poj 2796 Feel Good

最新推荐文章于 2022-01-14 15:26:41 发布

原创最新推荐文章于 2022-01-14 15:26:41 发布 · 402 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#poj #单调栈

单调栈专栏收录该内容

7 篇文章

订阅专栏

本文介绍了一种计算人类生活期间情绪价值的方法，通过将每个日子分配一个非负整数值来量化其情绪价值，并通过特定公式计算出一段时间内的总情绪价值。文章提供了一个C++实现示例，用于找出具有最大情绪价值的生活阶段。

Description

Bill is developing a new mathematical theory for human emotions. His recent investigations are dedicated to studying how good or bad days influent people's memories about some period of life.

A new idea Bill has recently developed assigns a non-negative integer value to each day of human life.

Bill calls this value the emotional value of the day. The greater the emotional value is, the better the daywas. Bill suggests that the value of some period of human life is proportional to the sum of the emotional values of the days in the given period, multiplied by the smallest emotional value of the day in it. This schema reflects that good on average period can be greatly spoiled by one very bad day.

Now Bill is planning to investigate his own life and find the period of his life that had the greatest value. Help him to do so.

题意是给出一串数，求给定的范围内最小值乘该范围内所有值的和的最大值

方法和poj 2559相同，仍然是假设某一点的数值是最小的，求左右两端可到达的最远的位置

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAX_N 100000
using namespace std;
int n,a[MAX_N+5],l[MAX_N+5],r[MAX_N+5];
int main()
{
    while(~scanf("%d",&n))
    {
        long long sum[MAX_N+5];
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        l[0]=0,r[n+1]=0;
        for(int i=1;i<=n;i++)
        {
            int k=i-1;
            while(k>0&&a[i]<=a[k]) k=l[k]-1;
            l[i]=++k;
        }
        for(int i=n;i>=1;i--)
        {
            int k=i+1;
            while(k<n+1&&a[i]<=a[k]) k=r[k]+1;
            r[i]=--k;
        }
        sum[0]=0,sum[1]=a[1];
        for(int i=2;i<=n;i++)
            sum[i]=sum[i-1]+a[i];
        long long maxn=-1;
        int maxi;
        for(int i=1;i<=n;i++)
        {
            long long q=sum[r[i]]-sum[l[i]-1];
            q=a[i]*q;
            if(q>maxn) { maxn=q; maxi=i;}
        }
        printf("%lld\n",maxn);
        printf("%d %d\n",l[maxi],r[maxi]);
    }
    return 0;
}