UVA11021 Tribles[离散概率 DP]

Posted 2020-08-16 Candy?

UVA - 11021 Tribles

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 first line of input gives the number of cases, N. N test cases follow. Each one starts with a line containingn(1≤n≤1000),k(0≤k≤1000)andm(0≤m≤1000). Thenextnlineswillgivethe 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

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每只麻球相互独立，求一只就可以了
f[i]表示i天后1只麻球及后代全死亡的概率
f[i]=p[0]+p[1]*f[i-1]+p[2]*f[i-1]^2+........
边界f[0]=0

//
//  main.cpp
//  uva11021
//
//  Created by Candy on 26/10/2016.
//  Copyright © 2016 Candy. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int N=1005;
int T,n,m,k;
double f[N],p[N];
inline double fp(double a,int b){
    double ans=1.0;
    for(;b;b>>=1,a*=a)
        if(b&1) ans*=a;
    return ans;
}
void dp(){
    f[0]=0;
    for(int i=1;i<=m;i++){
        f[i]=p[0];
        for(int j=1;j<n;j++) f[i]+=p[j]*fp(f[i-1],j);
    }
}
int main(int argc, const char * argv[]) {
    scanf("%d",&T);
    for(int cas=1;cas<=T;cas++){
        scanf("%d%d%d",&n,&k,&m);
        for(int i=0;i<n;i++) scanf("%lf",&p[i]);
        dp();
        printf("Case #%d: %.7lf\n",cas,fp(f[m],k));
    }
    
    return 0;
}