2013
12-09

# Minimum Inversion Number

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.

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.

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

10
1 3 6 9 0 8 5 7 4 2

16

#include<stdio.h>
#define maxn 5010
#define lson l,m,rt << 1
#define rson m+1,r,rt << 1 | 1
int tree[maxn<<2],x[maxn];
int min(int a,int b)
{
return a<b?a:b;
}
void PushUP(int rt)
{
tree[rt]=tree[rt<<1]+tree[rt<<1|1];
}
void build(int l,int r,int rt)
{
tree[rt]=0;
if(l==r)
return ;
int m=(l+r)>>1;
build(lson);
build(rson);
}
int query(int L,int R,int l,int r,int rt)
{
if(L<=l&&R>=r)
return tree[rt];
int m=(l+r)>>1;
int ret=0;
if(L<=m)
ret+=query(L,R,lson);
if(R>m)
ret+=query(L,R,rson);
return ret;
}
void update(int p,int l,int r,int rt)
{
if(l==r)
{
tree[rt]++;
return ;
}
int m=(l+r)>>1;
if(p<=m)
update(p,lson);
else
update(p,rson);
PushUP(rt);
}
int main()
{
int i,n,sum;
while(scanf("%d",&n)!=EOF)
{
build(0,n-1,1);
sum=0;
for(i=0;i<n;i++)
{
scanf("%d",&x[i]);
sum+=query(x[i],n-1,0,n-1,1);
update(x[i],0,n-1,1);
}
int ret=sum;
for(i=0;i<n;i++)
{
sum+=n-x[i]-1-x[i];
ret=min(ret,sum);
}
printf("%d\n",ret);
}
return 0;
}

1. 第2题，TCP不支持多播，多播和广播仅应用于UDP。所以B选项是不对的。第2题，TCP不支持多播，多播和广播仅应用于UDP。所以B选项是不对的。