Submission #304803

#	Time	Username	Problem	Language	Result	Execution time	Memory
304803		NaimSS	Bank (IZhO14_bank)	C++14	100 / 100	539 ms	15608 KiB

bank

#include <bits/stdc++.h>
#define ld long double
#define endl "\n"
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define pb(x) push_back(x)
#define mp(a,b) make_pair(a,b)
#define ms(v,x) memset(v,x,sizeof(v))
#define all(v) v.begin(),v.end()
#define ff first
#define ss second
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define per(i, a, b) for(int i = b-1; i>=a ; i--)
#define trav(a, x) for(auto& a : x)
#define allin(a , x) for(auto a : x)
#define Unique(v) sort(all(v));v.erase(unique(all(v)),v.end());
#define sz(v) ((int)v.size())
//#define int long long
using namespace std;
typedef vector<int> vi;
#define y1 abacaba
#define left sefude
#define db(x) cerr << #x <<" == "<<x << endl;
#define db2(x,y) cerr<<#x <<" == "<<x<<", "<<#y<<" == "<<y<<endl;
#define db3(x,y,z) cerr << #x<<" == "<<x<<", "<<#y<<" == "<<y<<", "<<#z<<" == "<<z<<endl; 
#define prl cerr<<"called: "<< __LINE__ << endl;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
inline ll mod(ll n, ll m ){ ll ret = n%m; if(ret < 0) ret += m; return ret; }
ll gcd(ll a, ll b){return (b == 0LL ? a : gcd(b, a%b));}
ll exp(ll a,ll b,ll m){
    if(b==0LL) return 1LL;
    if(b==1LL) return mod(a,m);
    ll k = mod(exp(a,b/2,m),m);
    if(b&1LL){
        return mod(a*mod(k*k,m),m);
    }
    else return mod(k*k,m);
}




const int N = 20;
int a[N],b[N];
vi here[N];

bool good[N][(1<<N) + 1];

int32_t main(){
  fastio;
  int n,m;
  cin >> n >> m;
  for(int i=0;i<n;i++)cin >> a[i];
  for(int i=0;i<m;i++)cin >> b[i];
  
  for(int msk=1;msk<(1<<m);msk++){
    int tot=0;
    for(int j=0;j<m;j++)if((1<<j)&msk){
      tot+=b[j];
    }
    for(int i=0;i<n;i++)if(a[i] == tot){
      here[i].pb(msk);
    }

  }

  for(int i=0;i<n;i++){
    
    for(int j=0;j<(1<<m);j++){
      if((i == 0 and j == 0) || (i!=0 and good[i-1][j])){

      }else continue;
      for(auto v : here[i]){

        if((v&j) == 0){
            good[i][j^v]=1;
        }
      }

    }
  }
  
  bool ok = 0;
  for(int j=0;j<(1<<m);j++)ok|=good[n-1][j];
  cout << (ok ? "YES" : "NO")<<endl;


  // math -> gcd it all
  // Did u check N=1? Did you switch N,M?
}

• Subtask 1: 19 / 19
• Subtask 2: 25 / 25
• Subtask 3: 27 / 27
• Subtask 4: 29 / 29