这题要注意到每只麻球的后代是独立存活的，所以如果某只麻球在某种情况下死亡的概率是P，那么k只麻球全部死亡的概率是P^k

设f[x]=每只麻球在x天后全部死亡的概率

f[i]=P₀+P₁f(i-1)+P₂f(i-1)²+……+P_n-1f(i-1)^n-1

最后由于有k个麻球，ans = f[m]^k

#include 
     
      #include 
      
       #define db doubleusing namespace std;const int maxn = 1005;int kase, n, k, m;db p[maxn], f[maxn];db ksm(db a, int b){    db ans = 1, base = a;    while (b)    {        if (b & 1) ans *= base;        base *= base;        b >>= 1;    }    return ans;}int main(){    scanf("%d", &kase);    for (int KASE = 1; KASE <= kase; KASE++)    {        scanf("%d%d%d", &n, &k, &m);        for (int i = 0; i < n; i++)            scanf("%lf", &p[i]);        f[0] = 0; f[1] = p[0];        for (int i = 2; i <= m; i++)        {            f[i] = 0;            for (int j = 0; j < n; j++)                f[i] += p[j] * ksm(f[i - 1], j);        }        printf("Case #%d: %.7f\n", KASE, ksm(f[m], k));    }    return 0;}