3316: JC loves Mkk

Description

題解：

看到環就先搞兩遍，最大值得話就二分，注意精度（long double）

然後跑一下單調隊列，

假設當前答案爲x，那麼所有值都減x，尋找l，r之間和大於0的，

然後就奇數開個單調隊列，偶數開個單調隊列

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<cstring>
#define eps 1e-7
#define ll long long
using namespace std;
const ll N=201000;
ll n,L,R;
long double l,r;
ll a[N];
long double sum[N];
ll l1[3],r1[3],p[3][N];
ll anszi,ansmu;
bool check(long double x)
{
	sum[0]=0.0;
	for(ll i=1;i<=n;i++) sum[i]=sum[i-1]+a[i]-x;
	l1[0]=l1[1]=1;r1[0]=r1[1]=0;
	for(ll i=L;i<=n;i++)
	{
		ll yu=i&1;
		//prllf("%d\n",i);
		while(l1[yu]<=r1[yu]&&p[yu][l1[yu]]<i-R) l1[yu]++;
		while(l1[yu]<=r1[yu]&&sum[i-L]<sum[p[yu][r1[yu]]]) r1[yu]--;
		p[yu][++r1[yu]]=i-L;
		if(sum[i]-sum[p[yu][l1[yu]]]>0)
		{
			ansmu=i-p[yu][l1[yu]];
			return true;
		}
	}
	return false;
}
ll gcd(ll x,ll y)
{
	if(y==0) return x;
	return gcd(y,x%y);
}
int main()
{
	scanf("%lld%lld%lld",&n,&L,&R);
	for(ll i=1;i<=n;i++){scanf("%lld",&a[i]);a[i+n]=a[i];if(a[i]>r) r=a[i];}
	if(L&1) L++;if(R&1) R--;n=n*2;
	l=0.00;long double mid;
	while(r-eps>l)
	{
		mid=(l+r)/2.000;
		if(check(mid)){l=mid;}
		else r=mid;
	}
	long double ans=(l+r)/2.0;anszi=(ll)(ans*ansmu+0.5);
	ll yu=gcd(anszi,ansmu);
	anszi/=yu;ansmu/=yu;
	if(ansmu==1)printf("%lld",anszi);
	else printf("%lld/%lld",anszi,ansmu); 
}