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;
}