oj.uz

Submission #980911

# Submission time Handle Problem Language Result Execution time Memory
980911 2024-05-12T15:40:51 Z Hadi_Alhamed Bank (IZhO14_bank) C++17
100 / 100
 99 ms 8792 KB 
//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

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
19.0 / 19.0

# Verdict Execution time Memory Grader output
1 Correct 0 ms 344 KB Output is correct
2 Correct 0 ms 348 KB Output is correct
3 Correct 1 ms 348 KB Output is correct
4 Correct 1 ms 604 KB Output is correct
5 Correct 75 ms 8636 KB Output is correct
6 Correct 0 ms 344 KB Output is correct
7 Correct 1 ms 348 KB Output is correct
8 Correct 2 ms 8540 KB Output is correct
9 Correct 75 ms 8576 KB Output is correct

Subtask #2
25.0 / 25.0

# Verdict Execution time Memory Grader output
1 Correct 0 ms 344 KB Output is correct
2 Correct 0 ms 388 KB Output is correct
3 Correct 1 ms 344 KB Output is correct
4 Correct 1 ms 344 KB Output is correct
5 Correct 1 ms 348 KB Output is correct
6 Correct 0 ms 348 KB Output is correct
7 Correct 0 ms 348 KB Output is correct
8 Correct 0 ms 344 KB Output is correct
9 Correct 0 ms 344 KB Output is correct
10 Correct 0 ms 344 KB Output is correct

Subtask #3
27.0 / 27.0

# Verdict Execution time Memory Grader output
1 Correct 2 ms 348 KB Output is correct
2 Correct 2 ms 348 KB Output is correct
3 Correct 1 ms 348 KB Output is correct
4 Correct 2 ms 344 KB Output is correct
5 Correct 2 ms 348 KB Output is correct
6 Correct 1 ms 348 KB Output is correct
7 Correct 2 ms 456 KB Output is correct
8 Correct 1 ms 348 KB Output is correct
9 Correct 1 ms 472 KB Output is correct
10 Correct 1 ms 472 KB Output is correct
11 Correct 1 ms 348 KB Output is correct

Subtask #4
29.0 / 29.0

# Verdict Execution time Memory Grader output
1 Correct 0 ms 344 KB Output is correct
2 Correct 0 ms 348 KB Output is correct
3 Correct 1 ms 348 KB Output is correct
4 Correct 1 ms 604 KB Output is correct
5 Correct 75 ms 8636 KB Output is correct
6 Correct 0 ms 344 KB Output is correct
7 Correct 1 ms 348 KB Output is correct
8 Correct 2 ms 8540 KB Output is correct
9 Correct 75 ms 8576 KB Output is correct
10 Correct 0 ms 344 KB Output is correct
11 Correct 0 ms 388 KB Output is correct
12 Correct 1 ms 344 KB Output is correct
13 Correct 1 ms 344 KB Output is correct
14 Correct 1 ms 348 KB Output is correct
15 Correct 0 ms 348 KB Output is correct
16 Correct 0 ms 348 KB Output is correct
17 Correct 0 ms 344 KB Output is correct
18 Correct 0 ms 344 KB Output is correct
19 Correct 0 ms 344 KB Output is correct
20 Correct 2 ms 348 KB Output is correct
21 Correct 2 ms 348 KB Output is correct
22 Correct 1 ms 348 KB Output is correct
23 Correct 2 ms 344 KB Output is correct
24 Correct 2 ms 348 KB Output is correct
25 Correct 1 ms 348 KB Output is correct
26 Correct 2 ms 456 KB Output is correct
27 Correct 1 ms 348 KB Output is correct
28 Correct 1 ms 472 KB Output is correct
29 Correct 1 ms 472 KB Output is correct
30 Correct 1 ms 348 KB Output is correct
31 Correct 83 ms 8540 KB Output is correct
32 Correct 92 ms 8540 KB Output is correct
33 Correct 90 ms 8540 KB Output is correct
34 Correct 82 ms 8536 KB Output is correct
35 Correct 86 ms 8540 KB Output is correct
36 Correct 90 ms 8540 KB Output is correct
37 Correct 78 ms 8540 KB Output is correct
38 Correct 77 ms 8540 KB Output is correct
39 Correct 2 ms 8536 KB Output is correct
40 Correct 79 ms 8540 KB Output is correct
41 Correct 83 ms 8536 KB Output is correct
42 Correct 99 ms 8536 KB Output is correct
43 Correct 88 ms 8540 KB Output is correct
44 Correct 84 ms 8536 KB Output is correct
45 Correct 3 ms 8540 KB Output is correct
46 Correct 89 ms 8540 KB Output is correct
47 Correct 84 ms 8536 KB Output is correct
48 Correct 2 ms 8536 KB Output is correct
49 Correct 78 ms 8540 KB Output is correct
50 Correct 79 ms 8540 KB Output is correct
51 Correct 85 ms 8536 KB Output is correct
52 Correct 85 ms 8536 KB Output is correct
53 Correct 85 ms 8540 KB Output is correct
54 Correct 79 ms 8540 KB Output is correct
55 Correct 82 ms 8544 KB Output is correct
56 Correct 83 ms 8540 KB Output is correct
57 Correct 77 ms 8792 KB Output is correct
58 Correct 78 ms 8540 KB Output is correct