One day, sssworld and DDD play games together, but there are some special rules in this games.
They both have their own HP. Each round they dice respectively and get the points P1 and P2 (1 <= P1, P2 <= 6). Small number who, whose HP to reduce 1, the same points will remain unchanged. If one of them becomes 0 HP, he loses. 
As a result of technical differences between the two, each person has different probability of throwing 1, 2, 3, 4, 5, 6. So we couldn’t predict who the final winner.

 

There are multiple test cases.
For each case, the first line are two integer HP1, HP2 (1 <= HP1, HP2 <= 2000), said the first player sssworld’s HP and the second player DDD’s HP. 
The next two lines each have six floating-point numbers per line. The jth number on the ith line means the the probability of the ith player gets point j. The input data ensures that the game always has an end. 

 

 

One float with six digits after point, indicate the probability sssworld won the game.

 

 

 

 

 #include <cstdio>

 #include <cstring>

 #include <algorithm>

 #include <iostream>

 #include <cmath>

 using namespace std;

 const double eps = 0.00000001;

 double d[][];

 double dp[][];

 int main()

 {

     int h1, h2;

     while(~scanf("%d%d", &h2, &h1)) { //题目输入反了

         double win = , lose = , ave = ;

         for(int i = ; i < ; i++)

             for(int j = ; j < ; j++)

                 scanf("%lf", &d[i][j]);

         for(int i = ; i < ; i++) {

             for(int j = ; j < ; j++) {

                 if(i == j) ave += d[][i] * d[][j];

                 else if(i < j) lose += d[][i] * d[][j];

                 else win += d[][i] * d[][j];

             }

         }

 //        printf("SIZE : %d\n", sizeof(dp));

         double ans = ;

         if(eps >= fabs( - ave)) {

             printf("%.6f\n", ans);

             continue;

         }

         for(int i = ; i <= h1+; i++)

             for(int j = ; j <= h2+; j++)

                 dp[i][j] = ;

         // dp[h1+1][h2-1] = dp[h1-1][h2+1] = dp[h1+1][h2] = dp[h1][h2+1] = 0;

         dp[h1][h2] = ;

         for(int i = h1; i >= ; i--) {

             for(int j = h2; j >= ; j--) {

                 if(i == h1 && j == h2) continue;

                 dp[i][j] = (dp[i+][j] * lose + dp[i][j+] * win) / ((double) - ave);

             }

         }

         for(int i = ; i <= h1; i++)

             ans += dp[i][] * win / ((double) - ave);

         printf("%.6f\n", ans);

     }

     return ;

 }