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CodeForces 546B C(Contest #1)

2015-07-19 21:24 267 查看
DescriptionColonel 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 InputInput
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\n",sum2-sum1);
  }return 0;
}

                                            
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