제출 #1188027

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1188027		versesrev	수열 (APIO14_sequence)	C++20	50 / 100	2096 ms	24136 KiB

sequence

#include <iostream>
#include <vector>
#include <limits>
#include <numeric>
#include <algorithm>

int main() {
  int n, k;
  std::cin >> n >> k;
  std::vector<int> xs(n);
  for (int& x : xs) std::cin >> x;
  
  std::vector<long long> sum(n);
  std::partial_sum(xs.begin(), xs.end(), sum.begin());
  
  long long INF = std::numeric_limits<long long>::max();
  std::vector<std::vector<long long>> dp(k+1, std::vector<long long>(n, INF));
  std::vector<std::vector<int>> back(k+1, std::vector<int>(n, -1));
  for (int i = 0; i < n; ++i) {
    dp[0][i] = sum[i] * sum[i];
    back[0][i] = -1;
  }
  for (int ik = 1; ik <= k; ++ik) {
    for (int i = 0, p = 0; i < n; ++i) {
      long long target = sum[i] / (ik + 1);
      //int p = std::distance(sum.begin(), std::lower_bound(sum.begin(),
      //    sum.begin() + i,
      //    sum[i] - target));
      while (p < i and (sum[i]-sum[p])*(ik+1) > sum[i]) ++p;
      for (int j = std::max(p+1, ik); j <= i; ++j) {
        long long r = sum[i] - sum[j-1];
        if (r * (ik+1) > sum[i]) continue;
        long long v2 = sum[j-1]*sum[j-1]/ik + r*r;
        if (v2 >= dp[ik][i]) break;
        long long v = dp[ik-1][j-1] + r*r;
        if (dp[ik][i] > v) {
          dp[ik][i] = v;
          back[ik][i] = j - 1;
        }
      }
      for (int j = std::max(p+1, ik); j >= ik; --j) {
        long long r = sum[i] - sum[j-1];
        if (r * (ik+1) < sum[i]) continue;
        long long v2 = sum[j-1]*sum[j-1]/ik + r*r;
        if (v2 >= dp[ik][i]) break;
        long long v = dp[ik-1][j-1] + r*r;
        if (dp[ik][i] > v) {
          dp[ik][i] = v;
          back[ik][i] = j - 1;
        }
      }
    }
  }
  std::vector<int> ans;
  for (int p = n - 1, ik = k; ik; --ik) {
    p = back[ik][p];
    ans.push_back(p + 1);
  }
  std::ranges::reverse(ans);
  std::cout << ((sum[n-1] * sum[n-1] - dp[k][n-1]) / 2) << "\n";
  for (int v : ans) std::cout << v << " ";
  std::cout << "\n";
}

• Subtask 1: 11 / 11
• Subtask 2: 11 / 11
• Subtask 3: 11 / 11
• Subtask 4: 17 / 17
• Subtask 5: 0 / 21
• Subtask 6: 0 / 29