# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
90852 | MohamedAhmed0 | Vudu (COCI15_vudu) | C++14 | 714 ms | 44212 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.
/* Author : Mohamed Ahmed (CF : MohamedAhmed04)
since we need to know number of subarrays that their average is >= p
then let l be length of subarray then l * sum >= p then if we subtract
p from all elements then we need to know number of sub arrays that sum >= 0
so now problem transformed to know number of prev subarrays that their sum <= current sum
so we can use segment tree or Binary Indexed tree (fenwick tree) to find sum of elements
but we have one problem....values can be very large so we need to compress values so first we
will compress and after that everytime we will add to answer number of sub arrays <= arr[i] and
update in tree arr[i] (increase number of occurences of it).
*/
#include <bits/stdc++.h>
using namespace std;
const int MAX = 1e6+5 ;
long long tree[4*MAX] ;
void update(int node , int l , int r , int idx)
{
if(l > idx || r < idx)
return ;
if(l == r)
{
tree[node]++;
return ;
}
long long mid = (l + r) >> 1 ;
update(node << 1 , l , mid , idx);
update(node << 1 | 1 , mid + 1 , r , idx) ;
tree[node] = tree[node << 1] + tree[node << 1 | 1] ;
}
long long query(int node , int l , int r , int from , int to)
{
if(from > r || to < l)
return 0 ;
if(l >= from && r <= to)
return tree[node] ;
long long mid = (l + r) >> 1 ;
long long x = query(node << 1 , l , mid , from , to) ;
long long y = query(node << 1 | 1 , mid + 1 , r , from , to);
return x+y ;
}
int main()
{
ios::sync_with_stdio(0);
cin.tie(0);
long long n ;
cin>>n ;
long long arr[n+1] ;
for(int i = 1 ; i <= n ; ++i)
cin>>arr[i] ;
long long p ;
cin>>p ;
long long ans = 0 , sum = 0;
vector< pair<long long , long long> >vp ;
vp.push_back({0 , 0});
for(int i = 1 ; i <= n ; ++i)
{
sum += (arr[i] - p) ;
vp.push_back({sum , i});
}
sort(vp.begin() , vp.end());
long long idx = 0 ;
for(int i = 0 ; i < vp.size() ; ++i)
{
if(i != 0)
{
if(vp[i].first == vp[i-1].first)
--idx;
}
arr[vp[i].second] = idx;
idx++;
}
update(1 , 0 , n , arr[0]);
for(int i = 1 ; i <= n ; ++i)
{
ans += query(1 , 0 , n , 0 , arr[i]) * 1ll;
update(1 , 0 , n , arr[i]);
}
return cout<<ans , 0 ;
}
Compilation message (stderr)
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