%%%%http://blog.csdn.net/actoy/article/details/9366485

（看上去好麻烦的样子啊）

1 #include
     
       2 #include
      
         3 #include
       
         4 #include
        
          5 #define LL long long 6 #define LD long double   7 #define  N 100005 8 using namespace std; 9 inline int ra()10 {11     int x=0,f=1; char ch=getchar();12     while (ch<'0' || ch>'9') {
   if (ch=='-') f=-1; ch=getchar();}13     while (ch>='0' && ch<='9') {x=x*10+ch-'0'; ch=getchar();}14     return x*f;15 }16 double dp[1010][1010];17 int main()18 {19     int n,s;20     double p1,p2,p3,p4;21     while (scanf("%d%d",&n,&s)!=EOF)22     {23     for (int i=0; i<=n; i++) for (int j=0; j<=s; j++) dp[i][j]=0;24     for (int i=n; i>=0; i--)25         for (int j=s; j>=0; j--)26         {27             if (i==n && j==s) continue;28             p1=(double)i/n*j/s*1.0;29             p2=(double)(n-i)/n*j/s*1.0;30             p3=(double)i/n*(s-j)/s*1.0;31             p4=(double)(n-i)/n*(s-j)/s*1.0;32             dp[i][j]=(p2*dp[i+1][j]+p3*dp[i][j+1]+p4*dp[i+1][j+1]+1.0)/(1.0-p1);33         }34     printf("%.4f\n",dp[0][0]);35     }36     return 0;37 }