codes/D_2_Magic_Powder_2.cpp at main · hereparvezali/codes · GitHub

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
#include <bits/stdc++.h>
#include <iostream>
using namespace std;

typedef long long ll;
typedef pair<ll, ll> pll;
typedef vector<bool> vbl;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<string> vstr;
typedef vector<pll> vpll;

#define endl '\n'
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
constexpr ll mod = 1000000007;

inline void Parvez() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
}
template <class T> inline istream &operator>>(istream &in, vector<T> &v) {
    for (T &x : v) {
        in >> x;
    }
    return in;
}
template <class T> inline ostream &operator<<(ostream &out, vector<T> &v) {
    if (v.empty())
        return out;
    for (ll i = 0; i < v.size() - 1; i++) {
        out << v[i] << ' ';
    }
    out << v[v.size() - 1];
    return out;
}

void solution(ll &T) {
    ll n, k;
    cin >> n >> k;
    vll needed(n), has(n);
    cin >> needed >> has;

    ll l = 0, r = 1e10, m, ans = 0;
    while (l <= r) {
        m = (l + r) / 2LL;
        ll kcopy = k;
        bool fulfilled = true;

        for (ll i = 0; i < n; i++) {
            if (m * needed[i] > has[i]) {
                if (m * needed[i] <= has[i] + kcopy) {
                    kcopy -= (m * needed[i] - has[i]);
                } else {
                    fulfilled = false;
                    break;
                }
                if (kcopy < 0) {
                    fulfilled = false;
                    break;
                }
            }
        }
        if (fulfilled) {
            ans = m;
            l = m + 1;
        } else {
            r = m - 1;
        }
    }
    cout << ans << endl;
}

signed main() {
    Parvez();
    ll TT = 1;
    // cin>>TT;
    for (ll T = 1; T <= TT; T++)
        solution(T);
    return 0;
}