题解:AWC 0004
A
ll n, s, t;
ll sum;
int main(){cin>>n>>s>>t;for(int i=1;i<=n;i++){ll x;cin>>x;sum+=x;}if((t-s)*60>=sum){cout<<"Yes\n";}else cout<<"No\n";
}
B
ll n, t;
ll sum;
int main(){cin>>n>>t;for(int i=1;i<=n;i++){ll x, y;cin>>x>>y;sum+=max(x-t*y, 0ll);}cout<<sum;
}
C
ll n, h[N], sum;
int main(){cin>>n;for(int i=1;i<=n;i++){cin>>h[i];}sort(h+1, h+1+n);ll nx=0;for(int i=1;i<=n;i++){sum+=abs(nx-h[i]);nx=h[i];}sum+=abs(nx);cout<<sum;
}
D
ll n, m;
pll p[N];
int main(){cin>>n>>m;vector<vector<int>> st(n+1);for(int i=1;i<=m;i++){cin>>p[i].fi>>p[i].se;st[p[i].fi].push_back(p[i].se);}priority_queue<ll,vector<ll>,greater<ll> >q;int pos=0;for(int i=1;i<=n;i++){for(auto v:st[i]){q.push(v);}while(!q.empty()&&q.top()<i){q.pop();}if(!q.empty()){q.pop();pos++;}}if(pos==m){cout<<"Yes\n";}else cout<<"No\n";
}
E
ll n, k, a[N], pre[N];
int main(){cin>>n>>k;for(int i=1;i<=n;i++){cin>>a[i];pre[i]=pre[i-1]+a[i];}unordered_map<ll,ll>mp;ll ans=0;mp[0]=1;for(int i=1;i<=n;i++){if(mp.find(pre[i]-k)!=mp.end())ans+=mp[pre[i]-k];mp[pre[i]]++;}cout<<ans;
}
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