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 ai (1 ≤ ai ≤ 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.
MD水题，用错了方法
题意是给出一个序列数，要求在每个数加一下，知道这个序列每个数都不相同
思路是 先排序，然后从第二个遍历到最后 每个数如果和前面的相同 则加1 如果前面的数小于后面的数 则后面的数等于前面的数加1
我自己写的是如果有相同的数，则这个数变成离它最近且序列不存在的数 .....当这个序列很大的时候这个方法就不行了
代码如下：
<span style="font-size:14px;">#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define N 3000
int a[N];
int main()
{
int n,i,j,m;
while(scanf("%d",&n)!=EOF)
{
memset(a,0,sizeof(a));
int s1=0,s2=0;
for(i=1;i<=n;i++)
{
cin>>a[i];
s1+=a[i];
}
sort(a+1,a+1+n);
s2=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]+1;
s2+=a[i];
}
cout<<s2-s1<<endl;
}
return 0;
}</span>
