感受
又被卡精度了，看来要理解出题组的意思。

两种求法，求解每一道题解出的概率
图片说明

求解完之后，就愉快地dfs

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
double p[15];
double ans[15], a[15], b[15], c[15];
double solve(double a, double b, double c){
    return 1.0 - (1.0 - a) * (1.0 - b) * (1.0 - c);
    //return a + (1.0 - a) * b + (1.0 - a) * c - (1.0 - a) * b * c;被卡精度了
}
void dfs(int k, double tmp, int rig_num){
    if(k == 13){
        ans[rig_num] += tmp;
        return ;
    }
    dfs(k + 1, tmp * p[k], rig_num + 1);
    dfs(k + 1, tmp * (1.0 - p[k]), rig_num);
}
int main(){
    for(int i = 1; i <= 12; i++){
        scanf("%lf", &a[i]);
    }
    for(int i = 1; i <= 12; i++){
        scanf("%lf", &b[i]);
    }
    for(int i = 1; i <= 12; i++){
        scanf("%lf", &c[i]);
    }
    for(int i = 1; i <= 12; i++){
        p[i] = solve(a[i], b[i], c[i]);
    }
    dfs(1, 1.0, 0);
    for(int i = 0; i <= 12; i++){
        printf("%.6f\n", ans[i]);
    }
    return 0;
}