传送门：https://www.luogu.org/problemnew/show/P3317
这道题的推导公式还是比较好理解的，但是由于这个矩阵是小数的，要注意高斯消元方法的使用；
#include <algorithm> #include <iterator> #include <iostream> #include <cstring> #include <cstdlib> #include <iomanip> #include <bitset> #include <cctype> #include <cstdio> #include <string> #include <vector> #include <stack> #include <cmath> #include <queue> #include <list> #include <map> #include <set> #include <cassert> /* ⊂_ヽ ＼＼ Λ＿Λ 来了老弟 ＼('ㅅ') > ⌒ヽ / へ＼ / / ＼＼ ﾚ ノ ヽ_つ / / / /| ( (ヽ | |、＼ | 丿 ＼ ⌒) | | ) / 'ノ ) Lﾉ */ using namespace std; #define lson (l , mid , rt << 1) #define rson (mid + 1 , r , rt << 1 | 1) #define debug(x) cerr << #x << " = " << x << "\n"; #define pb push_back #define pq priority_queue typedef long long ll; typedef unsigned long long ull; typedef long double ld; //typedef __int128 bll; typedef pair<ll ,ll > pll; typedef pair<int ,int > pii; typedef pair<int,pii> p3; //priority_queue<int> q;//这是一个大根堆q //priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q #define fi first #define se second //#define endl '\n' #define boost ios::sync_with_stdio(false);cin.tie(0) #define rep(a, b, c) for(int a = (b); a <= (c); ++ a) #define max3(a,b,c) max(max(a,b), c); #define min3(a,b,c) min(min(a,b), c); const ll oo = 1ll<<17; const ll mos = 0x7FFFFFFF; //2147483647 const ll nmos = 0x80000000; //-2147483648 const int inf = 0x3f3f3f3f; const ll inff = 0x3f3f3f3f3f3f3f3f; //18 const int mod = 1e9; const double esp = 1e-8; const double PI=acos(-1.0); const double PHI=0.61803399; //黄金分割点 const double tPHI=0.38196601; template<typename T> inline T read(T&x){ x=0;int f=0;char ch=getchar(); while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar(); while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); return x=f?-x:x; } inline void cmax(int &x,int y){if(x<y)x=y;} inline void cmax(ll &x,ll y){if(x<y)x=y;} inline void cmin(int &x,int y){if(x>y)x=y;} inline void cmin(ll &x,ll y){if(x>y)x=y;} /*-----------------------showtime----------------------*/ const int maxn = 109; double mp[maxn][maxn]; double a[maxn][maxn]; int n; double Gauss(int n){ for(int i=1; i<n; i++){ int mx = i; for(int j=i+1; j<n; j++){ if(fabs(a[mx][i]) < fabs(a[j][i])) mx = j; } swap(a[i], a[mx]); for(int j=i+1; j<n; j++){ double tmp = a[j][i]/a[i][i]; for(int k=i+1; k<n; k++) a[j][k] -= a[i][k] * tmp; } } double ans = 1; for(int i=1; i<n; i++) ans *= a[i][i]; return fabs(ans); } int main(){ scanf("%d", &n); double tmp = 1; for(int i=1; i<=n; i++) { for(int j=1; j<=n; j++) { scanf("%lf", &mp[i][j]); if(fabs(1.0 - mp[i][j]) < esp) mp[i][j] = 1.0- esp; if(i < j)tmp = tmp * (1.0-mp[i][j]); mp[i][j] = mp[i][j] / (1.0 - mp[i][j]); } } for(int i=1; i<=n; i++){ for(int j=i+1; j<=n; j++){ a[i][j] = a[j][i] = -mp[i][j]; a[i][i] += mp[i][j]; a[j][j] += mp[i][j]; } } printf("%.10f\n", Gauss(n) * tmp); return 0; }