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[0..1010,0..35,0..35] of double;
    p:array[0..1010,0..35] 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=0) and (t=0) and (m=0) then break;
    for i:=1 to t do
    begin
      for j:=1 to n do
        read(p[i,j]);
      readln;
    end;
    fillchar(f,sizeof(f),0);
    for i:=1 to t do
    begin
      f[i,0,0]:=1;
      for j:=1 to n do
      begin
        f[i,j,0]:=f[i,j-1,0]*(1-p[i,j]);
        for k:=1 to j do
          f[i,j,k]:=f[i,j-1,k-1]*p[i,j]+f[i,j-1,k]*(1-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:=1;
    for i:=1 to t do
      ans:=ans*(1-f[i,n,0]);
    tmp:=1;
    for i:=1 to t do
    begin
      s:=0;
      for j:=1 to m-1 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:0:3);
  end;
end.