Hearthstone

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.

題意：

給你N張奧數智慧，M張火球術（不打6） 打幾下面給出。求出摸1張牌打出>=P傷害的概率。

POINT：

遍歷各種摸牌可能。DP[i]代表當前i狀態的數量。詳細解釋在程式碼裡。

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <stack>
#include <algorithm>
#include <math.h>
using namespace std;
#define ll long long
ll f[26],dp[1<<20];
int P,N,M;
int atk[20];
void init()
{
    f[1]=1;
    f[0]=1;
    for(int i=2;i<=20;i++) f[i]=f[i-1]*i;
}
int check(int now)
{
    int sum=0;
    now=now>>N;
    for(int i=0;i<M;i++)
    {
        if(now&(1<<i))
        {
            sum+=atk[i];
        }
    }
    if(sum>=P) return 1;
    return 0;
}
ll gcd(ll a,ll b)
{
    return a==0?b:gcd(b%a,a);
}
int main()
{
    init();
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d %d %d",&P,&N,&M);
        memset(dp,0,sizeof dp);
        for(int i=0;i<(N+M);i++) dp[1<<i]=1;
        for(int i=0;i<M;i++)
        {
            scanf("%d",&atk[i]);
        }
        int amark=(1<<N)-1;
        int tot=(1<<(N+M))-1;
        ll ans=0;
        for(int i=0;i<=tot;i++)
        {
            int have = __builtin_popcount(i);
            int left = N+M-have;
            int havea= __builtin_popcount(i&amark);
            if(check(i))
            {
                ans+=dp[i]*f[left];//當前的狀態的數量*之後的全排序。
                continue;
            }
            if(havea*2+1-have<=0) continue;//無牌可摸
            for(int j=0;j<N+M;j++)
            {
                if(!((i>>j)&1))
                {
                    dp[i|(1<<j)]+=dp[i];//摸一張牌，加上之前的數量。
                }
            }
        }
        ll d=gcd(ans,f[N+M]);
        printf("%lld/%lld\n",ans/d,f[N+M]/d);
    }
    
}