CF 546 B. Soldier and Badges

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                               ***B. Soldier and Badges***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.InputFirst 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.OutputOutput single integer — minimum amount of coins the colonel has to pay.

Sample test(s)
input
4
1 3 1 4
output
1
input
5
1 2 3 2 5
output
2

题目大意：就是让所有的数不相等，所需要的操作数

样例解析：

Input:

4
1 3 1 4

Output:
1

先排一下序：1 1 3 4
所以不相等的话只需要第二个数+1
变为： 1 2 3 4
所以操作数 == 1

具体见代码：
上代码：

/*Date : 2015-09-03 晚上Author : ITAKMotto :今日的我要超越昨日的我，明日的我要胜过今日的我；以创作出更好的代码为目标，不断地超越自己。*/#include <iostream>#include <cstdio>#include <algorithm>using namespace std;int arr[3005];int main(){    int n;    while(~scanf("%d", &n))    {        for(int i=0; i<n; i++)            scanf("%d", &arr[i]);        sort(arr, arr+n);        int sum = 0;        for(int i=0; i<n; i++)        {            if(arr[i-1] >= arr[i])            {                sum += arr[i-1] - arr[i] + 1;                arr[i] += arr[i-1] -arr[i] + 1;            }        }        printf("%d\n",sum);    }    return 0;}