**根据题目样例解释得到每种卡牌拥有状态之间的关系，然后转换成等式，高斯消元是2^(3n) **
80分超时代码：

#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <queue>
#define ls  (u<<1)
#define rs  (u<<1|1)
#define mid (l+r>>1)
using namespace std;
typedef long long LL;
typedef pair<int,int> PII;
const int N = 13, mod = 1e9 + 7;double p[20][20];
double s[1<<N][1<<N]; 
int n, m;// 高斯消元  
int gauss(double a[][1<<N], int n) { // 0 鏈夎В ; 1 鏃犵┓瑙?; 2 鏃犺В 
const double eps = 1e-6;	int c, r;for (c = 0, r = 0; c < n; ++c) {        int t = r;for (int i = r; i < n; ++i)         if (fabs(a[i][c]) > fabs(a[t][c]))t = i;if (fabs(a[t][c]) < eps) continue;  for (int i = c; i < n + 1; ++i) swap(a[t][i], a[r][i]);for (int i = n; i >= c; --i) a[r][i] /= a[r][c];for (int i = r + 1; i < n; ++i)if (fabs(a[i][c]) > eps)for (int j = n; j >= c; --j)a[i][j] -= a[r][j] * a[i][c];++r;}if (r < n) {for (int i = r; i < n; ++i)if (fabs(a[i][n]) > eps) return 2;return 1; }for (int i = n - 1; i >= 0; --i)for (int j = i + 1; j < n; ++j)a[i][n] -= a[j][n] * a[i][j];return 0;
}void calc(int x)
{vector<pair<int,double> >a, b;a.clear(); b.clear();;for(int i = 0;i < n;i ++)if(x>>i&1)a.push_back({i, 0});else b.push_back({i, 0});double l = 0, r = 0;for(int i = 0;i < a.size();i ++)for(int j = 0;j < b.size();j ++){int x = a[i].first, y = b[j].first;l += p[x][y]; r += p[y][x]; a[i].second += p[x][y];b[j].second += p[y][x];}for(int i = 0;i < a.size();i ++)for(int j = 0;j < b.size();j ++){int z = a[i].first, y = b[j].first;s[x][x^1<<z] -= p[y][z]*a[i].second*b[j].second/l/r;s[x][x|1<<y] -= p[z][y]*a[i].second*b[j].second/l/r;}s[x][x] = 1;
}int main() 
{scanf("%d%d", &n, &m);for(int i = 0;i < n;i ++)for(int j = i+1;j < n;j ++)scanf("%lf", p[i]+j), p[j][i] = 1-p[i][j];s[(1<<n)-1][1<<n] = 1;for(int i = 0;i < 1<<n;i ++)calc(i);gauss(s, 1<<n);while(m --){int x = 0;for(int i = 0, j;i < n;i ++)scanf("%d", &j) ,j?x |= 1<<i:0;printf("%.5lf\n", s[x][1<<n]);}return 0;
}