# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
598042 | LIF | Arranging Shoes (IOI19_shoes) | C++14 | 0 ms | 0 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include<bits/stdc++.h>
#include<vector>
using namespace std;
int n;
int a[200005];
vector<int> v[300005];
long long int sum[300005];
bool vis[300005];
int lowbit(int x)
{
return x&(-x);
}
int add(int x,int val)
{
while(x<=n)
{
sum[x] +=val;
x+=lowbit(x);
}
}
long long int getsum(int x)
{
long long int ans = 0;
while(x>0)
{
ans += sum[x];
x-=lowbit(x);
}
return ans;
}
int main()
{
cin>>n;
n<<=1;
for(int i=1;i<=n;i++)
{
vis[i] = false;
cin>>a[i];
int x = a[i];
v[x+n].push_back(i);
add(i,1);
}
long long int ans = 0;
for(int i=n;i>=1;i--)//from the last to begin."Reserve".
{
if(vis[i] == true)continue;
vis[i] = true;
int x = a[i];
v[a[i]+n].pop_back();
int res = v[n-a[i]].back();
v[n-a[i]].pop_back();
vis[res] = true;
ans = ans + getsum(i-1)-getsum(res); //we only need to push the "res" to i-1.
if(a[i]<0) //since we collect pair in every time, we only need to observe the odd. if the odd is <0,that means it should swap one more time.
{
ans+=1;
}
add(res,-1);
}
//Although we haven't do one time in really life, we have try it in the brain.
cout<<ans<<endl;
return 0;
}