题目链接

题意：




题解：




















AC代码

/*
    Author:zzugzx
    Lang:C++
    Blog:blog.csdn.net/qq_43756519
*/
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define endl '\n'
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int mod=1e9+7;
//const int mod=998244353;
const double eps = 1e-10;
const double pi=acos(-1.0);
const int maxn=1e6+10;
const ll inf=0x3f3f3f3f;
const int dir[4][2]={{0,1},{1,0},{0,-1},{-1,0}};

double a[20],b[20],c[30],d[20];
double dp[20][20];

int main()
{
    //ios::sync_with_stdio(false);
    //cin.tie(0);cout.tie(0);
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    for(int i=1;i<=12;i++)cin>>a[i];
    for(int i=1;i<=12;i++)cin>>b[i];
    for(int i=1;i<=12;i++)cin>>c[i];
    for(int i=1;i<=12;i++)d[i]=a[i]+(1-a[i])*(b[i]+c[i]-b[i]*c[i]);
    dp[0][0]=1;
    for(int i=1;i<=12;i++)
        for(int j=0;j<=i;j++){
            dp[i][j]=dp[i-1][j-1]*d[i]+(1-d[i])*dp[i-1][j];
        }
    for(int i=0;i<=12;i++)
        printf("%.6f\n",dp[12][i]);
    return 0;
}