Merge pull request encrypted-def#465 from syoh0708/0x14_20366

encrypted-def · web-flow · commit bfb4486c1f93 · 2024-01-18T22:56:55.000+09:00
Create 20366.cpp
diff --git a/0x14/solutions/20366.cpp b/0x14/solutions/20366.cpp
@@ -1,11 +1,41 @@
-// Authored by : BaaaaaaaaaaarkingDog
+// Authored by : syoh0708
 // Co-authored by : -
-// http://boj.kr/****************
+// http://boj.kr/3589e6590845469191b7873b71550403
 #include <bits/stdc++.h>
+
 using namespace std;
 
-int main(void){
+const int MAX = 1'000'000'000;
+int n, ans = MAX;
+int a[605];
+
+int main() {
   ios::sync_with_stdio(0);
   cin.tie(0);
-  
-}
+
+  cin >> n;
+
+  for (int i = 0; i < n; i++) cin >> a[i];
+
+  sort(a, a + n);
+
+  for (int i1 = 0; i1 < n; i1++) {
+    for (int i2 = i1 + 3; i2 < n; i2++) {
+      int j1 = i1 + 1, j2 = i2 - 1;
+
+      while (j1 < j2) {
+        ans = min(ans, abs(a[i1] + a[i2] - a[j1] - a[j2]));
+
+        if (a[i1] + a[i2] <= a[j1] + a[j2]) j2--;
+        else j1++;
+      }
+    }
+  }
+
+  cout << ans;
+}
+/*
+답을 구성하는 눈사람이 a,b,c,d(a <= b <= c <= d)라고 할 때 (a, d)가 한 묶음이고
+(b, c)가 한 묶음인건 자명하기 때문에 인덱스를 i1 < j1 < j2 < i2로 둘 수 있다.
+a[i1]+a[i2]와 a[j1]+a[j2]가 근접한 (j1, j2) 쌍을 구하면 된다.
+*/
