UV-11021 tribles (recurrence + probability)

Description

Problem
Tribbles
Input:Standard Input

Output:Standard output

Gravitation,N. "The tendency of all bodies to approach one another with a strength Proportion to the quantity of matter they contain -- the quantity Matter they contain being ascertained by the strength of their tendency To approach one another. This is a lovely and edifying authentication How science, having made a proof B, makes B the proof of ."

Ambrose Bierce

You have a populationKTribbles. This participates species of tribbles live for exactly one day and then die. Just before death, a single tribble has the probabilityPiOf giving birthIMore tribbles. What is the probability that afterMGenerations, every tribble will be dead?

Input
The first line of input gives the number of cases,N.NTest Cases Follow. Each one starts with a line containingN(1 <=N<= 1000 ),K(0 <=K<= 1000) andM(0 <=M<= 1000). The nextNLines will give the probabilitiesP0,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	Sample output
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	Case #1: 0.3300000 Case #2: 0.4781370 Case #3: 0.6250000 Case #4: 0.3164062

Problemsetter: Igor naverniouk, EPS

Special thanks: Joachim Wulff

Q: K of them have only poll, and each living day will die. Some new poll may be generated before the death. Specifically, the probability of I being generated is pi, given m, calculate the probability that all the pods will die after M days. Note that the total death is not counted in M days.

Idea: the death of each ball does not affect each other. Therefore, you only need to find the probability F (m) of the total death of only one ball in the first place and M days. The formula of the full probability includes:

F (I) = P0 + p1 * F (I-1) + p2 * F (I-1) ^ 2 + .... pn-1 * F (I-1) ^ n-1, where PJ * F (I-1) ^ J indicates that the poll produced J offspring and all of them died after I-1, note that the death of J offspring is independent of each other, and the probability of each death is F (I-1), so according to the multiplication formula, the probability of all death of J offspring is F (I-1) ^ J. Finally, calculate K. Only the ball is needed.

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int maxn = 1010;int n, k, m;double p[maxn], f[maxn];int main() {int t, cas = 1;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, 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;}

UV-11021 tribles (recurrence + probability)