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## Discussion of Problem 1044. Lucky Tickets. Easy!

.001 AC solution without memoization .
Posted by Rafat Islam 2 Nov 2016 20:29
#include <bits/stdc++.h>
using namespace std ;

#define DEBUG(x) cout << '>' << #x << ':' << x << endl;
#define mem(x,val) memset((x),(val),sizeof(x))
#define all(x) x.begin(),x.end()
#define pb push_back
#define mp make_pair
#define PI acos(-1.0)

const int INF = 1 << 29 ;
typedef long long ll ;

int N(int n  , int k ) {
if(n == 1) return k<10?1:0 ;

int sum = 0 ;
for(int i = 0 ;i<= 9 && i<=k ; i++){
sum+=N(n-1 , k-i) ;
}
return sum  ;
}

int main() {
int n ,sum = 0 , temp  ;

scanf("%d" ,&n) ;
n/= 2 ;

for(int i = 0 ; i<=n*9 ;i++){
temp = N(n,i) ;
sum+=(temp*temp) ;
}

printf("%d\n" , sum) ;

return 0 ;
}
Re: .001 AC solution without memoization .
Posted by Alex 6 Aug 2017 22:17
Thx a lot!
I see it the same method as in http://ega-math.narod.ru/Quant/Tickets.htm
its a pity, but i dont understand it(