Can anyone tell me what is "Clar" in "About the New Judge"? I hardly found posts about it in Bing.

AtCoder Beginner Contest Pi!

I don't know if I can do D this time ):

Wow! New judge!

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omg , my java codes are working in 40ms , but why did you removed java 8 , it's the best version of the best language on this earth

Good contest I was pretty scared that they mentioned it will be harder than usual but I didn't felt this contest is way too different compared to typical atcoder rounds

yes

too funny :) I submitted 4 wrong codes on D, such a easy problem.

Whyyyyyyyyy both E and F are expected value based question. Guess its time to practice expected value :(

Pretty shocked to see that people are discussing ongoing contest in the comment box , some even telling solutions

how to solve E?

orz chromate00 for the first solve of Ex...

How to Solve D?

Using simulated annealing and passed task Ex, maybe the data is too weak?

Weak tests in D.
https://atcoder.jp/contests/abc314/submissions/44532834 should fail on 7 Atcoder 3 2 0 a 1 3 a 1 3 B

This is a tragedy! link

Can anyone talk more about problem E? I can hardly understand the editorial. Thank you so much.

Hi, could someone please take a look at my solution for problem E and point out problems with the approach? I have added comments on the DP state and transition. Would really appreciate your help. The code snippet is not long.

    vector<vector<double> > dp(n+1, vector<double>(2*m+1, 20000000000.0));
    dp[0][0]=0.0;
    for(ll i=0; i<n; i++){
        for(ll j=0; j<=2*m; j++){
            // dp[i][j] = expected money paid for getting j points using first i wheels
            
            // We do not rotate the i'th wheel
            dp[i+1][j]=min(dp[i+1][j], dp[i][j]);
            
            for(ll k=1; k<=R; k++){
                for(ll l=0; l<=P; l++){
                    
                    // We rotate the i'th wheel k times to get l points
                    // a[i][k][l] = probability to get l points after rotating i'th wheel k times
                    if(j+l<=2*m && a[i][k][l]>0.0){
                        
                        // cost to rotate i'th wheel k times
                        double tmp=k*cost[i];
                        
                        // multiply the probability to get l points by rotating i'th wheel k times
                        tmp*=a[i][k][l];
                        
                        // multiply the probability to choose i'th wheel
                        tmp/=(double)n;
                        
                        // dp state transition
                        dp[i+1][j+l]=min(dp[i+1][j+l], dp[i][j]+tmp);
                    }
                }
            }
        }
    }
    
    cout << dp[n][m] << endl;

Can you give me a data that makes my code in question D wrong? Thanks!

#include <bits/stdc++.h>
#define int long long
using namespace std;
inline int read(){
    int x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-f;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}
    return x*f;
}
inline void write(int x){
    if(x<0){
        putchar('-');
        x=-x;
    }
    if(x>9)write(x/10);
    putchar(x%10+'0');
}
int n;
string s;
int q;
int ALL_UP=0,ALL_LO=0;
map<int,pair<int,int>>a;
signed main(){
    n=read();
    cin>>s;
    s=' '+s;
    q=read();
    for(int i=1;i<=n;i++){
        if(s[i]>='a'&&s[i]<='z')a[i]={0,-1};
        else a[i]={1,-1};
    }
    for(int i=1;i<=q;i++){
        int t=read();
//        cout<<t<<endl;
        if(t==1){
            int x=read();
            char c;
            cin>>c;
            s[x]=c;
            if(c>='a'&&c<='z')a[x]={0,i};
            else a[x]={1,i};
        }else if(t==2){
            read();
            char c;
            cin>>c;
            ALL_LO=i;
        }else{
            read();
            char c;
            cin>>c;
            ALL_UP=i;
        }
    }
    int UPORLO,t;
    if(ALL_UP>ALL_LO)UPORLO=1;
    else UPORLO=0;
    t=max(ALL_LO,ALL_UP);
//    cout<<UPORLO<<' '<<t<<' '<<s<<endl;
//    for(int i=1;i<=n;i++){
//        cout<<a[i].first<<' '<<a[i].second<<endl;
//    }
    for(int i=1;i<=n;i++){
        if(UPORLO){
            if(UPORLO==a[i].first){
                if(s[i]>='a'&&s[i]<='z')cout<<(char)toupper(s[i]);
                else cout<<s[i];
            }else{
                if(a[i].second>t)cout<<s[i];
                else{
                    if(s[i]>='a'&&s[i]<='z')cout<<(char)toupper(s[i]);
                    else cout<<s[i];
                }
            }
        }else{
            if(UPORLO==a[i].first){
                if(s[i]>='A'&&s[i]<='Z')cout<<(char)tolower(s[i]);
                else cout<<s[i];
            }else{
                if(a[i].second>t)cout<<s[i];
                else{
                    if(s[i]>='A'&&s[i]<='Z')cout<<(char)tolower(s[i]);
                    else cout<<s[i];
                }
            }
        }
    }
    
    return 0;
}

New judge seems much faster?

/* It can be proved that the sought expected value is always rational. Also, the constraints of this problem guarantee that if the sought expected value is expressed as an irreducible fraction x y ​ , then x is not divisible by 998244353.

Now, there is a unique integer z between 0 and 998244352, inclusive, such that xz≡y(mod998244353). Report this z. */

can anyone show me the code to achieve that within timelimit?