Codeforces Round #467 (Div. 2) A. Olympiad[输入一组数,求该数列合法的子集个数]-优快云博客

本文介绍了一道计分赛题目，任务是确定合法子集的数量。合法子集需满足不含0分选手、非空及所有分数不大于选定分数的要求。文章通过使用STL set进行非零分数去重，实现了简洁高效的解决方案。

The recent All-Berland Olympiad in Informatics featured n participants with each scoring a certain amount of points.

As the head of the programming committee, you are to determine the set of participants to be awarded with diplomas with respect to the following criteria:

At least one participant should get a diploma.
None of those with score equal to zero should get awarded.
When someone is awarded, all participants with score not less than his score should also be awarded.

Determine the number of ways to choose a subset of participants that will receive the diplomas.

Input

The first line contains a single integer n (1 ≤ n ≤ 100) — the number of participants.

The next line contains a sequence of n integers a₁, a₂, ..., a_n (0 ≤ a_i ≤ 600) — participants' scores.

It's guaranteed that at least one participant has non-zero score.

Output

Print a single integer — the desired number of ways.

Examples

Input

Copy

4
1 3 3 2

Output

Input

Copy

3
1 1 1

Output

Input

Copy

4
42 0 0 42

Output

Note

There are three ways to choose a subset in sample case one.

Only participants with 3 points will get diplomas.
Participants with 2 or 3 points will get diplomas.
Everyone will get a diploma!

The only option in sample case two is to award everyone.

Note that in sample case three participants with zero scores cannot get anything.

[题意]:输入一组数,求该数列合法的子集个数.合法是指:不含0 && 非空子集 && 选定某个数且其他数<=该数

[分析]:先想要用标记数组,后来发现直接用set对非零数去重求set的size就可以了

[代码]:

#include<bits/stdc++.h>
#define MOD 1000000007
const int maxn = 1000;
using namespace std;
typedef long long ll;

int main()
{
    set<int> s;
    int n,a[maxn];
    cin>>n;
    for(int i=0;i<n;i++){
        cin>>a[i];
        if(a[i]!=0){
        s.insert(a[i]);
        }
    }
    cout<<s.size()<<endl;
}