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

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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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