Atcoder Beginner Content 143 B - TAKOYAKI FESTIVAL 2019

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# Problem

https://atcoder.jp/contests/abc143/tasks/abc143_b

# Time complexity

O(N2)O(N^2)

# Explanation

You can use a double loop as N is enough small.
Just sum up all the products of possible combinations.

# 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 51

Int N;
vector<Int> D(MAX_N, 0);

Int solve() {
  Int sum = 0;
  loop(i,0,N-1) {
    loop(j,i+1,N) {
      sum += D[i] * D[j];
    }
  }
  return sum;
}

int main() {
  cin >> N;
  loop(i,0,N) cin >> D[i];
  cout << solve() << endl;
}

Shun

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