#include<iostream> using namespace std; int main() {     long x=0;     long y=0;     cin>>x>>y;     if(y%x==0) cout<<x+y<<endl;     else&nbs...

#include<iostream> using namespace std; int main() {     int m,n;     cin>>m>>n;     if(m%n==0)cout<<"YES"<<endl;     else cout<<"NO"<<endl; }

#include<iostream> using namespace std; int main() {     int a,b;     cin>>a>>b;     if(a==b) cout<<"Tacit！"<<endl;     else cout<<"No Tacit!"<&lt...

#include<iostream> using namespace std; int main() {     int n;     cin>>n;     if(n%2!=0)cout<<3*n+1<<endl;     else cout<<n/2<<endl; }

#include<iostream> using namespace std; int main() {     int a,b;     cin>>a>>b;     if(a>b)cout<<">"<<endl;     else if(a==b)cout<<"="<<endl; &nbs...

#include<iostream> using namespace std; int main() {     long long a,b;     cin>>a>>b;     cout<<a*b-a-b; }

#include<iostream> using namespace std; #include<algorithm> int main() {     int a,b,c,x,y;     cin>>a>>b>>c;        x=min(b,c);     if(b+c<=a) y=0...

#include<iostream> using namespace std; #include<cmath> int main() {    int n;     double H,S1,V,L,K;     int a;     cin>>H>>S1>>V>>L>>K>>n; &...

#include<iostream> using namespace std; int main() {     int n;     cin>>n;     cout<<(n-1)/2<<endl; }

#include<bits/stdc++.h> using namespace std; int main() {     int n;     cin>>n;     if(n%3==0)cout<<n/3*2<<endl;     else cout<<n/3*2+1<<endl;   &nb...

#include<iostream> using namespace std; #include<algorithm> int main() {     int a,b,c,d,e,f;     cin>>a>>b>>c>>d>>e>>f;     cout<<min(a,e)+min(b,f)+min(c,d)<<e...

#include<iostream> using namespace std; int main() {     int a;     cin>>a;     if(a%2==0) cout<<a/2<<endl;     else cout<<(a+1)/2<<endl; }

#include<iostream> using namespace std; int main() {     long long a;     cin>>a;     cout<<a+1<<endl; } 显然问题为找到p,k,满足:p+k最大,且pk=n. ans=p+k=p+logpn. 当p增大时,logpn的减少量很小,因此尽量让p大,所以当p=n,k...

#include<bits/stdc++.h> using namespace std; int main() {     double a;     cin>>a;     double b=cbrt(a);     cout<<fixed<<setprecision(3)<<3*b<<endl; }