codeforce round #467(div.2)

A. Olympiad

给出n个数，让你找出有几个非零并且不重复的数

所以用stl的set

//#define debug
#include<stdio.h>
#include<math.h>
#include<cmath>
#include<queue>
#include<stack>
#include<string>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<functional>
#include<iomanip>
#include<map>
#include<set>
#define pb push_back
using namespace std;
typedef long long ll;
pair<ll,ll>PLL;
pair<int,ll>Pil;
const ll INF = 0x3f3f3f3f;
const double inf=1e8+100;
const ll maxn =1e5+100;
const int N = 1e4+10;
const ll mod=1000007;
int n,a[maxn];
set<int>s;
set<int>::iterator it;
void solve() {
	int i,j,t=1;
//	cin>>t;
	while(t--){
		cin>>n;
		while(n--){
			int so;
			cin>>so;
			if(so>0)
			s.insert(so);
		}
		cout<<s.size()<<endl;
		s.clear();
	}
}

int main() {
	ios_base::sync_with_stdio(false);
#ifdef debug
	freopen("in.txt", "r", stdin);
	freopen("out.txt","w",stdout);
#endif
	cin.tie(0);
	cout.tie(0);
	solve();

#ifdef debug
	fclose(stdin);
	fclose(stdout);
	system("out.txt");
#endif
	return 0;
}

B. Vile Grasshoppers

给定一个[p,y]区间，找出其中最大的素数

#define debug
#include<stdio.h>
#include<math.h>
#include<cmath>
#include<queue>
#include<stack>
#include<string>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<functional>
#include<iomanip>
#include<map>
#include<set>
#define pb push_back
using namespace std;
typedef long long ll;
pair<ll,ll>PLL;
pair<int,ll>Pil;
const ll INF = 0x3f3f3f3f;
const double inf=1e8+100;
const ll maxn =1e5+100;
const int N = 1e4+10;
const ll mod=1000007;
int p,y;
bool prime(int x) {
	for(int i=2; i*i<=x&&i<=p; i++) {
		if(x%i==0)
			return 0;
	}
	return 1;
}
void solve() {
	int i,j,t=1;
//	cin>>t;
	while(t--) {
		cin >> p >> y;
		for(i=y; i>p; i--) {
			if(prime(i)) {
				cout<<i<< endl;
				return;
			}
		}
		cout <<-1<< endl;
	}
}

int main() {
	ios_base::sync_with_stdio(false);
#ifdef debug
	freopen("in.txt", "r", stdin);
	freopen("out.txt","w",stdout);
#endif
	cin.tie(0);
	cout.tie(0);
	solve();

#ifdef debug
	fclose(stdin);
	fclose(stdout);
	system("out.txt");
#endif
	return 0;
}

C. Save Energy!

一个炉子打开可以烧k时间，julia每d时间去厨房看一趟，一只鸡在炉子一直在烧的时候，烧熟需要t时间，否则需要2t

分析：实际上可以用时间来代表一只鸡烧熟需要的能量(2*t)，所以炉子开着时产生的能量就为2*k,因此当

①k%d==0时，所花的时间就为t

②k%d!=0时，我们需要求一次循环的时间：d=(k/d+1)*d（包含k时间）；循环的能量：circle=2*k+d-k；循环几次：ans=2*t/circle；剩余能量：t=2*t%circle。最后判断剩余的能量在哪一个位子：（I）t/2<=k，ans=ans*d+t/2 （II）t/2>k，ans=ans*d+k+t-2*k(设最后一段所需要的时间为tt，则t=tt-k+2*k即tt=t+k-2*k)

#define debug
#include<stdio.h>
#include<math.h>
#include<cmath>
#include<queue>
#include<stack>
#include<string>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<vector>
#include<functional>
#include<iomanip>
#include<map>
#include<set>
#define pb push_back
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll>PLL;
typedef pair<int,ll>Pil;
const ll INF = 0x3f3f3f3f;
const double inf=1e8+100;
const ll maxn =1e4+100;
const int N = 1e4+10;
const ll mod=1000007;
const int ml=1e6;
ll k,d,t,ans=0;
void solve() {
	int i,j,tt=1;
//	cin>>t;
	while(tt--){
		ll x;
		cin>>k>>d>>t;
		if(k%d==0){
			cout<<t<<endl;
		}
		else{
			d=(k/d+1)*d;
			t*=2;
			x=d+k;
			ans=(t/x)*d;
			t%=x;
			if(t<=2*k){
				cout<<fixed<<setprecision(2)<<(double)ans+t*0.5<<endl;
			}
			else{
				t-=2*k;
				ans+=k;
				cout<<ans+t<<endl;
			}
		}
	}
}

int main() {
	ios_base::sync_with_stdio(false);
#ifdef debug
	freopen("in.txt", "r", stdin);
//	freopen("out.txt","w",stdout);
#endif
	cin.tie(0);
	cout.tie(0);
	solve();
#ifdef debug
	fclose(stdin);
	fclose(stdout);
	system("out.txt");
#endif
	return 0;
}