Submission #980911

#	Time	Username	Problem	Language	Result	Execution time	Memory
980911		Hadi_Alhamed	Bank (IZhO14_bank)	C++17	100 / 100	99 ms	8792 KiB

bank

//to live is to die
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

typedef long long int ll;
typedef unsigned long long ull;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef vector<int> vi;
typedef vector<long long> vl;
typedef vector<pair<int, int>> vpi;
typedef vector<pair<ll, ll>> vpl;
#define Clear(a, n)              \
    for (int i = 0; i <= n; i++) \
    {                            \
        a[i] = 0;                \
    }
#define clearMat(a, n, m, d)         \
    for (int i = 0; i <= n; i++)     \
    {                                \
        for (int j = 0; j <= m; j++) \
            a[i][j] = d;             \
    }
#define YES cout << "YES\n"
#define NO cout << "NO\n"
#define PB push_back
#define PF push_front
#define MP make_pair
#define F first
#define S second
#define rep(i, n) for (int i = 0; i < n; i++)
#define repe(i, j, n) for (int i = j; i < n; i++)
#define SQ(a) (a) * (a)
#define rep1(i, n) for (int i = 1; i <= n; i++)
#define Rrep(i, start, finish) for (int i = start; start >= finish; i--)
#define db(x)  cerr << #x <<" "; _print(x); cerr << endl;

#define forn(i, Start, End, step) for (int i = Start; i <= End; i += step)
#define rforn(i, Start, End, step) for (int i = Start; i >= End; i -= step)
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
// ll arr[SIZE];
/*
how to find n % mod ; n < 0?
x = (n+mod)%mod
if(x < 0) x += mod;
*/
void _print(int x)
{
    cerr << x;
}
void _print(ll x)
{
    cerr << x;
}
void _print(string x)
{
    cerr << x;
}
void _print(char x)
{
    cerr << x;
}
void _print(double x)
{
    cerr << x;
}
void _print(ull x)
{
    cerr << x;
}
void _print(vl x)
{
    for(auto e : x)
    {
        cerr << e << " ";
    }
    cerr << "\n";
}
void print(vpi x)
{
    for(auto e : x)
    {
        cerr << e.F << " " << e.S << "\n";
    }
    cerr << "\n";
}
void _print(vi x)
{
    for(auto e : x)
    {
        cerr << e << " ";
    }
    cerr << "\n";
}

void _print(deque<ll>x)
{
    for(auto e : x)
    {
        cerr << e << " ";
    }
    cerr << "\n";
}

//order_of_key(k): # of elements less than k (which is the index of x = k)
//find_by_order(k); iterator of the k-th element


template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T>
using ordered_multiset = tree<T, null_type,  less_equal <T>, rb_tree_tag, tree_order_statistics_node_update>;
template<class T> bool ckmin(T& a, const T& b)
{
    return b<a?a=b,1:0;
}
template<class T> bool ckmax(T& a, const T& b)
{
    return a<b?a=b,1:0;
}
template<typename T> istream& operator>>(istream& in, vector<T>& a)
{
    for(auto &x : a) in >> x;
    return in;
};
template<typename T> ostream& operator<<(ostream& out, vector<T>& a)
{
    for(auto &x : a) out << x << ' ';
    return out;
};

// priority_queue<data type , the container that would hold the values , greater<pair<int,int>>>
// greater means that we want the smallest value on top
// less means that we want the largest
// x ^ (n) mod m = ( (x mod m)^(n) ) mod m
char to_char(int num)
{
    return (char)(num + '0');
}

ll const MAX = 1e18+1;
ll const oo = 1e18 + 1;
ll const INF = 1e9 + 10;
const ll MOD = 1e9 + 7;
ll const SIZE = 2e5 + 900;
const int LOG = 20;
template<typename T, typename T2>
void add(T & X, T2 Y)
{
    X = (X + Y + MOD)%MOD;
}

template<typename T, typename T2>
T mult(T X, T2 Y)
{
    return X * Y % MOD;
}

void setIO(string s)
{
    freopen((s + ".in").c_str(), "r", stdin);
    freopen((s + ".out").c_str(), "w", stdout);
}
//x & (-x) give me the minBit of x
// x & (x - 1) turns off rightmost bit
const ll MAX_N = 1000006;
ll const P = 10000003;
ll const _P = 998244353;
ll const M = 1e9 + 9;

void solve()
{

    int N, M;
    cin >> N >> M;
    vi A(N);
    vi B(M);
    cin >> A;
    cin >> B;
    sort(all(A));
    vpi dp((1 << M) + 10);
    for(int Set = 0 ; Set < (1 << M) ; Set++)
    {
        for(int bill = 0; bill < M ; bill++)
        {
            if(Set & (1 << bill))
            {
                if(dp[Set ^ (1 << bill)].S + B[bill] == A[  dp[Set ^ (1 << bill)].F])
                {
                    //adding this bill will cause an increase in covered elements
                    dp[Set] = max(dp[Set] , {dp[Set ^ (1 << bill)].F + 1 , 0});
                }else
                {
                    //no increase in covered elements but
                    //we might be able to enhance the available sum
                    dp[Set] = max(dp[Set] , { dp[Set ^ (1 << bill)].F , dp[Set ^ ( 1 << bill)].S +  B[bill] } );

                }
            }
        }
        if(dp[Set].F == N)
        {
            YES;
            return;
        }
    }
    NO;

}

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
//    setIO("movie");

    int T = 1;
//    cin >> T;

    while(T--)
    {
        solve();
    }
    return 0;
}

/* stuff you should look for
 * WRITE STUFF DOWN,  ON PAPER
 * BFS THEN DFS
 * int overflow, array bounds
 * special cases (n=1?)
 * do sm th instead of nothing and stay organized
 * DON'T GET STUCK ON ONE APPROACH
 * (STUCK?)******** Try to simplify the problem(keeping in mind the main problem), ():
 * 1- problem to subProblem
 * 2- from simple to complex: start with a special
 *    problem and then try to update the solution for general case
 *    -(constraints - > solve it with none , one,two ... of them till you reach the given problem
      -(no constraints - > try to give it some)
      -how a special case may be incremented
 * 3-Simplification by Assumptions
 * REVERSE PROBLEM
 * PROBLEM ABSTRACTION
 * SMALL O BSERVATIONS MIGHT HELP ALOT
 * WATCH OUT FOR TIME
 * RETHINK YOUR IDEA,BETTER IDEA, APPROACH?
 * CORRECT IDEA, NEED MORE OBSERVATIONS
 * CORRECT APPROACH, WRONG IDEA
 * WRONG APPROACH
 * THINK CONCRETE THEN SYMBOL,
 * having the solution for the first m state , can we solve it for m + 1 ?
 * in many cases incremental thinking needs data sorting
 */

Compilation message (stderr)

bank.cpp: In function 'void setIO(std::string)':
bank.cpp:165:12: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
  165 |     freopen((s + ".in").c_str(), "r", stdin);
      |     ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
bank.cpp:166:12: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)' declared with attribute 'warn_unused_result' [-Wunused-result]
  166 |     freopen((s + ".out").c_str(), "w", stdout);
      |     ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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