Yay AC.

I realised that the OJ server is laggy, is that your experience as well?

I don't know if it is just me or not... but almost every time I have to wait several minutes to submit my code :(

Problem B mostly same as https://atcoder.jp/contests/abc190/tasks/abc190_d

just divide the final answer by 2

Congratulations Errichto on winning the round
Eagerly waiting for your screencast (to know how I messed up today)

How to solve A test case 2?

Since analysis doesn't seem to have been posted yet, how to solve final subtask of P3?

I managed to get an expected value of 16000 with the following but no clue how to improve:

For E = W, random gets you 20 expected wins and 20 expected draws, I believe this is optimal and can't be improved.

For all other cases, find the best constants A, B, C, D where A + B + C + D = 60 such that playing Rock A times, then Scissors B times, Paper C times and Rock again D times produces the best possible answer for such a method.

Turns out such constants can get you an expected win rate of $$$28.13 \times E$$$ for $$$W = 0$$$, $$$28.71 \times E$$$ for $$$W = \frac{E}{10}$$$ and $$$31.51 \times E$$$ for $$$W = \frac{E}{2}$$$, so a total expected score of $$$500 \times \frac{28.13 + 28.71 + 31.51 + 40}{4}$$$ or $$$16000$$$.

I did not find the round interesting also C was really weird. Managed to solve the first 2 and get a rank of 1400'ish. Here are the ideas,

//Think simple yet elegant.
#include <bits/stdc++.h>
using namespace std;
#define fast ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define ll  long long
#define all(v) v.begin(),v.end()
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define pi pair<int,int>
#define REP(i,n) for(int i=0;i<n;i++)
const int N = 3e5+10;
const ll mod = 1e9+7;


ll bin(ll a, ll b) {
	a%=mod;
    ll res = 1;
    while (b > 0) {
        if (b & 1)
            res = res * a % mod; //mod
        a = a * a % mod; //mod
        b >>= 1;
    }
    return res;
}

ll add(ll a,ll b){
	return (a+b+mod)%mod;
}
ll mul(ll a,ll b){
	return ((a%mod)*(b%mod))%mod;
}
ll cnt = 0;
void rec(int n,int k,string s,string now){
	if(s.length()==n){
		string t1 = s;
		reverse(all(s));
		if(t1==s and t1<now)
			cnt = add(cnt,1);
		return;
	}
	for(int i=0;i<k;i++){
		s+=(char)('a'+i);
		rec(n,k,s,now);
		s.pop_back();
	}
}

void run_case(){
	cnt = 0;
	ll n,k,i,j;
	cin >> n >> k;
	string s;
	cin >> s;
	
	ll ans = 0,rem;
	if(n%2)
		rem = n/2;
	else
		rem = n/2 - 1;
	for(i=0;i<n/2;++i){
		//<s[i]
		ans = add(ans,mul((int)(s[i]-'a'),bin(k,rem)));
		--rem;
	}
	
	if(n%2==0){
		string tt1 = s.substr(0,n/2);
		reverse(all(tt1));
		string tt = s.substr(0,n/2) + tt1;
		if(tt<s)
			ans = add(ans,1LL);
	}
	else{
		string tt;
		if(n>1){
			string tt1 = s.substr(0,n/2);
			reverse(all(tt1));
			 tt = s.substr(0,n/2) + s.substr(n/2,1) + tt1;
			 if(tt<s)
				ans = add(ans,1LL);
		}
		ans = add(ans,(int)(s[n/2]-'a'));
	
	}
	
//	rec(n,k,"",s);
	cout<<ans<<"\n";
}	

int main(){
	fast;
	int t;
	cin >> t;
	int cn = 1;
	while(t--){
		cout<<"Case #"<<cn<<": ";
		run_case();
	
		++cn;
	}
}

//Think simple yet elegant.
#include <bits/stdc++.h>
using namespace std;
#define fast ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define all(v) v.begin(),v.end()
#define F first
#define S second
#define ll long long
#define pb push_back
#define mp make_pair
#define pi pair<int,int>
#define REP(i,n) for(int i=0;i<n;i++)
const int N = 2e5+10;
ll mod = 1e9+7;


void run_case(){
	ll g,i,j;
	cin >> g;
	vector<pi> p;
	for(i=1;i*i<=2*g;++i){
		if((2*g)%i==0)
			p.pb({i,2*g/i});
	}
	set<ll> h;
	for(auto [k1,k2] : p){
		ll n1 = k1-1;
		if(k2-n1>=0 and (k2-n1)%2==0 and (k2-n1)/2 > 0)
			h.insert((k2-n1)/2);
		n1 = k2-1;
		if(k1-n1>=0 and (k1-n1)%2==0 and (k1-n1)/2 > 0)
			h.insert((k1-n1)/2);
	}
	cout<<(int)h.size()<<"\n";
}

int main(){
	fast;
	int q;
	cin >> q;
	int cn = 1;
	while(q--){	
			cout<<"Case #"<<cn<<": ";
		run_case();
	
		++cn;
	}
}
 

Expected value of probability of asking a problem from "Probability Theory" in Google Kickstart or Codejam =1.

Can someone help me to provide a counter-case for A, please?

#include<bits/stdc++.h>
using namespace std;

#define M1 1000000007
#define forl(i,n) for(int i=0;i<n;i++)
#define fore(i,n) for(int i=1;i<=n;i++)

int powM(int x,int y,int m)
{
    int ans=1,r=1;
    x%=m;
    while(r>0&&r<=y)
    {
        if(r&y)
        {
            ans*=x;
            ans%=m;
        }
        r<<=1;
        x*=x;
        x%=m;
    }
    return ans;
}
void ans(){
    int t;cin>>t;
    fore(l,t){
        cout<<"Case #"<<l<<": ";
        int n,k;string s;cin>>n>>k>>s;
        int f=0;
        if(n==1){
            int x=(s[0]-'a');
            cout<<x;
        }
        else if(n==2){
            int x=(s[0]-'a');
            if(s[0]<s[1]){
                x++;
                
            }
            cout<<x;
        }
        else{
            if(n%2){
                int c=(s[(n-1)/2]-'a');int g=0;
                for(int i=(n-3)/2;i>=0;i--){
                    if(s[i]<s[n-1-i]){
                        g=1;break;
                    }
                    else if(s[i]>s[n-1-i]){
                        break;
                    }
                }
                if(g){
                    c++;
                }
                forl(i,(n-1)/2){
                    if(s[i]!='a'){
                        int y=(s[i]-'a');
                        int r=(n-1)/2;
                        int z=powM(k,r-i,M1);
                        int w=(y*z)%M1;
                        c=(c+w)%M1;
                    }
                }
                cout<<c%M1;
            }
            else{
                int c=s[(n-1)/2]-'a';
                if(s[(n-1)/2]<s[n/2]){
                    c++;
                }
                int g=0;
                for(int i=(n-3)/2;i>=0;i--){
                    if(s[i]<s[n-1-i]){
                        g=1;break;
                    }
                    else if(s[i]>s[n-1-i]){
                        break;
                    }
                }
                if(g){
                    c++;
                }
                forl(i,(n-1)/2){
                    if(s[i]!='a'){
                        int y=(s[i]-'a');
                        int r=(n-1)/2;
                        int z=powM(k,r-i,M1);
                        int w=(y*z)%M1;
                        c=(c+w)%M1;
                    }
                }
                cout<<c%M1;
            }
        }
        cout<<endl;
    }
}
int32_t main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    ans();
    return 0;
}

Could somebody share how they came up with a good enough hash function for D? I came up with functions $$$(a \cdot RAND + b) \% mod1$$$ and $$$(a + RAND1) \cdot (b + RAND2) \% mod2$$$ pretty fast, but they didn't pass the second case and I was having trouble creating a better function.

for problem B . After analysing on first 100 values of 'g' . I found out that answer is total number of odd unique factors of 'g' . And finding factors can be done in O(sqrt(N)) for given value of N. I need proof for this :).

Me after reading C:
"I will never play rock paper scissors ever again!"

I can't understand what problem C meant they gave so confusing explanation too
If anyone has good explanation please link it here

I solved A subtask 1
B complete question
and D subtask 1 but I used randomisation and python large numbers

The transitions in the editorial for C are wrong.

Should be: \begin{split} v[r][p][s] & = & \max & (v[r — 1][p][s] + \frac{p}{n — 1} \times \mathbf{W} + \frac{s}{n — 1} \times \mathbf{E}, \ & & & v[r][p — 1][s] + \frac{s}{n — 1} \times \mathbf{W} + \frac{r}{n — 1} \times \mathbf{E}, \ & & & v[r][p][s — 1] + \frac{r}{n — 1} \times \mathbf{W} + \frac{p}{n — 1} \times \mathbf{E}) \ \end{split}

Between oversleeping and some mix of panic/annoyance at the lagging judge, I ended up chipping away at smol solves and settling in on D which (EVENTUALLY) worked out... thought I was being smurt by putting in an intentional TLE (infinite loop) to diagnose an RE bug, but somehow managed to both fix the latter (break rather than keep trying to parse an input that got shortened) by adding something unnecessary (account for (0)->0) while also adding an actual infinite loop bug (attempting to get a unique randint from too narrow of a range)... tl;dr crappy rainboy impression was actually crappy...?

272nd is the best I've done in these sorts of things, not sure how to feel about it... maybe slightly relieved after failrun at codejam round 2...?

Problem D is really easy to implement in Python. We don't even need to parse the expression manually, but rather we can let Python's built-in eval() take care of it.

The idea is to override the | operator (or any other unused operator) and then replace all existing occurrences of # with our overridden operator. The eval() function can then automatically parse and evaluate the resulting expression. We only need to provide the hash function.

from hashlib import blake2b

# Hack to define custom infix operators
# Reference - https://code.activestate.com/recipes/384122/
class Infix:
    def __init__(self, func):
        self.func = func
    
    def __or__(self, other):
        return self.func(other)

    def __ror__(self, other):
        return Infix(lambda x, self=self, other=other: self.func(other, x))

h = lambda x: int(blake2b(str(x).encode()).hexdigest(), 16)
op = Infix(lambda x, y: h((x, y)))

for i in range(1, int(input()) + 1):
    E = [input() for _ in range(int(input()))]
    V = [eval(expr.replace('#', '|op|')) for expr in E]

    num = dict()
    for v in V:
        num[v] = num.get(v, len(num) + 1)

    res = ' '.join(str(num[v]) for v in V)
    print('Case #{}: {}'.format(i, res))

In the "Rock Paper Scissor" problem, for the DP transactions there is no proof provided that the optimal answer for a particular state is resolvable from one of the exact previous state. I don't see why that would be the case always.

The problems were quite good, but the system works worse and worse everyday. I faced numerous issues with loading problem statements just after beginning at Kick Start Round B and Code Jam Round 2. It took almost minute for me to load at least one problem statement. Today at Kick Start Round C I spent even more time trying to load problem statement. At some point a saw the "Something went wrong, please try again in 30 seconds" message. And I was not deceived, I had to wait for more than 30 seconds before the next attempt to load a problem statement. And yes, all these delays between submitting and displaying the queued attempt left me with tons of doubts:

• Is the issue on my side?

• How much time a have to wait before doing something?

• Should I submit the solution again?

• Should I refresh the page?

And so on... Please, consider updating the contest system. As of today, it is causing too much inconvenience to the participants.

What will be the rating of problems A, B,C, D in terms of CF question ratings

Google kickstart

Smaller Strings

Getting TLE for Test Case2

(Following official editorial's method)

`Can someone suggest any improvements please?``` ~~~~~ Your code here...

#include <bits/stdc++.h>
using namespace std;

int exp(int a,int b){
    int res=1;
    while(b){
        if(b&1)
            res=res*a;
        a=a*a;
        b>>=1;
    }
    return res;
}
int basek(string s,int k){
    int ans=0,n=s.length();
    for(int i=n-1;i>=0;i--){
        ans=ans+(s[i]-'a')*exp(k,n-i-1);
    }
    return ans;
}
string generate(string s,int l,int v,int k){
    while(v>0){
        s+='a'+(v%k);
        v=v/k;
    }
    while(s.length()<l){
        s+='a';
    }
    reverse(s.begin(),s.end());
    return s;
}
int main() {
	// your code goes here
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	int t,l=1;
	cin>>t;
	while(t--){
		int n,k;
		cin>>n>>k;
		string s;
		cin>>s;
		int ans=0,h=ceil(n/2.0);
		string t="";
		

    for(int i=0;i<h;i++){
        t+=s[i];
    }


    int total=0;
    total=basek(t,k);
    for(int i=0;i<total;i++){
        string temp=generate("",h,i,k);
        if(temp<t){
            ans+=1;
        }
    }
        int nextSt=0;
       string r=t;
       reverse(r.begin(),r.end());
       if((n&1)){
           nextSt+=1;
       }
       for(int i=nextSt;i<r.length();i++){
           t+=r[i];
       }
       if(t<s){
           ans+=1;
       }   
    cout<<"Case #"<<l<<": "<<ans<<endl;
    l++;
}
return 0;

} ~~~~~