Submission #691768

#	Time	Username	Problem	Language	Result	Execution time	Memory
691768		saayan007	Bank (IZhO14_bank)	C++17	19 / 100	187 ms	17736 KiB

bank

#include <bits/stdc++.h>
using namespace std;

using ll = long long;
using pi = pair<int, int>;
using pl = pair<long long, long long>;
using vi = vector<int>;
using vl = vector<long long>;
using vpi = vector<pair<int, int>>;
using vpl = vector<pair<long long, long long>>;

#define fur(i, a, b) for(ll i = a; i <= (ll)b; ++i)
#define ruf(i, a, b) for(ll i = a; i >= (ll)b; --i)
#define fr first 
#define sc second
#define mp make_pair
#define pb emplace_back
#define nl "\n"
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()

#define no cout << "NO" << nl;return 0
#define yes cout << "YES" << nl;return 0

int main() 
{
    ios::sync_with_stdio(0);
    cin.tie(0);

    ll n, m;
    cin >> n >> m;
    ll a[n], b[m];
    fur(i, 0, n - 1) {
        cin >> a[i];
    }
    fur(i, 0, m - 1) {
        cin >> b[i];
    }

    ll M = 1 << m;
    pl sum[M];
    sum[0] = {0, 0};
    fur(i, 1, M - 1) {
        ll amt = 0;
        fur(j, 0, m - 1) {
            if(i & (1 << j))
                amt += b[j];
        }
        sum[i] = {amt, i};
    }

    sort(sum, sum + M);

    bool dp[n][M] = {};
    // base case dp[0][..] = 1;
    // sort(a, a + n);

    ll ptr = lower_bound(sum, sum + M, mp(a[0], -1ll)) - sum;
    if(ptr == M || sum[ptr].fr > a[0]) {
        no;
    }
    while(ptr < M && sum[ptr].fr == a[0]) {
        dp[0][sum[ptr].sc] = 1;
        ++ptr;
        if(n == 1) {
            yes;
        }
    }

    ptr = lower_bound(sum, sum + M, mp(a[0], -1ll)) - sum;
    fur(i, 0, n - 2) {
        fur(j, ptr, M - 1) {
            if(sum[j].fr > a[i])
                break;
            ll ptr2 = lower_bound(sum, sum + M, mp(a[i + 1], -1ll)) - sum;
            if(ptr2 == M || sum[ptr2].fr > a[i + 1]) {
                no;
            }
            fur(k, ptr2, M - 1) {
                if(sum[k].fr > a[i + 1])
                    break;

                ll b1 = sum[j].sc;
                ll b2 = sum[k].sc;
                if(b1 + b2 == (b1 | b2) && (b1|b2) == (b1^b2)) {
                    dp[i + 1][b1|b2] = 1;
                    if(i + 1 == n - 1) {
                        yes;
                    }
                }
            }
        }
    }

    no;
}

• Subtask 1: 19 / 19
• Subtask 2: 0 / 25
• Subtask 3: 0 / 27
• Subtask 4: 0 / 29