zoukankan      html  css  js  c++  java
  • CodeForces 546B C(Contest #1)

    Description

    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.

    Sample Input

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

    #include <map>
    #include <set>
    #include <vector>
    #include <math.h>
    #include <bitset>
    #include <list>
    #include <algorithm>
    #include <climits>
    using namespace std;

    #define lson 2*i
    #define rson 2*i+1
    #define LS l,mid,lson
    #define RS mid+1,r,rson
    #define UP(i,x,y) for(i=x;i<=y;i++)
    #define DOWN(i,x,y) for(i=x;i>=y;i--)
    #define MEM(a,x) memset(a,x,sizeof(a))
    #define W(a) while(a)
    #define gcd(a,b) __gcd(a,b)
    #define LL long long
    #define N 5000005
    #define INF 0x3f3f3f3f
    #define EXP 1e-8
    #define lowbit(x) (x&-x)
    const int mod = 1e9+7;
    #define LL __int64
    int n,a[3005];
    int main()
    {
      int i,j,ans;
      while(~scanf("%d",&n))
      {
        ans = 0;
        int sum1 = 0,sum2 = 0;
        for(i = 1; i<=n; i++)
        {
          scanf("%d",&a[i]);
          sum1+=a[i];
        }
        sort(a+1,a+1+n);
        sum2 = a[1];
        for(i = 2; i<=n; i++)
        {
          if(a[i] == a[i-1])
          a[i]++;
          else if(a[i]<a[i-1])
          a[i] +=(a[i-1]-a[i])+1;
          sum2+=a[i];
        }
        printf("%d ",sum2-sum1);
      }

    return 0;
    }



  • 相关阅读:
    Lotus iNotes 用户启用标识符保险库
    Domino NSD日志诊断/分析
    从 Domino 7.x 升级到 Domino 8.0.1 后服务器性能下降
    Domino服务器命令表
    源码:使用LotusScript发送mime格式邮件
    构架Domino CA中心之一
    如何在DNS中增加SPF记录
    构架Domino CA中心之二
    在Ubuntu 8.04上安装Domino R8.02
    内存陷阱 驯服C++中的野指针 沧海
  • 原文地址:https://www.cnblogs.com/xl1164191281/p/4659845.html
Copyright © 2011-2022 走看看