Topic Description
Farmer John is arranging his 
cows in a line to take a photo (
).
The height of the 
th cow in sequence is 
,
and the heights of all cows are distinct.
As with all photographs of his cows, FJ wants this one to come out looking as nice as possible. He decides that cow looks "unbalanced" if 
and 
differ
by more than factor of 2, where and are
the number of cows taller than on her left and right, respectively. That is, is
unbalanced if the larger of and is
strictly more than twice the smaller of these two numbers. FJ is hoping that not too many of his cows are unbalanced.
Please help FJ compute the total number of unbalanced cows.
FJ is arranging for his N cows to stand in a row to take pictures. (1<=N<=100,000) The height of the first cow in the sequence is h[i], and all cows in the sequence have different heights.
Like all his cow photographs, FJ wants this to look as good as possible. He believed that if the number of L[i] and R[i] were more than two times different, the first cow would be unbalanced. (L[i] and R[i] represent the number of cows on the left and right sides of the first cow higher than hers, respectively). If the number of the larger in L[i] and R[i] is more than double that of the smaller, the cow is also unbalanced. FJ doesn't want him to have too many cow imbalances.
Help FJ calculate the unbalanced cow population.
Input and output format
Input format:
The first line of input contains . The next lines
contain 
, each a nonnegative integer at most 1,000,000,000.
The first line is an integer N. The next N rows consist of H[1] to H[n], each of which is a non-negative integer (no more than 1,000,000,000).
Please output a count of the number of cows that are unbalanced.
Please export an unbalanced number of cows.
Input and Output Samples
7 34 6 23 0 5 99 2
3
This is essentially to ask you to reverse the order, but foolishly I went up and chose to fight violence... (So I'm really a scum)
Inverse order pair is just a tree array, discretize it, and then traverse it.
Sixty points of violence:
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
inline int read(void){
int x=0,f=1;
char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch)){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
const int maxn=100009;
typedef long long LL;
struct hehe{
int s; LL h;
}H[maxn];
inline bool cmp(const hehe&a,const hehe&b){
return a.h<b.h;
}
int main(){
// freopen("photo.in","r",stdin);
// freopen("photo.out","w",stdout);
int n=read(),ans=0;
for(int i=1;i<=n;i++){
H[i].h=read(); H[i].s=i;
}
sort(H+1,H+n+1,cmp);
for(int i=1;i<=n;i++){
int l=0,r=0;
for(int j=i+1;j<=n;j++){
if(H[i].s>H[j].s) l++;
if(H[i].s<H[j].s) r++;
}
// cout<<l<<' '<<r<<endl;
if((l==0&&r!=0)||(l!=0&&r==0)){ans++;continue;}
else if(l==0&&r==0) continue;
else{
int xx=max(l,r),yy=min(l,r);
if(xx/yy>2) ans++;
else if((xx/yy)==2&&yy*2!=xx) ans++;
}
}
cout<<ans<<endl;
return 0;
} //He did not use n^2.
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<queue>
using namespace std;
const int maxn=100009;
int n,a[maxn],b[maxn];
int l[maxn],r[maxn],f[maxn];
inline int lowbit(int x){
return x&-x;
}
inline void add(int x){
for(;x<=n;x+=lowbit(x))
f[x]++;
}
inline int sum(int x){
int ans=0;
for(;x;x-=lowbit(x))
ans+=f[x];
return ans;
}
int main(){
freopen("photo.in","r",stdin);
freopen("photo.out","w",stdout);
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]),b[i]=a[i];
sort(b+1,b+n+1);
for(int i=1;i<=n;i++)
a[i]=lower_bound(b+1,b+n+1,a[i])-b;
for(int i=1;i<=n;i++)
l[i]=i-1-sum(a[i]),add(a[i]);//Looking for an inverse pair.
memset(f,0,sizeof(f));
for(int i=n;i;i--)
r[i]=n-i-sum(a[i]),add(a[i]);//In reverse order.
int ans=0;
for(int i=1;i<=n;i++){
if(l[i]==0&&r[i]==0) continue;
if(!l[i]||!r[i]) ans++;
else{
int xx=max(l[i],r[i]),yy=min(l[i],r[i]);
if(xx/yy>2) ans++;
else if(xx/yy==2&&yy*2!=xx) ans++;
}
}
printf("%d",ans);
return 0;
}