SRM 646 div1 easy TheConsecutiveIntegersDivOne

考えたこと

• ソートしていい
• 全ての区間[l,l+k-1]についてi番目に合わせて連続するk個の列にするのにかかる操作量を計算
• 全部愚直にやってもO(nk^2)で余裕

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<ll> VL;
typedef vector<VL> VVL;
typedef pair<int, int> PII;

#define FOR(i, a, n) for (ll i = (ll)a; i < (ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(x) x.begin(), x.end()
#define IN(a, b, x) (a<=x&&x<b)
#define MP make_pair
#define PB push_back
const int INF = (1LL<<30);
const ll LLINF = (1LL<<60);
const double PI = 3.14159265359;
const double EPS = 1e-12;
const int MOD = 1000000007;
//#define int ll

template <typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template <typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }

int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};

class TheConsecutiveIntegersDivOne {
   public:
   int find(vector <int> numbers, int k)
  {
    int n = numbers.size();
    sort(ALL(numbers));

    int ans = INF;
    REP(l, n-k+1) {
      int r = l+k-1;
      // [l, r]
      FOR(j, l, r+1) {
        int ret = 0;
        FOR(k, l, r+1) {
          if(k < j) {
            int t = numbers[j] - (j-k);
            ret += abs(t-numbers[k]);
          } else if(j < k) {
            int t = numbers[j] + (k-j);
            ret += abs(t-numbers[k]);
          }
        }
        chmin(ans, ret);
      }
    }

    return ans;
  }
};