fzu1342

http://acm.fzu.edu.cn/problem.php?pid=1342

dp[i][j]  , i位尾巴为j的概率

const  int  maxn = 1008 ;
double  dp[maxn][12] ;

int  main(){
     int i , j , n , k  ;
     double  t  ;

     while(cin>>k>>n){

          for(i = 0 ; i <= k ; i++) dp[1][i] = 100.0 / double(k+1) ;

          for(i = 2 ; i <= n ; i++){
              dp[i][0] = (dp[i-1][0] + dp[i-1][1]) / double(k+1) ;
              for(j = 1 ; j < k ; j++)
                 dp[i][j] = (dp[i-1][j-1] + dp[i-1][j] + dp[i-1][j+1]) / double(k+1) ;
              dp[i][k] =  (dp[i-1][k-1] + dp[i-1][k]) / double(k+1) ;
          }

          for(t = 0.0 , i = 0  ; i <= k ; i++) t += dp[n][i] ;

          printf("%.5lf\n" , t) ;

     }
     return 0 ;
}