hihoCoder挑战赛29

多打打不同的比赛，找经验啊

题目4 : 不上升序列

时间限制:40000ms

单点时限:2000ms

内存限制:256MB

描述

给定一个长度为 n 的非负整数序列 a[1..n]。

你每次可以花费 1 的代价给某个 a[i] 加1或者减1。

求最少需要多少代价能将这个序列变成一个不上升序列。

输入

第一行一个正整数 n。

接下来 n 行每行一个非负整数，第 i 行表示 a[i]。

1 ≤ n ≤ 500000

0 < a[i] ≤ 109

输出

一个非负整数，表示答案。

样例解释

[5,3,4,5] -> [5,4,4,4]

样例输入

样例输出


hiho的代码

#include <bits/stdc++.h>
using namespace std;
long long c,n,k;
priority_queue<int,vector<int>, greater<int>  >a;
int main(){
    cin>>n;
    for(int i=0;i<n;i++){
        cin>>c;
        a.push(c);
        if(a.top()<c){
        k+=c-a.top();
        a.pop();
        a.push(c);
        }
    }
    cout<<k;
    return 0;
}

变成相反数，不用重载写起来舒服的

#include <bits/stdc++.h>
using namespace std;
long long c,n,k;
priority_queue<int>a;
int main()
{
    cin>>n;
    for(int i=0; i<n; i++)
    {
        cin>>c;
        c=-c,a.push(c);
        if(a.top()>c)k+=a.top()-c,a.pop(),a.push(c);
    }
    cout<<k;
    return 0;
}

大佬们都讲是原题，找了下这个题

Little Petya likes to play very much. And most of all he likes to play the following game:

He is given a sequence of N integer numbers. At each step it is allowed to increase the value of any number by 1 or to decrease it by 1. The goal of the game is to make the sequence non-decreasing with the smallest number of steps. Petya is not good at math, so he asks for your help.

The sequence a is called non-decreasing if a1 ≤ a2 ≤ ... ≤ aN holds, where N is the length of the sequence.

Input

The first line of the input contains single integer N (1 ≤ N ≤ 5000) — the length of the initial sequence. The following N lines contain one integer each — elements of the sequence. These numbers do not exceed 109 by absolute value.

Output

Output one integer — minimum number of steps required to achieve the goal.

Examples

input

5
3 2 -1 2 11

output

input

5
2 1 1 1 1

output

1
CF的这道题代码如下

#include <bits/stdc++.h>
using namespace std;
long long c,n,k;
priority_queue<int>a;
int main(){
    cin>>n;
    for(int i=0;i<n;i++){
        cin>>c;
        a.push(c);
        if(a.top()>c){
        k+=a.top()-c;
        a.pop();
        a.push(c);
        }
    }
    cout<<k;
    return 0;
}