3316: JC loves Mkk

Description

题解：

看到环就先搞两遍，最大值得话就二分，注意精度（long double）

然后跑一下单调队列，

假设当前答案为x，那么所有值都减x，寻找l，r之间和大于0的，

然后就奇数开个单调队列，偶数开个单调队列

#include<cstdio>#include<cstdlib>#include<iostream>#include<algorithm>#include<cstring>#define eps 1e-7#define ll long longusing 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); }