It's not about hope, You'll be gray if you participate.

Indian round ,Cool,,, Hopefully I won't be green this round :)

because IIT indore means The Indian Institutes of Technology Indore

Maybe time for getting a Shipping address in India too, and also ID proof.

PS.: Currently, shipping t-shirts would be tough in India too, shipping worldwide is almost impossible.

A blue can't set alone but he can do with orange or red. Orange/Red can only set contest.

Also if you have problemsetted in other juries or contests you can do it.

As an advisor; Try rewriting; as a contributor, in the beginning, that will surely stonk your contribution.

I have noticed you comment in almost each blog Maybe you should try to control your urge as it seems too irritating for others.

You got my upvote ;) good luck becoming positive

Naah, just population I believe xD

Anything else?

k

Yes in inventing viruses

Hello Do Lund Trump

I think it is just free points.

Funny thing is IMO can be read as International Math Olympiad:)

Why?

yes

what is it in cricket that you like?

LMAO

you can watch match recordings as well as participate virtually

You are initializing an array of size 2,00,001 in each testcase. Maybe that's the reason?

tujhe_kyon_janna_hai

Yes, someone please tell me what is the Pretest 3 in D.

• I got WA on Pretest 3.
• Then, I figured out I used to wrong iterator. After correcting it also, WA on pretest 3.
• Then I thought, maybe it is because of overflow. So, I used long long. Then also WA on Pretest 3.

I precomputed number of digits for 0 to 9 until 2e + 50 and then solved eatch testcase in log10(n) by simply adding values from the array.

its a series. https://oeis.org/A103377

Yes, let me try. I passed pretests of this problem after getting a TLE on the 2nd test.

Let $$$c[n][i]$$$ be the number of "i" $$$0\leq i\leq 9$$$ after $$$n$$$ operations. You build the relation between $$$c[n][i]$$$ and $$$c[n-1][i]$$$; which is a 10x10 matrix A; and $$$c[n][i]$$$ can be calculated via $$$c[0][i]$$$ and $$$A^n$$$.

So the goal is to calculate 2*10^5 matrices $$$A^k$$$ given the first matrix $$$A$$$(which is a sparse matrix).Since A is sparse, after calculating $$$A^{n-1}$$$, you can calculate $$$A^{n}$$$ in O(100) instead of O(1000). So the complexity to calculate all matrices is O(2*10^7); and I passed pretests in 888 ms.

I have just debugged my code :( But I think it will be AC. EDIT: It got AC: 112711140

We can calculate answer independently for every digit, so we can use precalculation.

Precalculate all the answers (200000) for 9. It will take about of 2*10^6 operations

Then, in every test case count how many times every digit is present in n. I have used cnt array for that.

Substract every number from 1..9 from 9 and check if it's less than m. If it is, just add calc[m — (10 — i)] * cnt[i] to the answer. Else, simply add cnt[i] to the answer.

Time Complexity: amortized O(1) per every test case.

Code:

#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#define all(x) (x).begin(), (x).end()
#define len(s) (int)((s).size())
#define str(ch) string(1, ch)
using namespace std;
using ll = long long;
using ull = unsigned long long;
using graph = vector<vector<int>>;

long long calc[200001];
void solve() {
    int n, m;
    cin >> n >> m;
    long long ans = 0;
    int cnt[10];
    memset(cnt, 0, sizeof(cnt));
    while(n) cnt[n % 10]++, n /= 10;
    for(int i = 0; i < 10; i++) {
        if((9 - i) < m)
            ans += calc[m - (10 - i)] * cnt[i], ans %= (int)1e9 + 7;
        else
            ans += cnt[i];
    }
    cout << ans << endl;
}   

int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    long long cnt[10];
    for(int i = 0; i < 10; i++)
        cnt[i] = 0;
    for(int i = 0; i <= 200000; i++)
        calc[i] = 0;
    cnt[9] = 1;
    for(int i = 0; i <= 200000; i++) {
        long long ncnt[10];
        memset(ncnt, 0, sizeof(ncnt));
        for(int j = 0; j < 10; j++) {
            if(j == 9) {
                ncnt[0] += cnt[9];
                ncnt[0] %= (int)1e9 + 7;
                ncnt[1] += cnt[9];
                ncnt[1] %= (int)1e9 + 7;
                continue;
            }
            ncnt[j + 1] += cnt[j];
            ncnt[j + 1] %= (int)1e9 + 7;
        }
        for(int j = 0; j < 10; j++) {
            cnt[j] = ncnt[j];
            calc[i] += cnt[j];
            calc[i] %= (int)1e9 + 7;
        }
    }
    int tt;
    cin >> tt;
    while(tt--) solve();
    return 0;
}

https://mirror.codeforces.com/contest/1513/submission/112705430

I am getting TLE On Pretest 3? What is wrong with my submission?

your code has "index out of bound" on line 51

I agree.