Description
Problem A
Tribbles
Input: Standard Input
Output: Standard Output
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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 kTribbles. 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 first 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.
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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
题意：有k只麻球，每只活一天就会死亡，临死前可能会生出一些新的麻球，具体来说，生i个的概率是Pi，给定m，求m天后所有麻球均死亡的概率，注意，不是m天时就已全部死亡的情况也算在内
思路：每只麻球死亡是互不影响的，所以只需求出一开始只有1只麻球，m天后全部死亡的概率f(m)，由全概率公式，有：
f(i) = p0+p1*f(i-1)+p2*f(i-1)^2+....pn-1*f(i-1)^n-1,其中pj*f(i-1)^j的含义是这个麻球生了j个后代，它们在i-1天后全部死亡，注意j个后代死亡是相互独立的，而每个死亡的概率都是f(i-1)，因此根据乘法公式，j个后代全部死亡的概率为f(i-1)^j，最后计算k只麻球就行了
#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;
}