#982889 (C++14) No.2701 A cans -> B cans

• 問題: No.2701 A cans -> B cans
• ユーザー: Keeby
• 提出日時: 2024-05-20 06:41:37
• 言語: C++14 (gcc 13.3.0 + boost 1.87.0)
• 結果: MLE
• 実行時間: -
• コード長: 1,713 bytes
• コンパイル時間: 1,847 ms
• コンパイル使用メモリ: 175,500 KB
• 実行使用メモリ: 1,179,264 KB
• 最終ジャッジ日時: 2024-12-20 17:41:58
• 合計ジャッジ時間: 418,932 ms
• ジャッジサーバーID （参考情報）: judge3 / judge2

ファイルパターン	結果
sample	AC * 1 MLE * 5
other	AC * 1 WA * 1 RE * 3 TLE * 31 MLE * 37

ソースコード

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#include <bits/stdc++.h>
//#include <atcoder/modint>
using namespace std;
//using namespace atcoder;
//using mint = modint998244353;
#define rep(i,n) for(long long i=0;i<n;i++)
#define rep1(i,n) for(long long i=1;i<=n;i++)
#define Rep(i,a,b) for(long long i=a;i<=b;i++)
#define all(a) (a).begin(), (a).end()
#define fst first
#define snd second
typedef unsigned long long ull;
typedef long long ll;
typedef vector<int> vec;
typedef vector<vector<int>> vvec;
typedef vector<long long> vecll;
typedef vector<vector<long long>> vvecll;
//#include<boost/multiprecision/cpp_int.hpp>
//using namespace boost::multiprecision;
const ll mod=998244353;
//const ll mod=1e9+7;
const ll INF=5e18;
const int INT_INF=1e9;
const double pi=3.14159265358979;
//xのn乗をpで割ったあまりを求める
ll modpow(ll x,ll n,ll p=mod){
  if(n==0) return 1;
  if(n%2==1) return (x*modpow(x,n-1,p))%p;
  else{
    ll t=modpow(x,n/2,p);
    assert(t>0);
    return (t*t)%p;
  }
}
//xの逆元(p:素数)
ll inv(ll x,ll p=mod){
  return modpow(x,p-2,p);
}
template <typename T>
int index(vector<T> &a, T x){
    return lower_bound(all(a),x)-a.begin();
}
struct State{
    ll c;
    ll z;
    ll i;
};
int main(){
    int n,m;
    cin >> n >> m;
    vector<vector<State>> dp(n,vector<State>(m+1,{0,0,0}));
    vecll a(n),b(n),c(n);
    rep(i,n) cin >> a[i] >> b[i] >> c[i];
    rep(i,n){
        rep1(j,m){
            State x=dp[i][j-1];
            x.z++;
            if(x.z+x.i>=a[i]){
                ll k=a[i];
                k-=x.i;
                x.i=max(x.i-k,0LL);
                x.z-=k;
                x.i+=b[i];
                x.c+=c[i]*b[i];
            }
            dp[i][j]=x;
        }
    }
    vecll ans(m+1,0);
    rep1(i,m){
        rep(j,n){
            if(a[j]<=i){
                ans[i]=max(ans[i],dp[j][i].c+ans[dp[j][i].z]);
            }
        }
    }
    rep1(i,m){
        cout << ans[i] << endl;
    }
    return 0;
}
 
 
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