Submission #1218671

#	Time	Username	Problem	Language	Result	Execution time	Memory
1218671		pauloaph	Knapsack (NOI18_knapsack)	C++17	17 / 100	1 ms	328 KiB

knapsack

#include<iostream>
#include<algorithm>
#include <cmath>
using namespace std;


int main(){

    int C;
    int n;

    cin >> C >> n;

    unsigned int V[n];
    unsigned int W[n];
    unsigned int B[n];
    

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

    vector<unsigned long long int> NW, NV;


    for (int i = 0; i < n; i++ ){
        if(B[i] > 1){
            int pos = 0;
            unsigned long long int x = 1;
            // procuramos o bit mais significativo
            while( B[i] > x){
                pos++;
                x = 1LL << pos;
            } 
            // aqui sabemos que (2 ^ pos) >= B[i]
            pos--;
            // agora 2 ^ pos < B[i], logo (2 ^ pos) - 1 <= B[i] 
            while ( pos >= 0 ){
                unsigned long long int x = 1LL << pos;
                NW.push_back(x * W[i]);
                NV.push_back(x * V[i]);
                B[i] -= x;
                pos--;
            }
            if ( B[i] > 0) { // se sobrou um resto
                NW.push_back(B[i] * W[i]);
                NV.push_back(B[i] * V[i]);
                B[i] = 0;
            }
        } else {
            NW.push_back(W[i]);
            NV.push_back(V[i]);
            B[i] = 0;
        }
        
    }

    // for(int i = 0; i < NW.size(); i++){
    //     cout << NW[i] << " " << NV[i] << endl;
    // }

    unsigned long long int dp[2][C+1];
    int prev = 0, act = 1;
    // caso base : quando nao temos mais itens
    for ( int c = 0; c <= C; ++c )
        dp[prev][c] = 0;


    for ( int i = NW.size() - 1; i >= 0; --i ){
        for ( int c = 0; c <= C; ++c ){
            dp[act][c] = dp[prev][c];
            if(NW[i] <= c ) {
                dp[act][c] = max( dp[act][c], dp[prev][c - NW[i]] + NV[i] );
            }
        }
        swap( act, prev );
    }
      
   cout << dp[prev][C] << endl;
   return 0;
}

• Subtask 1: 0 / 12
• Subtask 2: 17 / 17
• Subtask 3: 0 / 20
• Subtask 4: 0 / 24
• Subtask 5: 0 / 27