zoukankan      html  css  js  c++  java
  • Codeforces Round #371 (Div. 1) C. Sonya and Problem Wihtout a Legend 贪心

    C. Sonya and Problem Wihtout a Legend

    题目连接:

    http://codeforces.com/contest/713/problem/C

    Description

    Sonya was unable to think of a story for this problem, so here comes the formal description.

    You are given the array containing n positive integers. At one turn you can pick any element and increase or decrease it by 1. The goal is the make the array strictly increasing by making the minimum possible number of operations. You are allowed to change elements in any way, they can become negative or equal to 0.

    Input

    The first line of the input contains a single integer n (1 ≤ n ≤ 3000) — the length of the array.

    Next line contains n integer ai (1 ≤ ai ≤ 109).

    Output

    Print the minimum number of operation required to make the array strictly increasing.

    Sample Input

    7
    2 1 5 11 5 9 11

    Sample Output

    9

    Hint

    题意

    给你一个长度为n个序列,然后你每次操作可以使得一个数减小一,或者使得一个数增加一,问你最少多少次操作,可以使得这个序列为单增的。

    题解:

    我们让每一个数,一开始-=i,然后就可以把单增变成不降序列了。

    这下我们就可以贪心了,扔到一个堆里面去,每次拿出最大的元素,变成x就好了

    代码

    #include<bits/stdc++.h>
    using namespace std;
    int n;
    long long ans;
    priority_queue<int>s;
    int main()
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            int x;scanf("%d",&x);
            x-=i;
            s.push(x);
            if(s.top()>x)
            {
                ans+=s.top()-x;
                s.pop();
                s.push(x);
            }
        }
        cout<<ans<<endl;
    }
  • 相关阅读:
    计时器chronometer补充
    SpannableString富文本
    java数据结构整理(二)
    Java数据结构整理(一)
    graphical Layout调大一点
    eclipse快捷键
    弹出式菜单(下拉菜单)实现——PopupMenu
    二级横向菜单实现——ListView
    java求两个集合的差集
    Java中String的split()方法的一些需要注意的地方
  • 原文地址:https://www.cnblogs.com/qscqesze/p/5929195.html
Copyright © 2011-2022 走看看