UVA - 11021 - Tribles 递推概率

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GRAVITATION, n.
“The tendency of all bodies to approach one another with a strength
proportion to the quantity of matter they contain – the quantity of
matter they contain being ascertained by the strength of their tendency
to approach one another. This is a lovely and edifying illustration of
how science, having made A the proof of B, makes B the proof of A.”
Ambrose Bierce
You have a population of k Tribbles. This particular species of Tribbles live for exactly one day and
then die. Just before death, a single Tribble has the probability Pi of giving birth to i more Tribbles.
What is the probability that after m generations, every Tribble will be dead?
Input
The ?rst line of input gives the number of cases, N. N test cases follow. Each one starts with a line
containing n (1 ≤ n ≤ 1000), k (0 ≤ k ≤ 1000) and m (0 ≤ m ≤ 1000). The next n lines will give the
probabilities P0, P1, . . . , Pn−1.
Output
For each test case, output one line containing ‘Case #x:’ followed by the answer, correct up to an
absolute or relative error of 10−6
.
Sample Input
4
3 1 1
0.33
0.34
0.33
3 1 2
0.33
0.34
0.33
3 1 2
0.5
0.0
0.5
4 2 2
0.5
0.0
0.0
0.5
Sample Output
Case #1: 0.3300000
Case #2: 0.4781370
Case #3: 0.6250000
Case #4: 0.3164062

题意：给你 k个球，  一个球可以活一天，在它死的时候会有概率pi生出i个小球，（0<=i<n） 现在问你m天后 所有小球全部死亡的概率是多少

题解： 我们定义f[m] 为 一个小球 在活m天后死亡的概率 ，那么答案就是 f[m]^k

         对于f[i] = P0 + P1 * (f[i-1]^1) + P2 * (f[i-1] ^ 2) + ............

         递推得到答案

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std ;
typedef long long ll;

const int N=10000;

int main() {
    int T, cas = 1, n, m, k;
    double p[N],f[N];
    scanf("%d",&T);
    while(T--) {
        scanf("%d%d%d",&n,&k,&m);
        for(int i = 0; i < n ; i++) scanf("%lf",&p[i]);
        f[0] = 0;
        for(int i = 1; i <= m ; i++) {
            f[i] = 0.0;
            for(int j = 0; j < n ; j++) {
                f[i] += p[j] * pow(f[i-1],j);
            }
        }
        printf("Case #%d: %.7f\n",cas++, pow(f[m],k));
    }
    return 0;
}

代码

UVA - 11021 - Tribles 递推概率

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原文地址：http://www.cnblogs.com/zxhl/p/5118781.html