C - Minimum Inversion Number

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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

1 //线段树专题解题。2 //其实一开始没想通，后来手动模拟了一下整个树建立的过程就全部清楚了。3 //详细AC代码在下面：

1 #include"iostream" 2 #include"algorithm" 3 #include"cstdio" 4 #include"cstring" 5 #include"cmath" 6 #define max(a,b) a>b?a:b 7 #define min(a,b) a
           
            >1;21     Build(lson);//建立左节点22     Build(rson);//建立右节点23 }24 25 void UpData(int p,int l,int r,int rt) {26     if(r==l) {            //找到并更新目标点27         sum[rt]++;28         return ;29     }30     int m=(r+l)>>1;31     if(p<=m) UpData(p,lson);  //如果不是目标点向左右寻找32     if(p >m) UpData(p,rson);33     PushUp(rt);//将更新过的每个点的子节点的和更新。34 }35 36 int Query(int L,int R,int l,int r,int rt) {37     if(L<=l&&R>=r)         //大小超过整个范围38         return sum[rt];  //返回总数39     int m=(r+l)>>1;40     int ret=0;41     if(L<= m) ret += Query(L,R,lson);  //比x[i]大的树的左值和42     if(R > m) ret += Query(L,R,rson);  //比x[i]大的树的右值和43     return ret;44 }45 int x[MX];46 int main() {47     int n;48     int sums;49     char s[2];50     while(~scanf("%d",&n)) {51         sums=0;            52         Build(0,n-1,1);         //【这里应该从0~n-1比较好，从1~n的话0的位置不好放在哪里了。后面也就一样了。】 53         for(int i=0; i