Soldier and Badges(546B)

最新推荐文章于 2022-03-23 19:53:35 发布

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本文介绍了一个关于计算调整一组徽章的“酷值”所需最小花费的问题。为了确保每个士兵获得的徽章“酷值”各不相同，文章提供了一种算法来计算所需的最低成本。

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

Input

5
1 2 3 2 5

Output

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.

注：冷酷值只能上升，不能下降。

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <ctype.h>
int comp( const *a,const *b)
{
	return *(int*)a-*(int*)b; 
} 
int main()  
{	
	int n,i,m,sum;
	int s[3001];
	while(scanf("%d",&n)!=EOF)
	{
		m=0;
		for(i=0;i<n;i++)
		{
			scanf("%d",&s[i]);
			m+=s[i];
		}
		qsort(s,n,sizeof(int),comp); 
		sum=s[0];
		for(i=1;i<n;i++)
		{
			if(s[i]==s[i-1])
			{
				s[i]++;
			}
			else if(s[i]<s[i-1])
			{
				s[i]+=(s[i-1]-s[i])+1;
			}
			sum+=s[i];
		}
		printf("%d\n",sum-m);
	}
	
    return 0;  
}