why put B in contest, nice ACD problems though

Good and interesting problems, thanks for the good round!

After more than one hour dealing with mathematics, I finally get B accepted. Somehow I thought that I was attending a mathematics exam :D

Thank you for the round, interesting problems as always!

What's wrong with this code please help me ATCODER ARC 178 B !!!!

#include <bits/stdc++.h>
using namespace std;
#include <atcoder/modint>
using mint = atcoder::modint998244353;
mint ten = 10;

#define ll long long

void solve(){
    
    int a, b, c;
    cin >> a >> b >> c;

    if(a>b) swap(a,b);

    if(c!=b+1 && c!=b) {
      cout << 0 << endl;
      return;
    }

    mint ans, whole;


    // case 1: if b == c
    if(b==c){
      // subcase1: a==b
      if(a==b){
        ans = (8*ten.pow(a-1)) * (8*ten.pow(a-1) + 1)/2;
        cout << ans.val() << endl;
      }
      // subcase2: a!=b
      else {
        ans = 9*ten.pow(a-1) * (9*ten.pow(b-1) - (11*ten.pow(a-1) -1)/2 );
        cout << ans.val() << endl;
      }
    }
    // case 2: if b+1=c
    else {
        
      if(a==b) {
        
       whole = 81*ten.pow(a+b-2);
       ans = (8*ten.pow(a-1)) * (8*ten.pow(a-1) + 1)/2;
       cout << whole.val() - ans.val()   << endl;

      }
      else {
        whole = 81*ten.pow(a+b-2);
        ans = 9*ten.pow(a-1) * (9*ten.pow(b-1) - (11*ten.pow(a-1) -1)/2);
        cout << whole.val() - ans.val()   << endl;
      }
    }

       fflush(stdout);
       cout << flush;
    }

int main() {
    int t;
    cin >> t;
    for (int T = 1; T <= t; ++T)
    {
      
       solve();


    }

       return 0;
}

Good E but can't figure out the bug until contest end :(:(:(

upd: well i forgot a step for the solution (divide-conquer convolution)

For the problem B, can anyone figure out how the inclusion-exclusion principle is used to get that expression?

Problem C is very cool.

UPD: Got it.

Can anyone explain this part of the editorial of problem C?

I don't understand the last line. I tried to derive the 3rd equation from the 2nd but i only could derive up to this:

The official editorial of problem E noted:

Here, if $$$f(x)$$$ is a polynomial such that $$$[x^{i}]f(x)$$$ is the answer for $$$i$$$, the contribution of $$$j$$$ to $$$f(x)$$$ is $$$A_{j}x^{b+1}(1+x)^{j-b-1}$$$. However, if there exists a $$$j$$$ such that $$$j = b$$$, the contribution of $$$j$$$ to $$$f(x)$$$ is $$$A_{j}x^{j}$$$.

Therefore, if $$$c$$$ is the smallest integer satisfying $$$L \leq 2A_{i}$$$, and $$$B_{i}$$$ is the smallest integer satisfying $$$L - A_{b} \leq A_{i}$$$, the total contribution of all $$$j$$$ to $$$f(x)$$$ can be obtained by evaluating the following polynomial:

Since $$$B_{i}$$$ is monotonically decreasing with respect to $$$i$$$, and $$$i - B_{i}$$$ is monotonically increasing with respect to $$$i$$$, this expression can be evaluated in $$$O(N (\log{N})^{2})$$$ using divide-and-conquer and convolution.

Can anybody gives a clearer analysis of where the $$$f(x)$$$ comes from and how the formula is deduced? Additionally, what's the function of $$$c$$$? It seems that $$$c$$$ didn't appear in the following texts. Thanks a lot.