Covered Points Count CF1000C 思维 前缀和 贪心

 Covered Points Count

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given n segments on a coordinate line; each endpoint of every segment has integer coordinates. Some segments can degenerate to points. Segments can intersect with each other, be nested in each other or even coincide.

Your task is the following: for every li≤x≤ri.

Input

The first line of the input contains one integer 1≤n≤2⋅105) — the number of segments.

The next i-th segment.

Output

Print i.

Examples

input

Copy

3
0 3
1 3
3 8

output

Copy

6 2 1 

input

Copy

3
1 3
2 4
5 7

output

Copy

5 2 0 

Note

The picture describing the first example:

Points with coordinates [3] is covered by three segments.

The picture describing the second example:

Points [2,3] are covered by two segments and there are no points covered by three segments.

 

 

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 5e5 + 10;
const int mod = 1e9 + 7;
typedef long long ll;
struct node {
    ll first, second;
};
node a[maxn];
ll ans[maxn];
bool cmp( node p, node q ) {
    return p.first < q.first;
}
int main() {
    ll n;
    while( cin >> n ) {
        memset( ans, 0, sizeof(ans) );
        for( ll i = 1; i <= n; i ++ ) {
            ll l, r;
            cin >> l >> r;
            a[i*2-1].first = l, a[i*2].first = r+1;
            a[i*2-1].second = 1, a[i*2].second = -1;
        }
        ll cnt = 0;
        sort( a + 1, a + 2*n + 1, cmp );
        for( ll i = 1; i <= 2*n; i ++ ) {
            ans[cnt] += a[i].first - a[i-1].first;
            cnt += a[i].second;
        }
        for( ll i = 1; i <= n; i ++ ) {
            if( i != n ) {
                cout << ans[i] << " ";
            } else {
                cout << ans[i] << endl;
            }
        }
    }
    return 0;
}