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[Coding Challenge Tips] Heads I Win, Tails You Lose

To help your job search along, try participating in the TalentHub Coding Challenges. It’s a great addition to your resume and the perfect way to show off your coding skills to companies hiring programmers in Japan.

Go try one right now, or keep reading for tips on how to complete on of the previous challenges below!

Heads I Win, Tails You Lose

Difficulty: ★★★☆☆

Problem statement

The rivalry between Michael and Ray seems endless. These two boys even compete with each other by coin flipping.
Just now, Michael suggested a game with the following rules:
“Each person will flip a separate coin. If it lands on heads, the person who flipped the coin gets 1 point, otherwise no point is granted. After n flips, the person who has more points wins.”
Being fed up with such childish competition between his two friends, David comes to you in a need for support so that he can code the probability of Michael winning over Ray.

Input
– Enter n denoting the number of flips.

Output
– Print the number of cases where Michael wins over Ray mod 10^9+7, as the number can be very large.

Constraints:
– n <= 20

Sample Input
2

Sample Output
5

Answer

Suppose the number of times that one player has k point is cnt[k].
Then the result will be: sum of (cnt[k] * sum of (cnt[i]) with i < k).
We can use bitmask to compute cnt[k].

Here is sample code.

#include <iostream>
#include <string.h>
using namespace std;
#define MOD 1000000007
const int N = 21;
int n;
unsigned long long cnt[N];
int main(){
ios_base::sync_with_stdio(false);
cin >> n;
memset(cnt, 0, sizeof(cnt));
for(int i = 0; i < (1 << n); ++i) {
cnt[__builtin_popcount(i)]++;
}
int sum = 0, ans = 0;
for(int i = 0; i <= n; ++i){
ans = (ans + sum * cnt[i]) % MOD;
sum = (sum + cnt[i]) % MOD;
}
cout << ans << endl;
return 0;
}

 

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