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

4
5
3
4
5

样例输出

2

---

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=;i<n;i++){
        cin>>c;
        a.push(c);
        if(a.top()<c){
        k+=c-a.top();
        a.pop();
        a.push(c);
        }
    }
    cout<<k;
    return ;
}

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=; 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 ;
}

大佬们都讲是原题，找了下这个题

C. Sequence

time limit per test

1 second

memory limit per test

64 megabytes

input

standard input

output

standard output

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

4

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=;i<n;i++){
        cin>>c;
        a.push(c);
        if(a.top()>c){
        k+=a.top()-c;
        a.pop();
        a.push(c);
        }
    }
    cout<<k;
    return ;
}