Table of contents

• Problem
• Explanation
• Time complexity
• Solution

# Problem

https://atcoder.jp/contests/abc143/tasks/abc143_d

# Explanation

The range of the length of C is determined once the lengths of A and B are fixed.
Sort the edges according to their length.
Find the number of the edges which are included in the range by using binary search.

# Time complexity

$O(N^2)$ to fix A and B and $O(logN)$ to determind C with binary search.

$O(N^2logN)$

# Solution

// C++ 14
#include <iostream>
#include <string>
#include <vector>
#include <list>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <unordered_map>
#include <math.h>

#define ll long long
#define Int int
#define loop(x, start, end) for(Int x = start; x < end; x++)
#define loopdown(x, start, end) for(int x = start; x > end; x--)
#define rep(n) for(int x = 0; x < n; x++)
#define span(a,x,y) a.begin()+x,a.begin()+y
#define span_all(a) a.begin(),a.end()
#define len(x) (x.size())
#define last(x) (*(x.end()-1))

using namespace std;

#define MAX_N 2001

Int N;
vector<Int> L(MAX_N, -1);

void input() {
  cin >> N;
  loop(i,0,N) cin >> L[i];
}

void solve() {
  sort(span(L,0,N));  // sort edges by length asc
  
  Int count = 0;
  //O(N^2)
  loop(a,0,N-2) {
    loop(b,a+1,N) {
      Int min_c = L[a] - L[b] + 1;
      Int max_c = L[a] + L[b] - 1;
      // Binary search twice. O(2LogN)
      count += upper_bound(span(L,b+1,N), max_c) - lower_bound(span(L,b+1,N), min_c);
    }
  }
  cout << count << endl;
}

int main() {
  input();
  solve();
  return 0;
}