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. Diycode开源项目 TopicContentActivity分析

    1.效果预览以及布局分析 1.1.实际效果预览 左侧话题列表的布局是通过TopicProvider来实现的,所以当初分析话题列表就没有看到布局. 这里的话题内容不是一个ListView,故要自己布局. ...

  2. 使用chrome开发程序,自动刷新开发目录

    npm i livereload -g 在开发目录下: livereload 安装这个插件:https://chrome.google.com/webstore/detail/livereload/j ...

  3. Careercup - Microsoft面试题 - 5917873302142976

    2014-05-12 06:56 题目链接 原题: A link list contains following elements struct node{ int data; node* next; ...

  4. Python+Selenium框架设计篇之-简单介绍unittest单元测试框架

    前面文章已经简单介绍了一些关于自动化测试框架的介绍,知道了什么是自动化测试框架,主要有哪些特点,基本组成部分等.在继续介绍框架设计之前,我们先来学习一个工具,叫unittest.       unit ...

  5. Wordpress 通过 post id 获取文章 url

    global $post; echo get_permalink($post->ID); 函数详解: Codex - get_permalink() 注意:有些链接是通过 SEO 重定向的,比如 ...

  6. poj 3278 catch that cow BFS(基础水)

    Catch That Cow Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 61826   Accepted: 19329 ...

  7. Mybatis通过接口的方式实现增删改查

    导入jar包 [mybatis] [oracle] 生成数据库 1.添加Mybatis的配置文件mybatis-config.xml 在src目录下创建一个mybatis-config.xml文件,如 ...

  8. springboot的api测试

  9. 用Vundle管理Vim插件

    作为程序员,一个好用的Vim,是极其重要的,而插件能够使原本功能羸弱的Vim变得像其他功能强大的IDE一样好用.然而下载.配置插件的过程比较繁琐,大家往往需要自己进行下载/配置等操作,如果还涉及到更新 ...

  10. 【bzoj4386】[POI2015]Wycieczki 矩阵乘法

    题目描述 给定一张n个点m条边的带权有向图,每条边的边权只可能是1,2,3中的一种.将所有可能的路径按路径长度排序,请输出第k小的路径的长度,注意路径不一定是简单路径,即可以重复走同一个点. 输入 第 ...