Codeforces Round #467 (Div. 2) A. Olympiad[输入一组数,求该数列合法的子集个数]
1 second
256 megabytes
standard input
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.
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.
Print a single integer — the desired number of ways.
4
1 3 3 2
3
3
1 1 1
1
4
42 0 0 42
1
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 = ;
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
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