A题看数据很水直接暴力一遍,
#include<iostream> using namespace std; #define LL long long int main() { int k,l,m,n,d,ans=0; cin>>k>>l>>m>>n>>d; for(int i=1;i<=d;i++) { if(i%k==0||i%l==0||i%m==0||i%n==0)ans++; } cout<<ans<<endl; }
B题数据也很小,直接模拟他们要走的过程,不过要注意龙可能在不到一小时的时间就追上公主,所以要用double代码:
#include<iostream> using namespace std; #define LL long long int main() { double vp,vd,t,f,c; cin>>vp>>vd>>t>>f>>c; double n1=t*vp,n2=0; int ans=0; if(vp>vd) { cout<<0<<endl; return 0; } while(1) { if(n1>=c)break; double ti=n1/(vd-vp); //cout<<n1<<" "<<ti<<" "<<ans<<endl; if(ti*vp+n1>=c) { //cout<<ti*vp+n1; break; } else { ans++; n1=ti*vp+n1; n1=n1+(ti+f)*vp; } } cout<<ans<<endl; }
C题,如果b!=0就从2开始,记录sum和max每一次b>0都输出max+1,b=0之后再输出a遍max+1,其余的就输出max就行,注意b=0的情况,代码:
#include<bits/stdc++.h> using namespace std; vector<int>ans; int main() { ans.clear(); int n,a,b; cin>>n>>a>>b; if(a+b+1>n) { cout<<-1<<endl; return 0; } if(b==0) { if(a==0) { for(int i=0;i<n;i++)ans.push_back(15); } else if(a==n-1) { cout<<-1<<endl; return 0; } else { int sum=15,max=15,now=15; ans.push_back(15); ans.push_back(15); for(int i=2;i<n;i++) { if(a!=0) { ans.push_back(max+1); //cout<<" "<<max+1; max=max+1; a=a-1; } else { ans.push_back(max); //cout<<" "<<max; } } } } else { int sum=2,max=2; ans.push_back(2); for(int i=1;i<n;i++) { if(b!=0) { ans.push_back(sum+1); //cout<<" "<<sum+1; max=sum+1; sum=sum+sum+1; b--; } else if(a!=0) { ans.push_back(max+1); //cout<<" "<<max+1; max=max+1; a=a-1; } else { ans.push_back(max); //cout<<" "<<max; } } } if(a!=0||b!=0)cout<<-1<<endl; else { for(int i=0;i<ans.size();i++)cout<<ans[i]<<" "; } }