제출 #23166

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
23166		ruhanhabib39	수열 (APIO14_sequence)	C++14	0 / 100	13 ms	91876 KiB

sequence

#include <bits/stdc++.h>
using namespace std;

typedef long long Long;

const int MAXN = 1e5;
const int MAXK = 200;

int N, K;
Long A[MAXN + 10], S[MAXN + 10];

Long dp[2][MAXN + 10];
int pos[MAXK + 10][MAXN + 10];

struct dat {
   Long m, b; int i;
   Long f(Long x) const {
      return m*x + b;
   }
};

bool bad(dat l1, dat l2, dat l3) {
   return (l1.b - l2.b)*(l3.m - l1.m) > (l1.b - l3.b)*(l2.m - l1.m);
}

vector<dat> L;

void add(dat l) {
   L.push_back(l);
   while(L.size() >= 3 && bad(L[L.size()-3], L[L.size()-2], L[L.size()-1])) {
      L.erase(L.end() - 2);
   }
}

int ii = 0;

int get(Long x) {
   if(ii >= (int)L.size()) ii = L.size()-1;
   while(ii+1 < (int)L.size() && L[ii+1].f(x) > L[ii].f(x)) {
      ii++;
   }
   return L[ii].i;
}

void add(int x, int j) {
   Long m = S[j];
   Long b = - S[j] * S[j] + dp[x][j];
   add(dat{m, b, j});
}

void calc_dp() {
   for(int k = 1; k <= K; k++) {
      int x = k & 1;
      L.clear(); ii = 0;
      if(k > 0) add(x ^ 1, k-1);
      for(int i = k; i <= N; i++) {
         int j = get(S[i]);
         pos[k][i] = j;
         dp[x][i] = S[j]*S[i] - S[j]*S[j] + dp[x^1][j];
         add(x ^ 1, i);
      }
   }
}

int main() {
   scanf("%d%d", &N, &K);
   S[0] = 0;
   for(int i = 1; i <= N; i++) {
      scanf("%lld", &A[i]);
      S[i] = A[i] + S[i-1];
   }
   calc_dp();
   printf("%lld\n", dp[K&1][N]);
   int i = N, k = K;
   while(k > 0) {
      i = pos[k][i];
      printf("%d ", i);
      k--;
   }
   printf("\n");
}

컴파일 시 표준 에러 (stderr) 메시지

sequence.cpp: In function 'int main()':
sequence.cpp:66:25: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
    scanf("%d%d", &N, &K);
                         ^
sequence.cpp:69:27: warning: ignoring return value of 'int scanf(const char*, ...)', declared with attribute warn_unused_result [-Wunused-result]
       scanf("%lld", &A[i]);
                           ^

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