제출 #126480

#제출 시각아이디문제언어결과실행 시간메모리
126480hugo_pmHomecoming (BOI18_homecoming)C++17
0 / 100
63 ms33272 KiB
#include <bits/stdc++.h> #include "homecoming.h" using namespace std; #define int long long #define form2(i, a, b) for (int i = (a); i < (b); ++i) #define ford2(i, a, b) for (int i = (a-1); i >= (b); --i) #define form(i, n) form2(i, 0, n) #define ford(i, n) ford2(i, n, 0) #define chmax(x, v) x = max(x, (v)) #define chmin(x, v) x = min(x, (v)) #define fi first #define se second const long long BIG = 1000000000000000000LL; typedef long long ll; typedef long double ld; typedef pair<int, int> pii; const int PRIS=0, NON=1; const int borne = 2*1000*1000 + 5; int nbLivres, lenCyc; int rcp[borne]; int prix[borne]; int somPrix[borne]; int dp[borne][2]; int getSom(int a, int b) { if (a > b) return getSom(a, nbLivres - 1) + getSom(0, b); int rep = somPrix[b]; if (a) rep -= somPrix[a-1]; return rep; } long long solve(signed N, signed K, signed *A, signed *B) { nbLivres = N; lenCyc = K; form(i, nbLivres) { rcp[i] = A[i]; prix[i] = B[i]; somPrix[i] = prix[i]; if (i) somPrix[i] += somPrix[i-1]; } // Cas 1 : on prend le premier livre dp[0][NON] = -BIG; dp[0][PRIS] = rcp[0] - getSom(0, lenCyc-1); form2(iElem, 1, nbLivres) { int newAdd = (iElem + lenCyc - 1); // TODO PAS MOD // PRIS, et prev pris if (true) { int rep = dp[iElem-1][PRIS]; if (newAdd < nbLivres) rep -= prix[newAdd]; rep += rcp[iElem]; dp[iElem][PRIS] = rep; } // PRIS, et prev pas pris if (true) { int rep = dp[iElem-1][NON]; rep += rcp[iElem]; rep += getSom(iElem, min(newAdd, nbLivres-1)); chmax(dp[iElem][PRIS], rep); } // NON dp[iElem][NON] = max(dp[iElem-1][PRIS], dp[iElem-1][NON]); } int repCas1 = max(dp[nbLivres-1][PRIS], dp[nbLivres-1][NON]); // Cas 2 : on ne prend pas le premier livre dp[0][PRIS] = -BIG; dp[0][NON] = 0; form2(iElem, 1, nbLivres) { int newAdd = (iElem + lenCyc - 1) % nbLivres; // PRIS, et prev pris if (true) { int rep = dp[iElem-1][PRIS]; rep += rcp[iElem]; rep -= prix[newAdd]; dp[iElem][PRIS] = rep; } // PRIS, et prev pas pris if (true) { int rep = dp[iElem-1][NON]; rep += rcp[iElem]; rep -= getSom(iElem, newAdd); chmax(dp[iElem][PRIS], rep); } // NON dp[iElem][NON] = max(dp[iElem-1][PRIS], dp[iElem-1][NON]); } int repCas2 = max(dp[nbLivres-1][PRIS], dp[nbLivres-1][NON]); int wo = max(repCas1, repCas2); assert(wo >= 0); return wo; }
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