**Topic links**

The first solution is $o (n^3*p) $:

F[i][j][k] indicates that the first I person into J individual length of K has several schemes (arranged fixed to 123). N).

$f [i][j][k]=f[i-1][j][k]+f[i-1][j-1][k-a[i]]$

The outermost enumeration T represents the person being stuck. I=t not add f[i-1][j-1][k-a[i]].

$ans ={{(\sum f[n][j][k]*j*j!* (N-1-J)!) + (\sum f[n][n][k]*n)}}/(n!) $

< Span class= "Mjx-char mjxc-tex-main-r" >< Span class= "Mjx-char mjxc-tex-math-i" > < Span class= "MJX_ASSISTIVE_MATHML" >

to see this

#include <cstdio> #include <cstring> #define N 55int n,p,a[n];d ouble f[n][n][n],ans,c[n]= {1};int main () { scanf ("%d", &n); for (int i=1; i<=n; i++) { scanf ("%d", &a[i]); c[i]=c[i-1]*i; } scanf ("%d", &p); for (int t=0; t<=n; t++) { memset (f,0,sizeof f); F[0][0][0]=1; for (int i=1, i<=n; i++) for (int j=0; j<=i; j + +) for (int k=0; k<=p; k++) { f[i][j][k]=f[i-1][j][k ]; if (J>=1&&t!=i&&k>=a[i]) f[i][j][k]+=f[i-1][j-1][k-a[i]]; } for (int j=1; j<n; j + +) for (int k=1; k<=p; k++) if (a[t]+k>p) ans+=c[j]*c[n-1-j]/c[n]*f[n][j][k]*j; for (int k=1; k<=p; k++) ans+=f[n][n][k]*n; } printf ("%lf", ans);}

can also be more efficient, $O (N^2*P) $

Reference

F[i][j][k] represents the first I person **at least into J personal** This J person's length and for K has several schemes (ranked as 123). N).

So $ans= (\sum f[n][j][k]*j!* (n-j)!/(n!) $

I didn't quite understand why at least J was such a push, asked the next teammate, said because there is no card behind, so it is possible to enter more people.

In fact, the first method of F is also at least J person, and then add ans in the back when the third j+1 person.

And you can use a scrolling array to optimize to a 2-D array.

#include <cstdio> #define N 51int n,a[n],p;double fac[n]= {1},f[n][n],tol,ans;int main () { scanf ("%d", &n) ; for (int i=1; i<=n; i++) { scanf ("%d", a+i); Fac[i]=fac[i-1]*i; Tol+=a[i]; } scanf ("%d", &p); if (tol<=p) { printf ("%d", n); return 0; } F[0][0]=1; for (int i=1, i<=n; i++) for (int. j=n-1; j>=0; j--) for (int k=0; k+a[i]<=p; k++) f[j+1][k+a[i]]+= F[J][K]; for (int j=1; j<=n; j + +) for (int k=0; k<=p; k++) ans+=f[j][k]*fac[j]*fac[n-j]; Ans/=fac[n]; printf ("%lf", ans);}

"Codeforces 261B" Maxim and Restaurant (DP, expected)