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;

}

巴特西

codeforce round #467(div.2)

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