B - Soldier and Badges

Time Limit:3000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

 

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 a_i (1 ≤ a_i ≤ 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

Hint

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.

 

题意：

军官要给手下的n个士兵发奖章，给定n个整数，可在每个数上多次加1，要求最终结果所有n个数均不相等，求最少要加多少次1。

 

首先我们设一个数组a来储存所有n个数字，按升序排序。对a数组扫一遍，如果a[i]==a[i+1],则需要改变数的大小。那么我们怎么知道要加多少个一呢？在本题中我设了一个ans标记用来表示比a[i+1]大且不存在a中的最小整数，ans-a[i+1]即为我们要加1的个数。

注意：每次加1之后整个a数组就会发生改变，为此我定义了一个数组b来表示变化后的a数组，以便于求ans。

 

附AC代码：

#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

int a[3010],b[3010];

int main(){
    int n,t;
    cin>>n;
    for(int i=0;i<n;i++){
        cin>>a[i];
        b[i]=a[i];//b数组来表示变化后的a数组 
    }
    sort(a,a+n);//排序 
    sort(b,b+n);
    int ans=a[0];
    int sum=0;
    for(int i=0;i<n;i++){ 
        if(a[i]==ans){
            ans++;
        }
    }
    for(int j=0;j<n;j++){
        if(a[j]==a[j+1]){
            sum+=(ans-a[j+1]);
            b[j+1]=ans;//变化 
            ans++;
        }
        else if(a[j+1]>ans){//保证ans不小于a[j+1] 
            ans=a[j+1];
        }
        for(int i=0;i<n;i++){
        if(b[i]==ans){
            ans++;
        }
    }
    }
    cout<<sum;
    
    return 0; 
}