「板子」环形带限制的子段和

Posted 2020-12-12 614685877--aakennes

/*#!/bin/sh
dir=$GEDIT_CURRENT_DOCUMENT_DIR
name=$GEDIT_CURRENT_DOCUMENT_NAME
pre=${name%.*}
g++ -O2 $dir/$name -o $pre -g -Wall -std=c++11
if test $? -eq 0; then
    gnome-terminal -x bash -c "time $dir/$pre;echo;read;"
fi*/
#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=2e5+50;
int n,a[maxn],t,k;
int sum[maxn],q[maxn];
inline int read(){
	int s=0,w=1;
	char ch=getchar();
	while(ch<‘0‘||ch>‘9‘){
		if(ch==‘-‘)w=-1;ch=getchar();
	}
	while(ch>=‘0‘&&ch<=‘9‘)s=s*10+ch-‘0‘,ch=getchar();
	return s*w;
}
int main(){
    freopen("a.in","r",stdin);
	t=read();
	while(t--){
	    memset(q,0,sizeof(q));
        memset(sum,0,sizeof(sum));
	    n=read();k=read();
	    int ans=-INF;
	    for(int i=1;i<=n;i++)a[i]=read(),sum[i]=sum[i-1]+a[i];
	    for(int i=n+1;i<=2*n;i++)sum[i]=sum[i-1]+a[i-n];
	    int head=0,tail=0,l,r;
	    for(int i=1;i<=k+n;i++){
	        while(head<tail&&sum[i-1]<sum[q[tail-1]])tail--;
	        q[tail++]=i-1;
	        while(head<tail&&i-q[head]>k)head++;
	        if(sum[i]-sum[q[head]]>ans){
	            ans=sum[i]-sum[q[head]];
	            l=q[head]+1;r=i;
	        }
	    }
	    if(l>n)l-=n;
	    if(r>n)r-=n;
	    cout<<ans<<" "<<l<<" "<<r<<endl;
	}
	
	return 0;
}