f [ i , j ] 表示 i 个 分类 ， j 种 bug 的期望
方程很简单
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1e3 + 10 ;double f [N] [N] ;int main () {int s , n ;while ( scanf ( "%d%d" , & n , & s ) != EOF ) {memset ( f , 0 , sizeof f ) ;for ( int i = n ; i >= 0 ; -- i )for ( int j = s ; j >= 0 ; -- j ){if ( i == n && j == s ) continue ;double p1 = 1.0 * i / n * j / s ;double p2 = 1.0 * j / s * ( n - i ) / n ;double p3 = 1.0 * ( s - j ) / s * i / n ;double p4 = 1.0 * ( s - j ) / s * ( n - i ) / n ;f [i] [j] = ( f [i+1] [j] * p2 + f [i] [j+1] * p3 + f [i+1] [j+1] * p4 + 1.0 ) / ( 1.0 - p1 ) ;}printf ( "%.4f\n" , f [0] [0] ) ;}return 0 ;
}