HDU 5816 Hearthstone(多校第七场)

题目链接

Hearthstone

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 200    Accepted Submission(s): 73

Problem Description

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<X<=1000) values for the B-Cards.

 

Output

For each test case, output the probability as a reduced fraction (i.e., the greatest common divisor of the numerator and denominator is 1). If the answer is zero (one), you should output 0/1 (1/1) instead.

 

Sample Input

23 1 21 23 5 101 1 1 1 1 1 1 1 1 1

 

Sample Output

1/346/273

题解：这个题用dfs搜索即可过。

#include <cstring>#include <queue>#include <map>#include <set>#include <cstdio>#include <cstring>#include <cmath>#include <iostream>#include <algorithm>using namespace std;typedef unsigned long long ull;int n,m,p,a[25],vis[25];ull frac[25];ull gcd(ull a,ull b){    return a%b?gcd(b,a%b):b;}ull dfs(int pp,int nn,int tot,int cnt){    ull res=0;    if(pp<=0)return frac[tot];    if(cnt==0||tot==0)return 0;    if(nn)        res+=nn*dfs(pp,nn-1,tot-1,cnt+1);    for(int i=0;i<m;i++)    {        if(vis[i])continue;        vis[i]=1;        res+=dfs(pp-a[i],nn,tot-1,cnt-1);        vis[i]=0;    }    return res;}int main(){    frac[0]=1;    for(int i=1;i<=20;i++)        frac[i]=i*frac[i-1];    int T;    scanf("%d",&T);    while(T--)    {        memset(vis,0,sizeof(vis));        scanf("%d%d%d",&p,&n,&m);        for(int i=0;i<m;i++)            scanf("%d",&a[i]);        ull fx=dfs(p,n,n+m,1);        ull fy=frac[n+m];        if(fx)        {            ull z=gcd(fx,fy);            fx/=z;            fy/=z;        }        else            fy=1;        cout<<fx<<"/"<<fy<<endl;    }    return 0;}