Codeforces Round #422 (Div. 2) Archive

Codeforces Round #422 (Div. 2), problem: (D) My pretty girl Noora Solution In C/C++

Codeforces Round #422 (Div. 2), problem: (D) My pretty girl Noora Solution In C/C++

#include<bits/stdc++.h> using namespace std; long long f,pwr,mod=1e9+7,t,r,l,ans; main() { int i,j,k; cin>>t>>l>>r; pwr=1; //for(i=1;i<=r-l+1;i++)pwr=(1ll*pwr*t)%mod; for(i=0;i<=r;i++) f=1e18; f=0; for(i=1;i<=r;i++) { for(j=2*i,k=2;j<=r;j+=i,k++) { f=min(f,(1ll*i*(k-1)*k)/2+f); } } ans=0;long long pr=1; for(i=l;i<=r;i++) ans=(ans+((f)%mod*pr)%mod)%mod,pr=(pr*t)%mod; cout<<ans<<endl; }
Codeforces Round #422 (Div. 2), problem: (C) Hacker, pack your bags Solution In C/C++

Codeforces Round #422 (Div. 2), problem: (C) Hacker, pack your bags Solution In C/C++

#include<cstdio> #include<vector> #include<cstring> #include<algorithm> #define ll long long using namespace std; ll ans,c,n,x,L,R,cc; vector<pair<ll,ll> >l,r; int main(){ scanf(“%lld%lld”,&n,&x); for(int i=1;i<=n;i++){ scanf(“%lld%lld%lld”,&L,&R,&cc); l.push_back(make_pair(R-L+1,cc)); r.push_back(make_pair(R-L+1,cc)); } memset(c,0x3f3f,sizeof(c));ans=c; for(int i=1;i<200100;i++){ for(int j=0;j<l.size();j++)l.first>=x?0:ans=min(ans,c.first]+l.second); for(int j=0;j<r.size();j++)c.first]=min(c.first],r.second); } printf(“%lld”,ans==c?-1:ans); return 0; …