Problem Description
N sticks are arranged in a row, and their lengths are a1,a2,...,aN.
There are Q querys. For i-th of them, you can only use sticks between li-th to ri-th. Please output the maximum circumference of all the triangles that you can make with these sticks, or print −1 denoting no triangles you can make.
Input
There are multiple test cases.
Each case starts with a line containing two positive integers N,Q(N,Q≤1e5).
The second line contains N integers, the i-th integer ai(1≤ai≤1e9) of them showing the length of the i-th stick.
Then follow Q lines. i-th of them contains two integers li,ri(1≤li≤ri≤N), meaning that you can only use sticks between li-th to ri-th.
It is guaranteed that the sum of Ns and the sum of Qs in all test cases are both no larger than 4×1e5.
Output
For each test case, output Q lines, each containing an integer denoting the maximum circumference.
Sample Input
5 3
2 5 6 5 2
1 3
2 4
2 5
Sample Output
13
16
16
Solution:
可以发现，无法组成三角形的最劣条件便是斐波那契数列，而在本题数列最多到44项(ai<=1e9)
于是我们便可以用主席树来维护区间第K大，然后暴力询问就OK了
Code:
#include<cstdio>
#include<ctype.h>
#include<cstring>
#include<iostream>
#include<algorithm>
#define int long long
using namespace std;
const int N=1e5+1;
int n,q,cnt,a[N],b[N],rt[N];
struct SegTree{
int tot,ls[N*40],rs[N*40],sz[N*40];
void clear(){
tot=0;
memset(ls,0,sizeof(ls));
memset(rs,0,sizeof(rs));
memset(rt,0,sizeof(sz));
}
void build(int &q,int l,int r){
q=++tot;sz[q]=0;
if(l==r) return ;
int mid=l+r>>1;
build(ls[q],l,mid);
build(rs[q],mid+1,r);
}
void ins(int &q,int lst,int l,int r,int x){
q=++tot;
ls[q]=ls[lst],rs[q]=rs[lst];
sz[q]=sz[lst]+1;
if(l==r) return ;
int mid=l+r>>1;
if(mid>=x) ins(ls[q],ls[lst],l,mid,x);
else ins(rs[q],rs[lst],mid+1,r,x);
}
int query(int q,int p,int l,int r,int k){
if(l==r) return b[l];
int v=sz[ls[q]]-sz[ls[p]];
int mid=l+r>>1;
if(v>=k) return query(ls[q],ls[p],l,mid,k);
else return query(rs[q],rs[p],mid+1,r,k-v);
}
}T;
int read(){
int x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-f;ch=getchar();}
while(isdigit(ch)){x=x*10+ch-48;ch=getchar();}
return x*f;
}
void solve(){T.clear();
for(int i=1;i<=n;i++)
a[i]=read(),b[i]=a[i];
sort(b+1,b+n+1);
cnt=unique(b+1,b+n+1)-b-1;
T.build(rt[0],1,cnt);
for(int i=1;i<=n;i++){
int v=lower_bound(b+1,b+cnt+1,a[i])-b;
T.ins(rt[i],rt[i-1],1,cnt,v);
}
for(int i=1;i<=q;i++){
int l=read(),r=read(),u=r-l+1,flag=0;
if(u<=2){puts("-1");continue;}
int a=T.query(rt[r],rt[l-1],1,cnt,u);
int b=T.query(rt[r],rt[l-1],1,cnt,--u);
while(u){
int v=T.query(rt[r],rt[l-1],1,cnt,--u);
if(v+b>a){
printf("%lld
",a+b+v);
flag=1;break;
}a=b,b=v;
}if(!flag) puts("-1");
}
}
signed main(){
while(~scanf("%lld%lld",&n,&q)) solve();
return 0;
}