제출 #886585

#제출 시각아이디문제언어결과실행 시간메모리
886585ono_de206Self Study (JOI22_ho_t2)C++14
100 / 100
203 ms7624 KiB
#include<bits/stdc++.h>
using namespace std;

#define fast ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define in insert
#define all(x) x.begin(),x.end()
#define pb push_back
#define eb emplace_back
#define ff first
#define ss second

#define int long long

typedef long long ll;
typedef vector<int> vi;
typedef set<int> si;
typedef multiset<int> msi;
typedef pair<int, int> pii;
typedef vector<pii> vpii;
// typedef pair<int, int> P;

template<typename T, typename U>
ostream & operator << (ostream &out, const pair<T, U> &c) {
	out << c.first << ' ' << c.second;
	return out;
}

template<typename T>
ostream & operator << (ostream &out, vector<T> &v) {
	const int sz = v.size();
	for (int i = 0; i < sz; i++) {
		if (i) out << ' ';
		out << v[i];
	}
	return out;
}

template<typename T>
istream & operator >> (istream &in, vector<T> &v) {
	for (T &x : v) in >> x;
	return in;
}

template<typename T, typename U>
	istream & operator >> (istream &in, pair<T, U> &c) {
	in >> c.first;
	in >> c.second;
	return in;
}

template<typename T>
void mxx(T &a, T b){if(b > a) a = b;}
template<typename T>
void mnn(T &a, T b){if(b < a) a = b;}

const int mxn = 3e5 + 10, MXLOG = 22, mod = 1e9 + 7, P = 1181, D = 1523;
const long long inf = 2e18 + 10;
int a[mxn], b[mxn], x[mxn], n, m;

bool check(int mid) {
	int mn = 0, mx = 0;
	for(int i = 1; i <= n; i++) {
		if(x[i] < mid) mn += (mid - x[i] + b[i] - 1) / b[i];
		if(x[i] > mid) mx += (x[i] - mid) / a[i];
		mnn(mn, inf);
		mnn(mx, inf);
	}
	return mn <= mx;
}

void go() {
	cin >> n >> m;
	for(int i = 1; i <= n; i++) {
		cin >> a[i];
	}
	for(int i = 1; i <= n; i++) {
		cin >> b[i];
	}
	int l = 1e18 + 10, r = 1e18 + 10;
	for(int i = 1; i <= n; i++) {
		mxx(a[i], b[i]);
		x[i] = a[i] * m;
		mnn(l, x[i]);
	}
	while(r - l > 1) {
		int mid = (l + r) / 2;
		if(check(mid)) l = mid;
		else r = mid;
	}
	cout << l << '\n';
}

signed main() {
	fast;
	int t = 1;
	// cin >> t;
	while(t--) {
		go();
	}
	return 0;
}
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