백준 알고리즘 3273번 : 두 수의 합

링크

https://www.acmicpc.net/problem/3273

sol1) 투포인터(12ms)

#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#define X first
#define Y second
#define pb push_back
#define fastio cin.tie(0)->sync_with_stdio(0)
#define MAX(a, b) (((a) > (b)) ? (a) : (b))
#define sz(v) (int)(v).size()
#define all(v) v.begin(), v.end()
#define rall(v) (v).rbegin(), (v).rend()
#define compress(v) sort(all(v)), (v).erase(unique(all(v)), (v).end())
#define OOB(x, y) ((x) < 0 || (x) >= n || (y) < 0 || (y) >= m)
#define debug(x) cout << (#x) << ": " << (x) << '\n'
using namespace std;
using ll = long long;
using ull = unsigned long long;
using dbl = double;
using ldb = long double;
using pii = pair<int,int>;
using pll = pair<ll,ll>;
using vi = vector<int>;
using vs = vector<string>;
using tii = tuple<int,int,int>;
template<typename T> using wector = vector<vector<T>>;

const int MOD = 1e9 + 7;
const int INF = 1e9 + 7;
const ll LNF = 1e18 + 7;

int main() {
	fastio;
  int n,x; cin >> n;
  vi v(n);
  for(int i = 0; i < n; i++) cin >> v[i];
  cin >> x;
  sort(all(v));
  int cnt = 0;
  int st = 0,en = n - 1;
  while(st < en){
    if(st >= n) break;
    if(v[st] + v[en] == x) cnt++;
    if(v[st] + v[en] > x) en--;
    else st++;
  }
  cout << cnt << "\n";
	return 0;
}

입력 받은 배열을 오름차순으로 정렬해준 뒤에 투 포인터를 이용해서 쌍의 개수를 찾아주면 되는 문제였습니다.
st = 0, en = n - 1로 놓고 v[st] + v[en] > x이면 en--를, < 0 이면 st++를 해주어서 쌍의 갯수를 구해주면 됩니다.
비슷한 문제로 boj 2467번 용액문제가 있습니다.

sol2) 이분탐색(12ms)

#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#define X first
#define Y second
#define pb push_back
#define fastio cin.tie(0)->sync_with_stdio(0)
#define MAX(a, b) (((a) > (b)) ? (a) : (b))
#define sz(v) (int)(v).size()
#define all(v) v.begin(), v.end()
#define rall(v) (v).rbegin(), (v).rend()
#define compress(v) sort(all(v)), (v).erase(unique(all(v)), (v).end())
#define OOB(x, y) ((x) < 0 || (x) >= n || (y) < 0 || (y) >= m)
#define debug(x) cout << (#x) << ": " << (x) << '\n'
using namespace std;
using ll = long long;
using ull = unsigned long long;
using dbl = double;
using ldb = long double;
using pii = pair<int,int>;
using pll = pair<ll,ll>;
using vi = vector<int>;
using vs = vector<string>;
using tii = tuple<int,int,int>;
template<typename T> using wector = vector<vector<T>>;

const int MOD = 1e9 + 7;
const int INF = 1e9 + 7;
const ll LNF = 1e18 + 7;

int main() {
	fastio;
  int n,x,cnt = 0; cin >> n;
  vi v(n);
  for(int i = 0; i < n; i++) cin >> v[i];
  cin >> x;
  sort(all(v));
  for(int i = 0; i < n; i++) if(binary_search(all(v),x - v[i])) cnt++;
  cout << cnt/2 << "\n";
	return 0;
}

• 이분탐색으로도 풀 수 있습니다.