概率与数学期望简单题

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UVA - 11021 - Tribles

题目传送：Tribles

AC代码：

#include <map>#include <set>#include <list>#include <cmath>#include <deque>#include <queue>#include <stack>#include <bitset>#include <cctype>#include <cstdio>#include <string>#include <vector>#include <complex>#include <cstdlib>#include <cstring>#include <fstream>#include <sstream>#include <utility>#include <iostream>#include <algorithm>#include <functional>#define LL long long#define INF 0x7fffffffusing namespace std;int N;int n, k, m;//因为每只麻球都相互独立，所以只要先求一只麻球的即可double P[1005];//P[i]代表一只麻球生出i只麻球的概率double f[1005];//f[i]代表每一只麻球以及他的后代i天后全部死亡的概率，则答案为f[m]^kint main() {    scanf("%d", &N);    for(int cas = 1; cas <= N; cas ++) {        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] * pow(f[i-1], j);        }        printf("Case #%d: %.7lf\n", cas, pow(f[m], k));    }    return 0;}

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