每次都进行如上操作，那么最多进行n-1次后会与原序列重合；
这个序列的元素是从0---（n-1），当某个元素位于首位置时，其本身的数值就是后面的逆序个数；
每次将它移至末尾，原来的逆序即变成了正序，正序变逆；由此可得 sum=sum-a[i]+（n-a[i]-1）；
#include
#include
#include
#include
#include
using namespace std;
#define Min(a,b)(a