Card Game || Codeforces Educational Round 136 Div2 Problem C
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Card Game || Codeforces Educational Round 136 Div2 Problem C
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Thanks a lot
Thanks
Thank you brother:)
why have you done n-1Cn/2 when we have picked nth element for alex so it should be n-1Cn/2 -1
What is the expected rating of this problem?
Thank You very much for the editorial. Full supportโค
can you plz provide code?
can you please upload the editorial of Reset K Edges problem
๐
bro wt resource did u use to learn combinatorics topics.
Thank youuuu
Thanks for editorial.
but suppose , borris has the nth card and alex has the ( n – 1 ) th card , and its alex turn . now the question is why would he waste his ( n – 1 )th card knowing that the orther person already has the
n th card , instead he would try to make borris waste his nth card in the beginning itself by using ( n – 2 )nd or ( n – 3 )rd card and then when his turn comes , so he plays the ( n – 1 )th card and win th e game .
are they bound to use the maximum card they have ? , is that the exact statement they should have inserted ?
why we didn't need of memoization ?
nice explanation bhaiya thank you
thank you sir
can number of ways alex has the (nth card) cant be equal to n-1 choose (n/2)-1
i was thinking of Bitwise DP, can it be possible with that ?
i have a doubt in my code my code gives ans for <=56 but for 58 and 60 due to mod i think it giving wrong ans??
Good explanation ๐
I got all the observations that you have got, but failed at implementation(taking alex equivalent to boris or viceversa)
Can u suggest some more problems like this.
Nice explanation ๐
Dp means??
Great explanation for such a complex question๐
what is that ncr thing that you did in the code , how can i learn that , is that some efficiient approach to deal with factorial things , thank you sir
nice explanation mate
idk why but this is failing on n=60
#include <bits/stdc++.h>
using namespace std;
#define FastIO ios_base::sync_with_stdio(false); cin.tie(NULL);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define int long long
#define all(x) (x).begin(),(x).end()
//#define loop(i,n) for(int i=0; i<n ;i++)
typedef pair<int,int> ppi;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
// #define mod 998244353
int fact(int n){
if(n==0) return 1;
return n*fact(n-1);
}
int c(int n,int r){
if(n<r) return 0;
if(r == 0 || r==n) return 1;
return (fact(n))/fact(n-r) / fact(r);
}
void fun(int n , vvi &dp){
if(n==2) return;
fun(n-2,dp);
dp[n][0] = (c(n-1 , n/2) + dp[n-2][1] )% 998244353 ;
dp[n][1] = ( c(n-2 , n/2) + dp[n-2][0])% 998244353;
dp[n][2] = (dp[n-2][2])% 998244353;
}
signed main(){
FastIO
//cout<<fixed<<setprecision(10);
int z;
cin>>z;
while(z–){
int n;
cin>>n;
//is A have nth card he wins
//similarly is B have nth card plus (n-2)th card B wins
//else draw
vvi dp (n+1 , vector<int> (3,-1));
//as n is greater than 1
//base cases
dp[2][0] = 1;
dp[2][1] = 0;
dp[2][2] = 1;
fun(n,dp);
cout<<dp[n][0]<<" "<<dp[n][1]<<" "<<dp[n][2];
cout<<endl;
}
return 0;
}
great explanation sir!
please can you share the entire code. I am not getting the correct answer for n=60. I couldn't understand how you defined Invfact[ ] and modInverse( ) ๐ฅฒ
Thanku and Hare krishna Sir ๐๐
nice explanation
i did not understand the base case can you explain if possible