1 solutions
-
1
#include <bits/stdc++.h> using namespace std; long long n, a[100005], b[100005], mida, midb, ans; int main() { cin >> n; for (int i = 1;i <= n;i++) { cin >> a[i] >> b[i]; } sort (a + 1, a + n + 1); sort (b + 1, b + n + 1); mida = a[(1 + n) / 2]; midb = b[(1 + n) / 2]; for (int i = 1;i <= n;i++) { ans += abs(mida - a[i]) + abs(midb - b[i]); } cout << ans; return 0; }
- 1
Information
- ID
- 1068
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- 8
- Tags
- # Submissions
- 14
- Accepted
- 7
- Uploaded By