题解 [LNOI2022] 吃

传送门

考虑将一些乘法改成加法
那么就是求一个最大的 \(\frac{1+\sum b+i}{\prod a_i}\)
先将 \(a_i=1\) 的全都选了
剩下的发现分母是指数级的，所以最多选 log 个
那么每次贪心选能最大化分式的值的就好了
复杂度 \(O(n\log n)\)

点击查看代码

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 500010
#define fir first
#define sec second
#define ll long long
//#define int long long

char buf[1<<21], *p1=buf, *p2=buf;
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int n;
ll a[N], b[N];
const ll mod=1e9+7;
inline ll qpow(ll a, ll b) {ll ans=1; for (; b; a=a*a%mod,b>>=1) if (b&1) ans=ans*a%mod; return ans;}

namespace force{
	int rec;
	double ans;
	void solve() {
		int lim=1<<n;
		for (int s=0; s<lim; ++s) {
			double now=1;
			for (int i=1; i<=n; ++i) if (s&(1<<i-1)) now+=b[i];
			for (int i=1; i<=n; ++i) if (!(s&(1<<i-1))) now*=a[i];
			if (now>ans) rec=s;
			ans=max(ans, now);
		}
		cout<<(ll)fmod(ans, mod)<<endl;
		// for (int i=1; i<=n; ++i) if (rec&(1<<i-1)) cout<<i<<' '; cout<<endl;
	}
}

namespace task1{
	ll ans=1;
	pair<ll, ll> sta[N];
	inline ll gcd(ll a, ll b) {return !b?a:gcd(b, a%b);}
	void solve() {
		random_device seed;
		mt19937 rand(seed());
		for (int i=1; i<=n; ++i) sta[i]={a[i], b[i]};
		sort(sta+1, sta+n+1, [](pair<ll, ll> a, pair<ll, ll> b){
			pair<ll, ll> t1={a.fir/gcd(a.fir, a.sec), a.sec/gcd(a.fir, a.sec)};
			pair<ll, ll> t2={b.fir/gcd(b.fir, b.sec), b.sec/gcd(b.fir, b.sec)};
			if (t1==t2) return a.sec<b.sec;
			else return (double)a.sec/(double)a.fir>(double)b.sec/(double)b.fir;
		});
		// sort(sta+1, sta+n+1, [](pair<ll, ll> a, pair<ll, ll> b){return a.fir==b.fir?a.sec>b.sec:a.fir<b.fir;});
		cout<<"sta: "; for (int i=1; i<=n; ++i) cout<<"("<<sta[i].fir<<','<<sta[i].sec<<") "; cout<<endl;
		pair<double, ll> sumb={1, 1}, proda={1, 1};
		for (int i=1; i<=n; ++i) ans=ans*sta[i].fir%mod;
		for (int i=1; i<=n; ++i) if (sta[i].fir<=sta[i].sec) {
			if ((sumb.fir+sta[i].sec)/(proda.fir*sta[i].fir) > sumb.fir/proda.fir) {
				sumb.fir+=sta[i].sec, sumb.sec=(sumb.sec+sta[i].sec)%mod;
				proda.fir*=sta[i].fir, proda.sec=proda.sec*sta[i].fir%mod;
				cout<<"("<<sta[i].fir<<','<<sta[i].sec<<") ";
			}
		}
		// cout<<endl;
		printf("%lld\n", ans*qpow(proda.sec, mod-2)%mod*sumb.sec%mod);
	}
}

namespace task2{
	int top;
	ll ans=1;
	pair<ll, ll> sta[N];
	inline ll gcd(ll a, ll b) {return !b?a:gcd(b, a%b);}
	void solve() {
		pair<long double, ll> sumb={1, 1}, proda={1, 1};
		for (int i=1; i<=n; ++i) ans=ans*a[i]%mod;
		for (int i=1; i<=n; ++i)
			if (a[i]==1) sumb.fir+=b[i], sumb.sec=(sumb.sec+b[i])%mod;
			else sta[++top]={a[i], b[i]};
		// cout<<"sta: "; for (int i=1; i<=n; ++i) cout<<"("<<sta[i].fir<<','<<sta[i].sec<<") "; cout<<endl;
		while (1) {
			int maxi=0;
			pair<long double, ll> maxb=sumb, maxa=proda;
			for (int i=1; i<=top; ++i) if (sta[i].fir<=sta[i].sec) {
				if ((sumb.fir+sta[i].sec)/(proda.fir*sta[i].fir) > maxb.fir/maxa.fir) {
					maxb.fir=sumb.fir+sta[i].sec, maxb.sec=(sumb.sec+sta[i].sec)%mod;
					maxa.fir=proda.fir*sta[i].fir, maxa.sec=proda.sec*sta[i].fir%mod;
					maxi=i;
					// cout<<"("<<sta[i].fir<<','<<sta[i].sec<<") ";
				}
			}
			if (!maxi) break;
			else swap(sta[maxi], sta[top--]), sumb=maxb, proda=maxa; //, cout<<"once"<<endl, cout<<maxi<<endl;
		}
		// cout<<endl;
		printf("%lld\n", ans*qpow(proda.sec, mod-2)%mod*sumb.sec%mod);
	}
}

signed main()
{
	// freopen("food.in", "r", stdin);
	// freopen("food.out", "w", stdout);

	n=read();
	for (int i=1; i<=n; ++i) a[i]=read();
	for (int i=1; i<=n; ++i) b[i]=read();
	// force::solve();
	// task1::solve();
	task2::solve();

	return 0;
}