hdu5662 YJQQQAQ and the function (單調棧)

Problem Description

YJQQQAQ has an array A of length n. He defines a function fl,r,k where l,r,k are positive integers that satisfies l≤r and r×k≤n, and the value of the function equals to p×q×⌊k√⌋ where p equals to the sum value of Al×k,A(l+1)×k,...,Ar×k and q equals to the minimal value of them. YJQQQAQ wants to choose the positive integers l,r,k carefully to maximize the value of the function.

 

Input

The first line contains an integer T(1≤T≤3)——The number of the test cases. For each test case:
The first line contains an integers n(1≤n≤300,000).
The second line contains n integers describing the given array A, the ith integer is Ai(1≤Ai≤1,000,000). Between each two adjacent integers there is a white space separated.

 

Output

For each test case, the only line contains the only integer that is the maximum value of the function.

 

Sample Input

1 3 2 3 1

 

Sample Output

10

題意：給你n個數，你要找到3個數，l,r,k,l<=r，r*k<=n且p*q*sqrt(k)最小，其中p是A[l*k],A[(l+1)*k]...,A[r*k]的和，q是這些數的最小值。

思路：看到求一些數的和乘這些數中最小值的最小值，容易想到單調棧，我們只要先把在k確定的情況下找出所有符合條件的A[],然後再用單調棧找出以每一個數位最小值的左右邊界，然後更新ans的最大值就行了。ps：還是不習慣用單調棧，所以用兩遍單調隊列做了，時間複雜度會差一下，不過差不多。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<bitset>
#include<algorithm>
using namespace std;
#define lth th<<1
#define rth th<<1|1
typedef long long ll;
typedef long double ldb;
#define inf 99999999
#define pi acos(-1.0)
#define maxn 300050
int b[maxn],a[maxn];
ll sum[maxn];
int L[maxn],R[maxn];
int q[511111][2];

int main()
{
    int n,m,i,j,T,l,r,k;
    int front,rear;
    ll ans;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++){
            scanf("%d",&a[i]);
        }
        ans=0;
        for(k=1;k<=n;k++){
            int tot=0;
            sum[0]=0;
            for(i=1;i*k<=n;i++){
                tot++;
                b[tot]=a[i*k];
                sum[tot]=sum[tot-1]+b[tot];
            }
            front=1,rear=0;

            for(i=1;i<=tot;i++){
                while(front<=rear && q[rear][0]>=b[i] ){
                    rear--;
                }
                if(rear==0){
                    L[i]=1;
                }
                else{
                    L[i]=q[rear][1]+1;
                }
                rear++;
                q[rear][0]=b[i];q[rear][1]=i;

            }


            front=1,rear=0;

            for(i=tot;i>=1;i--){
                while(front<=rear && q[rear][0]>=b[i] ){
                    rear--;
                }
                if(rear==0){
                    R[i]=tot;
                }
                else{
                    R[i]=q[rear][1]-1;
                }
                rear++;
                q[rear][0]=b[i];q[rear][1]=i;
                ans=max(ans,(sum[R[i] ]-sum[L[i]-1 ])*b[i]*(int)sqrt((double)k)  );

            }
        }
        printf("%lld\n",ans);




    }
    return 0;





}