[hdu5985]概率题的推导

最新推荐文章于 2019-09-17 21:03:19 发布

aaron_1996

最新推荐文章于 2019-09-17 21:03:19 发布

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本文介绍了一种通过多次抛掷不同类型的硬币来确定哪种硬币最幸运的算法。通过计算每种硬币在每轮中存活的概率，最终得出每种硬币成为最幸运硬币的概率。

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题目描述

Bob has collected a lot of coins in different kinds. He wants to know which kind of coins is lucky. He finds out a lucky kind of coins by the following way. He tosses all the coins simultaneously, and then removes the coins that come up tails. He then tosses all the remaining coins and removes the coins that come up tails. He repeats the previous step until there is one kind of coins remaining or there are no coins remaining. If there is one kind of coins remaining, then this kind of coins is lucky. Given the number of coins and the probability that the coins come up heads after tossing for each kind, your task is to calculate the probability for each kind of coins that will be lucky.

算法思路

这一题在比赛的时候我想的太多了,切分的子问题太多反而导致求解变得十分困难,最后导致我们没有A下这一题,可以说责任在我。
思路的话十分简单,首先,我们先计算出到达第k步的时候硬币i死亡的概率
$k i l l [i] [j] = (1 - p [i] j) n u m [i]$ $kill[i][j]=(1-p[i]^j)^{num[i]}$
我们就可以计算出到达第i步之后i存活的概率
$r e c v [i] [j] = 1 - k i l l [i] [j]$ $recv[i][j]=1-kill[i][j]$
那么,我们就可以得到某一个硬币i成为lucky coins的概率
$a n s [i] = \sum j = 1 m a x (r e c v [i] [j] - r e c v [i] [j + 1) * \prod k = 0, k \neq i n k i l l [k] [j]$ $ans[i]=\sum_{j=1}^{max}(recv[i][j]-recv[i][j+1)*\prod_{k=0,k \neq i}^{n}kill[k][j]$
这个max是如何确定呢,我们知道所有的概率都在0.4-0.6之间,而总的硬币的个数在100000之内,我们就可以计算收敛的速度了。

代码

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

#define MAXN 75
#define MAXM 15

int t,n;
int num[MAXM];
double p[MAXM];
double killed[MAXM][MAXN+5];
double recv[MAXM][MAXN+5];
double ans[MAXM];

void Solve()
{
    int i,j,k;

    for(i=0;i<MAXM;i++)
        ans[i] = 0.0;

    if(n==1){
        printf("%.6f\n",1.0);
        return;
    }

    for(i=0;i<n;i++){
        double tmp = p[i];
        for(j=1;j<=MAXN;j++){
            killed[i][j] = pow(1-tmp,num[i]);
            recv[i][j] = 1 - killed[i][j];
            tmp *= p[i];
        }
    }

    for(i=0;i<n;i++){
        for(j=1;j<MAXN;j++){
            double tmp = 1.0;
            for(k=0;k<n;k++){
                if(k!=i)
                    tmp *= killed[k][j];
            }
            ans[i] += (recv[i][j]-recv[i][j+1])*tmp;
        }
    }

    for(i=0;i<n;i++)
        printf("%.6f%c",ans[i],(i==n-1)?'\n':' ');

    return;
}

int main()
{
    freopen("input","r",stdin);
    int i;

    scanf("%d",&t);

    while(t--){
        scanf("%d",&n);

        for(i=0;i<n;i++)
            scanf("%d%lf",&num[i],&p[i]);

        Solve();
    }
    return 0;
}