Submission #1218625

#TimeUsernameProblemLanguageResultExecution timeMemory
1218625pauloaphKnapsack (NOI18_knapsack)C++20
17 / 100
5 ms9544 KiB
#include<iostream>
#include<algorithm>
#include <cmath>
using namespace std;

const int MAXN =  100000;
const int MAXC = 1000;

int main(){

    int C;
    int n;

    cin >> C >> n;

    int V[n];
    int W[n];
    int B[n];
    long int total = n;

    for (int i = 0; i < n; i++){
        cin >> V[i] >> W[i] >> B[i];
        total += (long int)(log2(B[i]));
    }

    long int NW[MAXN];
    long int NV[MAXN];
    long int M = 0;


    // vamos criar os novos itens (pacotes)
    for ( int i = 0; i < n; ++i ){
        int pos = 0;
        // procuramos o bit mais significativo
        while ( (1 << pos) < B[i] )
            pos++;
        // aqui sabemos que (2 ^ pos) >= B[i]
        pos--;
        // agora 2 ^ pos < B[i], logo (2 ^ pos) - 1 <= B[i] 
        while ( pos >= 0 ){
            NW[M] = (1 << pos) * W[i];
            NV[M] = (1 << pos) * V[i];
            B[i] -= (1 << pos);
            M++;
            pos--;
        }
        if ( B[i] ){ // se sobrou um resto
            NW[M] = B[i] * W[i];
            NV[M] = B[i] * V[i];
            M++;
            B[i] = 0;
        }
    }

   long int dp[M + 1][C + 1];

   // caso base : quando nao temos mais itens
   for ( int c = 0; c <= C; ++c )
      dp[M][c] = 0;

   for ( int i = M - 1; i >= 0; --i )
      for ( int c = 0; c <= C; ++c ){
         dp[i][c] = dp[i + 1][c];
         if ( NW[i] <= c )
            dp[i][c] = max( dp[i][c], dp[i + 1][c - NW[i]] + NV[i] );
      }

   cout << dp[0][C] << endl;
   return 0;
}
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