Codeforces AIM Tech Round (Div. 2) B. Making a String

B. Making a String

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

You are given an alphabet consisting of n letters, your task is to make a string of the maximum possible length so that the following conditions are satisfied:

the i-th letter occurs in the string no more thanai times;
the number of occurrences of each letter in the string must be
distinct for all the letters that occurred in the string at least once.

Input
The first line of the input contains a single integer n (2  ≤  n  ≤  26) — the number of letters in the alphabet.

The next line contains n integers
ai (1 ≤ ai ≤ 109) —i-th of these integers gives the
limitation on the number of occurrences of thei-th character in the string.

Output
Print a single integer — the maximum length of the string that meets all the requirements.

Sample test(s)

Input

3
2 5 5

Output

11

Input

3
1 1 2

Output

3

Note
For convenience let's consider an alphabet consisting of three letters: "a", "b", "c". In the first sample, some of the optimal strings are:
"cccaabbccbb", "aabcbcbcbcb". In the second sample some of the optimal strings are: "acc", "cbc".

题意：26个字母组成的串，给你字母的数量和每个字母可以放的数量，组成串的时候有一定规则：

1.串中每个字母的数量不能多于给定的数量

2.串中每个字母出现的次数都不同

问组成的最长串是多少。

思路：直接排序，然后一个一个放，并检查所放数量之前是否出现过。。

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 100010
#define LL long long
#define ll __int64
#define INF 0x7fffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
#define eps 1e-10
using namespace std;
int gcd(int a,int b){return b?gcd(b,a%b):a;}
int lcm(int a,int b){return a/gcd(a,b)*b;}
LL powmod(LL a,LL b,LL MOD){LL ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
//head
ll num[30];
ll a[30];
int main()
{
int n,i,j;
while(scanf("%d",&n)!=EOF)
{
int cnt=0;
for(i=0;i<n;i++)
scanf("%I64d",&num[i]);
sort(num,num+n);
for(i=0;i<n;i++)
{
while(num[i])
{
int bz=0;
for(j=0;j<cnt;j++)
if(a[j]==num[i])
{
bz=1;
break;
}
if(bz==0)
break;
num[i]--;
}
if(num[i])
a[cnt++]=num[i];
}
ll sum=0;
for(i=0;i<cnt;i++)
sum+=a[i];
printf("%I64d\n",sum);
}
return 0;
}