hdu5662 YJQQQAQ and the function (单调栈)

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

题意：给你n个数，你要找到3个数，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是这些数的最小值。

思路：看到求一些数的和乘这些数中最小值的最小值，容易想到单调栈，我们只要先把在k确定的情况下找出所有符合条件的A[],然后再用单调栈找出以每一个数位最小值的左右边界，然后更新ans的最大值就行了。ps：还是不习惯用单调栈，所以用两遍单调队列做了，时间复杂度会差一下，不过差不多。

#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;

}

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hdu5662 YJQQQAQ and the function (单调栈)

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