2015 HUAS Provincial Select Contest #1 C

题目：

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.

题目大意：输入N个数，至少要加多少个1，才能使这N个数各不相同；

解题思路：用数组存储这N个数，用头文件algorithm中的sort函数排列，再用

                       for(int j=1;j<n;j++)
			{
				if(a[j]<=a[j-1])
				{
					l+=a[j-1]-a[j]+1;
					a[j]+=a[j-1]-a[j]+1;
					
				}
			}
求出答案。
代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int main()
{
    int n,l;
    int* a; 
    while(scanf("%d",&n)==1)
    {    
        if(n>=1&&n<=3000)
        {
              l=0;
            a=new int[n];
            for(int i=0;i<n;i++)
            {
                scanf("%d",&a[i]);
                if(a[i]<1||a[i]>n)
                  scanf("%d",&a[i]);
            }
            sort(a,a+n);
            for(int j=1;j<n;j++)
            {
                if(a[j]<=a[j-1])
                {
                    l+=a[j-1]-a[j]+1;
                    a[j]+=a[j-1]-a[j]+1;
                    
                }
            }
            printf("%d
",l);
        }
    }
    return 0;
}