zoukankan      html  css  js  c++  java
  • Soldier and Badges (集合)

    Colonel has n badges. He wants to give one badge to every of his n soldiers. Each badge has a coolness factor, which shows how much it's owner reached. Coolness factor can be increased by one for the cost of one coin.

    For every pair of soldiers one of them should get a badge with strictly higher factor than the second one. Exact values of their factors aren't important, they just need to have distinct factors.

    Colonel knows, which soldier is supposed to get which badge initially, but there is a problem. Some of badges may have the same factor of coolness. Help him and calculate how much money has to be paid for making all badges have different factors of coolness.

    Input

    First line of input consists of one integer n (1 ≤ n ≤ 3000).

    Next line consists of n integers ai (1 ≤ ai ≤ n), which stand for coolness factor of each badge.

    Output

    Output single integer — minimum amount of coins the colonel has to pay.

    Example

    Input
    4
    1 3 1 4
    Output
    1
    Input
    5
    1 2 3 2 5
    Output
    2

    Note

    In first sample test we can increase factor of first badge by 1.

    In second sample test we can increase factors of the second and the third badge by 1.

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<string>
    #include<set>
    #include<vector>
    #include<stack>
    #include<queue>
    #include<algorithm>
    #include<iostream>
    #include<cstdio>
    #include<algorithm>
    using namespace std;
    int main()
    {
    
        int n;
        cin >> n;
        set <int> s;
        int a[3005];
        int flag = 0;
        for (int i = 0; i<n; i++)
            cin >> a[i];
        s.insert(a[0]);
        set<int>::iterator it;
        for (int i = 1; i<n; i++)
        {
            it = s.find(a[i]);
            while (it != s.end())
            {
                a[i]++;
                flag++;
                it = s.find(a[i]);
            }
            s.insert(a[i]);
        }
        cout << flag << endl;
        return 0;
    }
  • 相关阅读:
    PerfDog WEB端使用手册
    PerfDog4.0探索,支持用户自建web云
    无AI不测试:人工智能时代背景下,如何发展与应用自动化测试?
    性能测试实践 | PerfDog助力微信小游戏/小程序性能调优
    mysql面向过程学习
    阿里云操作记录
    socket网络编程
    学习慕课广告系统
    xinhuadouxxx总结
    maven+springboot+mybatis快速搭建简单使用
  • 原文地址:https://www.cnblogs.com/edych/p/7213935.html
Copyright © 2011-2022 走看看