A. Olympiad
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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, ..., an (0 ≤ ai ≤ 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
3
Input

Copy
3
1 1 1
Output
1
Input

Copy
4
42 0 0 42
Output
1
Note

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

  1. Only participants with 3 points will get diplomas.
  2. Participants with 2 or 3 points will get diplomas.
  3. 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 = ;
using namespace std;
typedef long long ll; int main()
{
set<int> s;
int n,a[maxn];
cin>>n;
for(int i=;i<n;i++){
cin>>a[i];
if(a[i]!=){
s.insert(a[i]);
}
}
cout<<s.size()<<endl;
}

模拟/stl

Codeforces Round #467 (Div. 2) A. Olympiad[输入一组数,求该数列合法的子集个数]的更多相关文章

  1. Codeforces Round #467 (div.2)

    Codeforces Round #467 (div.2) 我才不会打这种比赛呢 (其实本来打算打的) 谁叫它推迟到了\(00:05\) 我爱睡觉 题解 A. Olympiad 翻译 给你若干人的成绩 ...

  2. Codeforces Round #467 (Div. 2) B. Vile Grasshoppers

    2018-03-03 http://codeforces.com/problemset/problem/937/B B. Vile Grasshoppers time limit per test 1 ...

  3. Codeforces Round #467 Div.2题解

    A. Olympiad time limit per test 1 second memory limit per test 256 megabytes input standard input ou ...

  4. Codeforces Round #467 (Div. 1) B. Sleepy Game

    我一开始把题目看错了 我以为是博弈.. 这题就是一个简单的判环+dfs(不简单,挺烦的一题) #include <algorithm> #include <cstdio> #i ...

  5. Codeforces Round #467 (Div. 1). C - Lock Puzzle

    #include <algorithm> #include <cstdio> #include <cstring> #include <iostream> ...

  6. Codeforces Round #467 Div. 1

    B:显然即相当于能否找一条有长度为奇数的路径使得终点出度为0.如果没有环直接dp即可.有环的话可以考虑死了的spfa,由于每个点我们至多只需要让其入队两次,复杂度变成了优秀的O(kE).事实上就是拆点 ...

  7. Codeforces Round #467 (Div. 2) E -Lock Puzzle

    Lock Puzzle 题目大意:给你两个字符串一个s,一个t,长度<=2000,要求你进行小于等于6100次的shift操作,将s变成t, shift(x)表示将字符串的最后x个字符翻转后放到 ...

  8. Codeforces Round #467 (Div. 2) B. Vile Grasshoppers[求去掉2-y中所有2-p的数的倍数后剩下的最大值]

    B. Vile Grasshoppers time limit per test 1 second memory limit per test 256 megabytes input standard ...

  9. Codeforces Round #346 (Div. 2) E - New Reform 无相图求环

    题目链接: 题目 E. New Reform time limit per test 1 second memory limit per test 256 megabytes inputstandar ...

随机推荐

  1. A1016 Phone Bills (25)(25 分)

    A1016 Phone Bills (25)(25 分) A long-distance telephone company charges its customers by the followin ...

  2. 洛谷P4231 三步必杀

    题目描述: $N$ 个柱子排成一排,一开始每个柱子损伤度为0. 接下来勇仪会进行$M$ 次攻击,每次攻击可以用4个参数$l$ ,$r$ ,$s$ ,$e$ 来描述: 表示这次攻击作用范围为第$l$ 个 ...

  3. Linux命令之---touch

    命令简介 linux的touch命令不常用,一般在使用make的时候可能会用到,用来修改文件时间戳,或者新建一个不存在的文件. 命令格式 touch [选项]... 文件... 命令参数 -a   或 ...

  4. mysql之处理金钱小数点后的多余0

    问题产生原因:我们在做基金项目   产生大量的金钱  在GP首页展示首页信息的时候要求去除多余的0   由于我们在数据库设计的时候查询返回数据 例如18.100000 这种形式  而我们需要将多余的0 ...

  5. mongoTemplate学习笔记

    mongoTemplate的andExpression表达式 Aggregation<Post> agg = Aggregation.newAggregation( Record.clas ...

  6. 史上最全的MSSQL笔记

    http://www.cnblogs.com/gameworld/archive/2015/09/08/4790881.html

  7. WebApi 跨域

    http://www.cnblogs.com/lori/p/3557111.html http://bbs.csdn.net/topics/391020576

  8. MongoDB快速入门学习笔记6 MongoDB的文档删除操作

    db.集合名称.remove({query}, justOne)query:过滤条件,可选justOne:是否只删除查询到的第一条数据,值为true或者1时,只删除一条数据,默认为false,可选. ...

  9. DataFrame的iloc与loc的区别是什么?

    对于一个DataFrame A,A.loc[k]是读取A中index为k的那一行.A.iloc[k]是读取A中的第k行.

  10. Kafka 1.0版本发布

    Kafka 1.0版本发布 1.0.0 2017年11月1日发布 源码下载: kafka-1.0.0-src.tgz(asc,sha512) 二进制下载: Scala 2.11 - kafka_2.1 ...