poj2151

求每只队伍都回答出题目，且至少有一只队伍回答出n道题的概率
存在性问题我们可以转化为任意性问题
用P(每支队伍都回答出题目)-P(每只队伍回答的题目数小于n)
然后我们可以递推求解

 var f:array[..,..,..] of double;
     p:array[..,..] of double;
     k,i,j,n,m,t:longint;
     s,ans,tmp:double;

 begin
   while not eof do
   begin
     readln(n,t,m);
     if (n=) and (t=) and (m=) then break;
     for i:= to t do
     begin
       for j:= to n do
         read(p[i,j]);
       readln;
     end;
     fillchar(f,sizeof(f),);
     for i:= to t do
     begin
       f[i,,]:=;
       for j:= to n do
       begin
         f[i,j,]:=f[i,j-,]*(-p[i,j]);
         for k:= to j do
           f[i,j,k]:=f[i,j-,k-]*p[i,j]+f[i,j-,k]*(-p[i,j]); //f[i,j,k]表示第i支队前j道题答出k题的概率
       end;
     end;
     ans:=;
     for i:= to t do
       ans:=ans*(-f[i,n,]);
     tmp:=;
     for i:= to t do
     begin
       s:=;
       for j:= to m- do
         s:=s+f[i,n,j];  //答出题目出数<n满足加法原理
       tmp:=tmp*s;  //每队都小于满足乘法原理
     end;
     ans:=ans-tmp;
     writeln(ans::);
   end;
 end.