HDU

Hearthstone is an online collectible card game from Blizzard Entertainment. Strategies and luck are the most important factors in this game. When you suffer a desperate situation and your only hope depends on the top of the card deck, and you draw the only card to solve this dilemma. We call this “Shen Chou Gou” in Chinese.

Now you are asked to calculate the probability to become a “Shen Chou Gou” to kill your enemy in this turn. To simplify this problem, we assume that there are only two kinds of cards, and you don’t need to consider the cost of the cards.
-A-Card: If the card deck contains less than two cards, draw all the cards from the card deck; otherwise, draw two cards from the top of the card deck.
-B-Card: Deal X damage to your enemy.

Note that different B-Cards may have different X values.
At the beginning, you have no cards in your hands. Your enemy has P Hit Points (HP). The card deck has N A-Cards and M B-Cards. The card deck has been shuffled randomly. At the beginning of your turn, you draw a card from the top of the card deck. You can use all the cards in your hands until you run out of it. Your task is to calculate the probability that you can win in this turn, i.e., can deal at least P damage to your enemy.

Input
The first line is the number of test cases T (T<=10).
Then come three positive integers P (P<=1000), N and M (N+M<=20), representing the enemy’s HP, the number of A-Cards and the number of B-Cards in the card deck, respectively. Next line come M integers representing X (0

#include<bits/stdc++.h>using namespace std;typedef long long ll;template <class T> inline void in(T &x) {    T f = 1; char c; while ((c = getchar()) < '0' || c > '9') if (c == '-') f = -1;    x = c - '0';    while ((c = getchar()) >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - '0'; x *= f;}int T,n,m,u=0,hp,tu[1005];ll dp[1<<20],jiecheng[21];ll gcd(ll q,ll w){    if(!w)return q;    return gcd(w,q%w);}int main(){    jiecheng[1]=jiecheng[0]=1;    for(int a=2;a<=20;a++)jiecheng[a]=jiecheng[a-1]*a;    in(T);    while(T--)    {        memset(dp,0,sizeof(dp));        in(hp),in(n),in(m);        for(int a=n;a<m+n;a++)in(tu[a]);        dp[0]=1;        int nm=n+m;        for(int a=0;a<(1<<nm);a++)        {            if(!dp[a])continue;            int as=0,hq=0,sh=0;            for(int b=n;b<nm;b++)            {                if(a&(1<<b))                {                    sh+=tu[b];                    hq++;                }            }            if(sh>=hp)continue;            for(int b=0;b<n;b++)            {                if(a&(1<<b))                {                    as++;                }            }            if(as-hq+1<=0)continue;            for(int b=0;b<nm;b++)            {                if(a&(1<<b))continue;                dp[a|(1<<b)]+=dp[a];            }        }        ll fz=0,fm=jiecheng[nm];        for(int a=0;a<(1<<nm);a++)        {            if(!dp[a])continue;            int as=0,hq=0,sh=0;            for(int b=n;b<nm;b++)            {                if(a&(1<<b))                {                    sh+=tu[b];                    hq++;                }            }            for(int b=0;b<n;b++)            {                if(a&(1<<b))                {                    as++;                }            }            if(sh<hp)continue;            fz+=dp[a]*jiecheng[nm-as-hq];        }        ll gc=gcd(fz,fm);        printf("%lld/%lld\n",fz/gc,fm/gc);    }}