AtCoder Beginner Contest 182 Tutorial

[utterances-bot]

AtCoder Beginner Contest 182 Tutorial | CP Wiki

A number can be divided by $3$ if and only if the sum of its digits can be divided by $3$.

https://cp-wiki.vercel.app/en/tutorial/atcoder/ABC182/

[swapno7064]

can you explain F a bit more??

[lucifer1004]

@swapno7064 I have added an example.

[heyuhhh]

why we need to ensure the current sum can be divided by a_{i+1}?

[lucifer1004]

@heyuhhh Because in the future, we can never use numbers less than $a_{i+1}$. So if the current value cannot be divided by $a_{i+1}$, it will never become 0.

[swapno7064]

from the above example how come 7 unique value can be generated means we have to find no of unique 'y' above 'X' . so how come 7 will be the answer?? i think only valid unique numbers are 192,216, 190 above 'X= 190'.?? can explain the transition.

[lucifer1004]

@swapno7064 Note that for each number in the intermediate transitions, there is a count of different ways to reach that number.

[swapno7064]

but the condition says we will use minimum coins like in the given example 108+(236)+(12)=192 only 1 way to reach 192...instead of 12 we cannot use (34)coins ??

[lucifer1004]

@swapno7064 The coefficients can be negative.

[Mastan1301]

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef vector<ll> vint;
typedef vector<vint> vvint;
typedef vector<bool> vbool;
typedef vector<vbool> vvbool;
typedef pair<ll, ll> iPair;
#define ff first
#define ss second

int main(){
	ll n, k = 0;
	cin >> n;
	vint cnt(3, 0);
	while(n){
		cnt[(n % 10) % 3]++;
		n /= 10;
		k++;
	}
	
	ll ans = abs(cnt[1] - cnt[2]) % 3;
	if(ans <= k - 1){
		cout << ans;
	}
	else
		cout << -1;


	return 0;
}

Can you tell me why this method won't work?

[Mastan1301]

This is for Problem-C.