E. Exam Results
题意：对于一门考试，及格线是最高分x乘以p%，给定n个学生在心态好与差的情况下的分数ai与bi，给定常数p，请你估算出，在最理想情况下有多少人及格。

思路：不断枚举最大值，这个过程中及格线也在不断变化，每次把淘汰的最大值先从及格区减去，换成他们发挥不好的情况，然后再去判断能否达到新的及格线。

代码偷了队友的，思路虽然有但是代码我真写不出来。

#include 
using namespace std;
#define endl '\n'
#define int long long
const int N = 3e6+100;
int t,n,p;
struct node
{int id,a,b;bool operator<(const node &c)const{return c.a>a;}
}stu[N];
mapmp;
vectorg[N],v;
sets;
signed main()
{cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);cin>>t;int ca=0;while(t--){cin>>n>>p;mp.clear();s.clear();v.clear();for(int i=1;i<=n;i++) g[i].clear();int cnt=0,maxx=0,ct=0;for(int i=1;i<=n;i++){cin>>stu[i].a>>stu[i].b,stu[i].id=i;if(!mp[stu[i].a]) mp[stu[i].a]=++cnt;g[mp[stu[i].a]].push_back(i);maxx=max(maxx,stu[i].b);}priority_queueq;for(int i=1;i<=n;i++){if(stu[i].a>=maxx) s.insert(stu[i].a);}s.insert(maxx);for(auto x:s) v.push_back(x);double jg=v.back()*p*1.0/100;int ans=0,res=0;for(int i=1;i<=n;i++){double grade=stu[i].a;//cout<<"grade="<=jg) res++;else q.push(stu[i]);}ans=max(ans,res);//cout<<"ans="<0){double now=v.back()*p*1.0/100;for(auto x:g[mp[las]]){double grade=stu[x].b;if(graderes--;stu[x].a=stu[x].b;q.push(stu[x]);}}while(!q.empty()){double grade=q.top().a;if(grade>=now) res++,q.pop();else break;}ans=max(res,ans);las=v.back();//cout<<"res="<

# include
using namespace std;
# define mod 1000000009typedef long long int ll;
mapmp;
int main ()
{int t;cin>>t;while(t--){int n;cin>>n;mp.clear();int cnt=0;for(int i=1;i<=n;i++){int x;cin>>x;mp[x]++;if(mp[x]<=2)cnt++;}int ans=cnt/2;if(cnt%2)ans++;cout<

#include 
using namespace std;
//#define int long long
#define endl '\n'
const int N = 2e5+100;
int a[N],a1[N],b[N];int getLIS(int a[],int n)
{b[1]=a[1];int len=1;for(int i=2;i<=n;i++){if(b[len]b[++len]=a[i];}else{int pos=lower_bound(b+1,b+len+1,a[i])-b;b[pos]=a[i];}}return len;
}
int getLDS(int a[],int n)
{b[1]=a[1];int len=1;for(int i=2;i<=n;i++){if(b[len]>=a[i])//>=与upper结合，代表非严格递减子序列{b[++len]=a[i];}else{int pos=upper_bound(b+1,b+len+1,a[i],greater())-b;b[pos]=a[i];}}return len;
}
signed main()
{cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);int t;for(cin>>t;t;t--){int n;cin>>n;for(int i=1;i<=n;i++){cin>>a[i];a1[n-i+1]=a[i];}int ans1=getLIS(a,n);int ans2=getLIS(a1,n);cout<

#include using namespace std;
#define endl '\n'
const int N = 2e6+100;
int a[N];
char s[N];
int main()
{cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);int t;for(cin>>t;t;t--){int n;cin>>n;int res=(1<<20)-1;for(int i=0;icin>>a[i];if(a[i]!=i)res&=i;}cout<

#include 
using namespace std;
#define endl '\n'
#define int long long
const int N = 3e6+100;
int a[N];
int ans[N];
signed main()
{cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);int t;for(cin>>t;t;t--){int n,k;cin>>n>>k;bool falg=false;for(int i=1;i<=n;i++){char ch;cin>>ch;a[i]=ch-'0';if(a[i])falg=true;}if(!falg){cout<<0<if(a[i]){fi=i;break;}}for(int i=n;i>=1;i--){if(a[i]){la=i;break;}}//cout<if(n-la<=k){swap(a[n],a[la]);}else if(la-1<=k){swap(a[fi],a[1]);}}else{if(n-la<=k){swap(a[n],a[la]);k-=(n-la);}if(fi-1<=k){swap(a[fi],a[1]);}}int ans=0;for(int i=1;i<=n;i++){//cout<if(i==1){ans+=10;}else if(i==n){ans+=1;}else{ans+=11;}}}//cout<