题意：

思路：

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

typedef long long ll;
ll dp[35][150];//dp(i,j)表示买i张彩票获得j元的方案数
int n;
ll init(){
    dp[0][0] = 1;
    for(int i = 1;i <= n;i++){
        for(int j = 1;j <= n*4;j++){
            if(j) dp[i][j] += dp[i-1][j-1];
            if(j-1) dp[i][j] += dp[i-1][j-2];
            if(j-2) dp[i][j] += dp[i-1][j-3];
            if(j-3) dp[i][j] += dp[i-1][j-4];
        }
    }
    ll res = 0;
    for(int i = 3*n;i <= 4*n;i++){
        res += dp[n][i];
    }
    return res;
} 
ll qpow(ll a,ll b){
    ll res = 1;
    while(b){
        if(b&1) res = res * a;
        a = a * a;
        b >>= 1;
    }
    return res;
}
int main(){
    scanf("%d",&n);
    ll all = qpow(4,n);
    ll res = init();
    ll d = __gcd(all,res);
    all/=d,res/=d;
    printf("%lld/%lld\n",res,all);
    return 0;
}