zoukankan      html  css  js  c++  java
  • C

    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.

    Input

    The first line of the input contains n - the number of days of Bill's life he is planning to investigate(1 <= n <= 100 000). The rest of the file contains n integer numbers a1, a2, ... an ranging from 0 to 10 6 - the emotional values of the days. Numbers are separated by spaces and/or line breaks.

    Output

    Print the greatest value of some period of Bill's life in the first line. And on the second line print two numbers l and r such that the period from l-th to r-th day of Bill's life(inclusive) has the greatest possible value. If there are multiple periods with the greatest possible value,then print any one of them.

    Sample Input

    6
    3 1 6 4 5 2
    

    Sample Output

    60
    3 5

    思路:找到区间的最小值,然后我们想办法去扩大这个区间。如果a[i]的值比最小值大肯定可以,如果比他小就不可以了

    这题要注意他的a[i]值可以是0    淦!

     1 #include <stdio.h>
     2 #include <algorithm>
     3 #include <iostream>
     4 #include <stdlib.h>
     5 #include <math.h>
     6 #include <stack>
     7 #include <queue>
     8 
     9 #define INF 0x3f3f3f3f
    10 #define MAXN 100005
    11 #define LL long long
    12 using namespace std;
    13 
    14 int L[MAXN],R[MAXN],a[MAXN];
    15 LL pre[MAXN];
    16 
    17 stack<int> stc;
    18 
    19 int main()
    20 {
    21     int n;
    22     scanf("%d",&n);
    23     pre[0] = 0;
    24     for (int i=1;i<=n;i++)
    25     {
    26         scanf("%d",&a[i]);
    27         pre[i] = pre[i-1]+a[i];
    28     }
    29     a[n+1] = -1;
    30     a[0] = -1;
    31     for (int i=1;i<=n+1;i++)
    32     {
    33         if (stc.empty() || a[i]>=a[stc.top()])
    34             stc.push(i);
    35         else
    36         {
    37             while (!stc.empty() && a[i]<a[stc.top()])
    38             {
    39                 R[stc.top()] = i-1;
    40                 stc.pop();
    41             }
    42             stc.push(i);
    43         }
    44     }
    45     while (!stc.empty())
    46         stc.pop();
    47     for (int i=n;i>=0;i--)
    48     {
    49         if (stc.empty() || a[i]>=a[stc.top()])
    50             stc.push(i);
    51         else
    52         {
    53             while (!stc.empty() && a[i]<a[stc.top()])
    54             {
    55                 L[stc.top()] = i+1;
    56                 stc.pop();
    57             }
    58             stc.push(i);
    59         }
    60     }
    61     LL ans = -1;
    62     int l,r;
    63     for (int i=1;i<=n;i++)
    64     {
    65         if ((pre[R[i]]-pre[L[i]-1])*a[i]>ans)
    66         {
    67             ans = (pre[R[i]]-pre[L[i]-1])*a[i];
    68             l = L[i];
    69             r = R[i];
    70         }
    71     }
    72     printf("%lld
    %d %d",ans,l,r);
    73     return 0;
    74 }
  • 相关阅读:
    常见26个jquery使用技巧详解(比如禁止右键点击、隐藏文本框文字等)
    禁用页面及页面所有frame内的右键菜单
    JS模态窗口返回值兼容问题解决方案
    winform登录时,在密码框按下回车,直接登陆
    C#中实现邮件发送功能
    CTFHub-彩蛋(持续更新~)
    CTFHub技能树-目录遍历
    免密钥方式登陆配置
    Ansible_Day1
    Python_Day2_共享你的代码
  • 原文地址:https://www.cnblogs.com/-Ackerman/p/11298184.html
Copyright © 2011-2022 走看看