2011 北京赛区 I bupt 246 Activation

     After 4 years' waiting, the game "Chinese Paladin 5" finally comes out. Tomato is a 
crazyfan,andluckilyhegotthefirstrelease.Nowheisathome,readytobeginhisjourney. 
Butbeforestartingthegame,hemustfirstactivatetheproductontheofficialsite.Thereare 
toomanypassionate fans thatthe activationserver cannot dealwith allthe requestsat the 
sametime,soalltheplayersmustwaitinqueue.Eachtime,theserverdealswiththerequest 
of the first player in the queue, and the result may be one of the following, each has a 
probability: 
     1. Activation failed 
     1. Activation failed 
     11.. AAccttiivvaattiioonn ffaaiilleedd: This happens with the probability of p1. The queue remains 
unchangedandtheserverwilltrytodealwiththesamerequestthenexttime. 
     2.Connectionfailed 
     2.Connectionfailed 
     22..CCoonnnneeccttiioonnffaaiilleedd:Thishappenswiththeprobabilityofp2.Somethingjusthappened 
andthefirstplayerinqueuelosthisconnectionwiththeserver.Theserverwill thenremove 
hisrequestfromthequeue.Afterthat,theplayerwill immediatelyconnecttotheserveragain 
andstartsqueuingatthetailofthequeue. 
     3.Activationsucceeded 
     3.Activationsucceeded 
     33..AAccttiivvaattiioonnssuucccceeeeddeedd: Thishappenswiththeprobability ofp3.Congratulations,the 
playerwill leavethequeueandenjoythegamehimself. 
     4. Service unavailable 
     4. Service unavailable 
     44.. SSeerrvviiccee uunnaavvaaiillaabbllee: This happens with the probability of p4. Something just 
happened and the server is down. The website must shutdown the server at once.All the 
requeststhatarestillinthequeuewill neverbedealt. 
     Tomatothinks it sucks if theserver is down while heis still waiting in the queue and 
thereareno morethanK-1guysbeforehim.Andhewantstoknowtheprobabilitythatthis 
uglythinghappens. 
     Tomakeit clear,wesaythreethingsmayhappentoTomato:hesucceededactivatingthe 
game;theserverisdownwhileheisinthequeueandtherearenomorethanK-1guysbefore 
him; the server is downwhile heis in the queue andthere areat least Kguys beforehim. 
Nowyouaretocalculatetheprobabilityofthesecondthing. 


Input 
Input 
IInnppuutt 


     Therearenomorethan40test cases.Eachcaseinoneline,containsthreeintegersand 
fourrealnumbers:N,M(1<=M<=N<=2000),K(K>=1),p1,p2,p3,p4(0<=p1,p2,p3, 
p4<=1,p1+p2+p3+p4=1),indicatingthereareNguysinthequeue(thepositionsare 
numbered from 1 to N), and at the beginning Tomato is at the Mth position, with the 

probabilityp1,p2,p3,p4mentionedabove. 

Output 
Output 
OOuuttppuutt 


    Arealnumberinonelineforeachcase,theprobabilitythattheuglythinghappens. 
    Theanswershouldberoundedto5digitsafterthedecimalpoint. 


SampleInput 
SampleInput 
SSaammpplleeIInnppuutt 


2210.10.20.30.4 
3210.40.30.20.1 
4230.160.160.160.52 


SampleOutput 
SampleOutput 
SSaammpplleeOOuuttppuutt 


0.30427 
0.23280 
0.90343 

//f[i][j]=p1*f[i][j]+p2*f[i][j-1]+p3*f[i-1][j-1]+p4*(j<=k)
//f[n][n]=...+p2*f[n][n-1]+..
//f[n][1]=...+p2*f[n][m]+..
//f[n][2]=...+p2*f[n][1]+..p3*f[n-1][1]
//f[n][n-1]=...+p2*f[n][n-2]+..+p3*f[n-1][n-3]
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int n,m,k;
double p1,p2,p3,p4;
double f[2100][2100];
//考虑1的情况
bool equ(double a,double b)
{
    return fabs(a-b)<1e-8;
}
int main()
{
    while(scanf("%d%d%d%lf%lf%lf%lf",&n,&m,&k,&p1,&p2,&p3,&p4)==7)
    {
        if(equ(p4,0))
        {
            printf("0.00000\n");
            continue;
        }
        memset(f,0,sizeof(f));
        f[1][1]=p4*(1<=k)/(1-p1-p2);//?
        for(int i=2;i<=n;i++)
        {
            double a=p2,b=p4*(1<=k);
            double tmp=p2/(1-p1);//?
            if(i==3&&b==0.1)
            {
                b*=tmp;b+=p3*f[2][1];
                cout<<b<<endl;
                b=p4*(1<=k);
            }
            for(int j=2;j<=i-1;j++)
            {
                a*=tmp,b*=tmp;b+=p3*f[i-1][j-1]+p4*(j<=k);
            }
            a*=tmp,b*=tmp,b+=p3*f[i-1][i-1]+p4*(i<=k);
            f[i][i]=b/(1-p1-a);//?
            f[i][1]=(p2*f[i][i]+p4*(1<=k))/(1-p1);
            for(int j=2;j<=i-1;j++)
            {
                f[i][j]=(p2*f[i][j-1]+p3*f[i-1][j-1]+p4*(j<=k))/(1-p1);//?
            }
        }
        printf("%.5lf\n",f[n][m]);
    }
    return 0;
}