A - Collecting Bugs

1  / * define dp [i] [j] As the number of days to complete i clock, j subsystems required, then it is clear dp [i] [j] is transferred from the four states to
 2  P1 = i * J / (n-* S ) 
 . 3  P2 = (Ni) * J / (n-* S) 
 . 4  P3 = I * (SJ) / (n-* S) 
 . 5  P4 = (Ni) * (SJ) / (n-* S)
 . 6  but for DP [ i] [j], this time apparently dp [i] [j] is 0, can not be transferred,
 7  at this time by the method of transposition, the dp [i] [j] to the left, can be obtained dp [i] [j] of the transfer equation 
 8  state four, * / 
. 9 #include <the iostream>
 10 #include <Queue>
 . 11 #include <CString>
 12 is  the using  namespace STD;
 13 is typedef Long  Long LL;
 14  const  int MAXN = 1E5 +5; 
15 double dp[1005][1005];
16 int main()
17 {
18     int n,s;
19     while(cin>>n>>s)
20     {
21         memset(dp,0,sizeof(dp));
22         for(int i=n;i>=0;i--)
23         {
24             for(int j=s;j>=0;j--)
25             {
26                 if(n==i&&j==s) continue;
27                 double p=i*1.0/n;double q=j*1.0/s;
28                 //cout <<p<<" "<<q<<endl;
29                 dp[i][j]=((1-p)*(1-q)*(dp[i+1][j+1])+(1-p)*q*(dp[i+1][j])+p*(1-q)*(dp[i][j+1])+1)/(1-p*q);
30             }
31         }
32         printf("%.4lf\n",dp[0][0]);
33     }
34     return 0;
35 }