Hearthstone

Posted 2020-09-13 ph = x

题意：

有$n$个无中生有，有$m$个不同的杀，第$i$个杀掉$X_i$滴血，敌人血量$P$，求问第一回合就将敌人杀死的概率是多少。

 

解法：

二进制枚举$A$类，$B$类卡的顺序，这样就确定了取了几个$B$卡，dp即可

$f(i,j)$表示选了$i$个卡，伤害和为$j$的方案数。

$ans = \sum {f(j,P)j!(m-j)!}$

总效率$O(n 2^{n+m})$

认真读题。

 

#include <iostream>
#include <cstdio>
#include <cstring>

#define LL unsigned long long
#define N 23
#define bit(x) (1<<(x))

using namespace std;

int P, n, m;
int X[N];
LL f[N][N][1010], comb[N][N], fac[N];

LL gcd(LL a, LL b) {
    if (b == 0) return a;
    return gcd(b, a % b);
}

int main() {
    comb[0][0] = 1;
    fac[0] = 1;
    for (int i = 1; i <= 20; i++) {
        fac[i] = fac[i-1] * i;
        comb[i][0] = 1;
        for (int j = 1; j <= i; j++)
            comb[i][j] = comb[i-1][j-1] + comb[i-1][j];
    }
    int T;
    scanf("%d", &T);
    while (T--)
    {
        memset(f, 0, sizeof(f));
        scanf("%d %d %d", &P, &n, &m);
        X[0] = 0;
        for(int i = 1; i <= m; i++) scanf("%d", &X[i]);
        f[0][0][0] = 1;
        for(int i = 0; i < m; i++)
            for(int j = 0; j <= i; j++)
                for(int k = P; k >= 0; k--)
                {
                    f[i+1][j][k] += f[i][j][k];
                    f[i+1][j+1][min(k+X[i+1],P)] += f[i][j][k];
                }
        LL ans0 = 0, ans1 = 0;
        for(int S=0;S<(1<<(n+m));S++)
        {
            int cnt=0,i,j=0;
            for(i=0;i<n+m;i++) if(bit(i)&S) cnt++;
            if(cnt!=m) continue;
            ans1 += fac[m];
            cnt=1;
            for(i=0;i<n+m && cnt;i++)
            {
                if(bit(i)&S) cnt--, j++;
                else cnt++;
            }
            ans0 += f[m][j][P] * fac[m-j] * fac[j];
        }
        if(ans0 == 0) {
            puts("0/1");
            continue;
        }
        LL d = gcd(ans0, ans1);
        cout << ans0/d << ‘/‘ << ans1/d << endl;
    }
    return 0;
}

View Code